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Chapter<br />
1<br />
Limits and Continuity<br />
Overview<br />
In Chapter 1, we explore limits and continuity, two basic concepts<br />
of calculus. Limits are the first of the four Big Ideas in<br />
the 2016–2017 AP ® Calculus Curriculum Framework.<br />
Big Idea 1: Limits<br />
Big Idea 2: Derivatives<br />
Big Idea 3: Integrals and the Fundamental Theorem of<br />
Calculus<br />
Big Idea 4: Series (AP ® Calculus BC only)<br />
The concept of limits is central to calculus. To understand<br />
calculus, it is crucial for students to know what it means for a<br />
function to have a limit and how to find the limit, when it exists.<br />
In Chapter 1, students learn what a limit is, how to find the<br />
limit of a function, and how to prove that limits exist using the<br />
definition of a limit. The chapter begins with numerical and<br />
graphical approaches to identifying limits, and students learn<br />
that these methods sometimes fail. Students then explore a variety<br />
of analytical techniques for finding limits. The precise<br />
definition of a limit, the so-called e–∂ (epsilon-delta) definition,<br />
is introduced at the end of the chapter.<br />
Building on what they’ve learned about limits, students<br />
learn the precise definition of continuity. Students often have<br />
an intuitive sense of the concept of continuity, because they<br />
have likely been tracing letters, numbers, and shapes from a<br />
young age. You might help students to make connections between<br />
these intuitive concepts and the formal definition of<br />
continuity in Section 1.3.<br />
As students work through Chapter 1, they will encounter<br />
the familiar concepts of horizontal and vertical asymptotes<br />
in the context of limits. From Chapter P, or their precalculus<br />
course, students should understand how to determine the domain<br />
and range of a function when viewing it graphically as<br />
well as when they are given an expression. They should also be<br />
able to identify horizontal and vertical asymptotes graphically<br />
and be able to find all asymptotes of a function analytically.<br />
Building on this knowledge, in this chapter, students will use<br />
the concept of a limit to find and verify values at which a function<br />
has an asymptote. This skill will be revisited when they<br />
learn about curve sketching in Chapter 4.<br />
Relationship to the AP® Calculus<br />
Curriculum Framework<br />
Chapter 1 covers topics from Big Idea 1: Limits. Here is a table<br />
that indicates the Essential Knowledge statements and Learning<br />
Objectives from the 2016–2017 AP ® Calculus Curriculum<br />
Framework that are covered in this chapter.<br />
Essential Knowledge<br />
Statements Learning Objectives Coverage<br />
EK 1.1A1 LO 1.1A(a) Section 1.1<br />
EK 1.1A2 LO 1.1A(b) Section 1.1<br />
EK 1.1A3 LO 1.1A(b) Section 1.1<br />
EK 1.1B1 LO 1.1B Section 1.1<br />
EK 1.1C1 LO 1.1C Section 1.2<br />
EK 1.1C2 LO 1.1C Section 1.2<br />
EK 1.1D1 LO 1.1D Section 1.5<br />
EK 1.1D2 LO 1.1D Section 1.5<br />
EK 1.2A1 LO 1.1D Section 1.3<br />
EK 1.2A2 LO 1.2A Section 1.3<br />
EK 1.2A3 LO 1.2A Section 1.3<br />
EK 1.2B1 LO 1.2 B Section 1.3<br />
1-2<br />
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Chapter 1 Section Topics<br />
Section 1.1: Limits of Functions Using<br />
Numerical and Graphical Techniques<br />
Students are introduced to the slope of a tangent line to a<br />
graph. Students should already be familiar with calculating<br />
the slope of a line given two distinct points. The concept of a<br />
limit is introduced through calculating the slope of the secant<br />
line on a curve based upon two points (x, f(x)) and (c, f(c)). If<br />
we continue to calculate the slope of the secant line with points<br />
that become closer and closer together (as x → c), the slope of<br />
the secant line approaches the slope of the tangent line at a<br />
given point. This work lays a foundation for the calculation of<br />
the slope of the tangent line as well as introduces the concept<br />
of the limit.<br />
Students investigate the concept of a limit first through<br />
the use of a table of numbers and then by using a graph. Both<br />
methods are handy for determining a limit, but both have<br />
drawbacks. Several examples are offered so that the students<br />
can gain an understanding of the pros and cons of these two<br />
methods for finding a limit of a function. Stress to students<br />
that a limit is the value that a function approaches as x approaches<br />
a particular value. Students often misinterpret the<br />
limit of a function as x → c with the value of f(c).<br />
Section 1.2: Limits of Functions Using<br />
Properties of Limits<br />
Through a series of examples, students explore finding limits using<br />
properties of limits and see that in many cases, a limit can<br />
be found by direct substitution or by substitution after some algebraic<br />
manipulation. The examples at the end of the section are<br />
of particular importance, as they introduce alternative strategies<br />
that can be used when substitution will not work. Students also<br />
are introduced to the average rate of change of a function in this<br />
section. This will provide the foundation they will use to learn<br />
about derivatives in Chapter 2.<br />
Section 1.3: Continuity<br />
Section 1.3 introduces the concept of continuity. Students are<br />
presented a formal definition of continuity at a point and use<br />
that definition to determine if a function is continuous at a<br />
given value. The section also covers classifications of discontinuities,<br />
in particular, removable, jump, and infinite discontinuities.<br />
The section concludes with the introduction of the<br />
Intermediate Value Theorem, which is applicable for functions<br />
that are continuous on a given closed interval.<br />
Section 1.4: Limits and Continuity of<br />
Trigonometric, Exponential, and<br />
Logarithmic Functions<br />
In Section 1.4, the first concept covered is finding a limit using<br />
the Squeeze Theorem. Students then explore finding limits<br />
and continuity of trigonometric, exponential, and logarithmic<br />
functions.<br />
Section 1.5: Infinite Limits; Limits at Infinity;<br />
Asymptotes<br />
We extend the language of limits in Section 1.5 to allow c to<br />
be ∞ or −∞ (limits at infinity) and to allow L to be ∞ or −∞<br />
(infinite limits). These limits can then be used for locating asymptotes<br />
that aid in graphing certain functions. Sometimes<br />
when we think about finding asymptotes, we find ourselves<br />
turning quickly to rational functions. Yet other functions (like<br />
exponential functions, logarithmic functions, and trigonometric<br />
functions) have asymptotes as well. See Example 10 for an application<br />
of finding a limit at infinity of an exponential function.<br />
Section 1.6: The e–d Definition of a Limit<br />
(Optional)<br />
As the chapter stresses throughout, we can be sure a limit is<br />
correct only if it was based on the e–d definition of a limit.<br />
In this section, we examine this definition and how to use it<br />
to prove that a limit exists, to verify the value of a limit, and<br />
to show if a limit does not exist. Although the content in this<br />
section is not directly tested on either the AB or BC version of<br />
the exam, students may benefit from coverage of this section.<br />
Promoting Good Habits and Skills<br />
Students may not have a great deal of experience in justifying<br />
their answers. This skill is fundamental to the course and to applying<br />
calculus to future fields of study. It is also a key part of the<br />
AP ® Calculus Curriculum Framework as addressed in MPAC 6.<br />
1. It is important to be able to justify solutions and to do so<br />
succinctly and completely. It helps to teach this skill by<br />
example. Write out solutions that are just enough.<br />
Explain to students why each step is sufficient, specific,<br />
and on point.<br />
2. Get in the habit of asking your students to explain<br />
their reasoning as they state answers to questions<br />
that you pose or problems you assign. You can build<br />
a justification step into assignments so that students<br />
expect to have to do so.<br />
Chapter 1 • Limits and Continuity<br />
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Chapter 1: Resources<br />
TRM Teacher’s Resource Materials<br />
The following resources can be found by clicking on the links in the<br />
Teacher’s e-Book (TE-book), logging into the Teacher’s Resource<br />
Center, or opening the Teacher’s Resource Flash Drive (TRFD).<br />
• Chapter 1 PD Video<br />
• Chapter 1 Bell Ringers The prepared Bell Ringer slide set is<br />
a great tool to help you make good use of the first few minutes<br />
of class. Designed to be used every day, each Bell Ringer set<br />
contains a set of short mental math questions that review prerequisite<br />
topics and keep the student engaged while practicing<br />
basic skills. The number and type of questions vary by chapter.<br />
• AP ® Calculus AB Exam Prep Flashcards Flashcards are<br />
a great tool for students to use in learning the essentials<br />
of calculus. Prepped to be printed and cut into 2 × 3–inch<br />
cards, the flashcards are designed to be used in class or on<br />
the go. Each card is labeled with the chapter number in<br />
which the topic is first introduced so they can be sorted and<br />
used throughout the course as a study aid.<br />
• Chapter 1 Alternate Examples All of the Chapter 1 Alternate<br />
Examples are also provided in PDF document format.<br />
Use these as additional examples in class, as the basis for assessments,<br />
or as additional practice for students.<br />
• Chapter 1 AP ® Calc Skill Builders All of the Chapter 1<br />
AP ® Calc Skill Builders are also provided in PDf format.<br />
Use these to provide instant practice of skills that are essential<br />
for success on the AP ® Exam.<br />
• Chapter 1 Skill Building Worksheets There are printable<br />
worksheets for each section, with solutions. Each one is<br />
referenced as point of use in the wraparound pages.<br />
Section 1.1 Worksheet 1<br />
Section 1.1 Worksheet 2<br />
Section 1.1 Worksheet 3<br />
Section 1.2 Worksheet 1<br />
Section 1.2 Worksheet 2<br />
Section 1.2 Worksheet 3<br />
Section 1.3 Worksheet 1<br />
Section 1.3 Worksheet 2<br />
Section 1.3 Worksheet 3<br />
Section 1.4 Worksheet 1<br />
Section 1.4 Worksheet 2<br />
Section 1.5 Worksheet 1<br />
Section 1.5 Worksheet 2<br />
Section 1.5 Worksheet 3<br />
Section 1.6 Worksheet<br />
• Chapter 1 Prepared Tests (Forms A and B) No need to worry<br />
about your students sharing exam information when you<br />
have two tests in two parallel versions to use for your various<br />
sections or as a makeup exam. Each test is 4 pages long and is<br />
designed to be scored out of a maximum of 50 points, making<br />
percentages simple.<br />
• Chapter 1 Teacher’s Solutions Manual Complete worked solutions<br />
to every problem in the book are found in the Teacher’s<br />
Solutions Manual, which may be downloaded as a PDF. The<br />
solutions for each set of Section Problems and for the Chapter<br />
Review Problems, AP ® Review Problems, and Tests at the ends<br />
of chapters are referenced at point of use in the Teacher’s Edition<br />
and may be downloaded as a smaller chunk of material for<br />
ease of use.<br />
• Additional Chapter 1 Resources We have created a list of<br />
third-party videos, Web sites, and other resources to support<br />
the content in this chapter. The Word document includes clickable<br />
URLs to help you access this external content. (Note: all<br />
of the URLs were live when this book was published.)<br />
Free-Response Questions from<br />
Previous AP® Exams<br />
Chapter 1 is too early to have students work with previous<br />
free-response questions. Limit and continuity problems seldom<br />
appear directly in free-response questions on the exam. It<br />
is possible, however, that a subpart of a free-response question<br />
may ask about a concept related to the limit or continuity.<br />
Students are more likely to encounter limit and continuity<br />
problems in the multiple-choice questions of the exam. On<br />
average, students can expect about 3 or 4 multiple-choice questions<br />
that cover limits. However, it’s likely that multiple-choice<br />
questions on limits will be framed so that they also require<br />
knowledge gained later in the year. For example, the exam<br />
may have questions on limits that incorporate more advanced<br />
content from later chapters, such as the limit of the difference<br />
quotient (Section 2.1) or L’Hôpital’s Rule (Section 4.5)<br />
or mixed in with problems related to the derivative, covered in<br />
Chapters 3 and 4.<br />
It is a good idea to start exploring the College Board<br />
Web site to become familiar with what is available to teachers.<br />
Released free-response questions can be found on the<br />
AP ® Central Web site: http://apcentral.collegeboard.com/apc/<br />
members/exam/exam_information/232050.html<br />
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Chapter<br />
1<br />
Free-response questions often draw on content that spans<br />
more than one chapter. As students prepare during the year,<br />
you can introduce students to the skills required to solve full<br />
FRQ on the exam. If there are appropriate questions or subparts,<br />
look under this headline in future chapters to see what<br />
might be helpful to use in class.<br />
As an AP ® Calculus teacher, you can gain access to complete<br />
released exams by logging into your Course Audit Account<br />
on the Course Home Page.<br />
AB Calculus:<br />
http://apcentral.collegeboard.com/apc/public/courses/<br />
teachers_corner/2178.html<br />
BC Calculus:<br />
http://apcentral.collegeboard.com/apc/public/courses/<br />
teachers_corner/2118.html<br />
You may want to keep these old exams on hand to select multiple-choice<br />
questions related to limits to present to students<br />
for further practice. Likewise, you may want to make a chart<br />
or find some other organizing tool that allows you to quickly<br />
group multiple-choice questions by topic so they may be used<br />
at appropriate times throughout the year.<br />
You may also want to direct your students to the following<br />
Kahn Academy site. College Board has partnered with Kahn<br />
Academy to provide support to students. Many released FRQs<br />
from previous exams are reviewed in these worked example<br />
videos:<br />
https://www.khanacademy.org/math/calculus-home/apcalc-topic<br />
Chapter 1: Pacing Guides, Objectives,<br />
and Suggested Assignments<br />
These pacing guides are based on a schedule with 125 standard<br />
45-minute classroom sessions before the exam. This 125-<br />
day course includes assessment days and allows about 3 weeks<br />
for review before the AP ® Calculus exam.<br />
The suggested homework assignments list odd-numbered<br />
problems, whenever possible, so students can check their answers<br />
against the back-of-the-book answers. If you would rather<br />
students not have access to the answers while doing homework,<br />
adding 1 to the exercise numbers usually will do the trick, because<br />
the homework problems typically are paired. The answers<br />
for the even-numbered problems are not in the Answer appendix.<br />
The authors observe pairing strictly in the Skill Building<br />
category. The Applications and Extensions usually are paired,<br />
although more advanced problems tend not to be paired. The<br />
AP ® Practice Problems are not paired.<br />
Calculus AB Pacing Guide<br />
Only topics that appear on the AP ® Exam are included.<br />
Day Topic <strong>Sullivan</strong>/Miranda Chapter Objectives Suggested Assignment<br />
1 Section 1.1<br />
1. Discuss the slope of the tangent line to the graph<br />
2. Investigate a limit using a table<br />
7, 9, 11, 17–28<br />
2 Section 1.1 3. Investigate a limit using a graph<br />
31, 35, 37, 38, 41, 42<br />
All AP ® Practice Problems<br />
3 Section 1.2<br />
1. Find the limit of a sum, a difference, and a product<br />
2. Find the limit of a power and the limit of a root<br />
3. Find the limit of a polynomial<br />
35, 37, 39, 41, 43, 47, 48, 51–59 odd<br />
4. Find the limit of a quotient<br />
4 Section 1.2<br />
5. Find the limit of an average rate of change<br />
61–65 odd, 67, 71, 73–79 odd<br />
6. Find the limit of a difference quotient<br />
All AP ® Practice Problems<br />
1. Determine whether a function is continuous at a number<br />
5 Section 1.3 2. Determine intervals on which a function is continuous<br />
13–18, 19–35 odd<br />
3. Use properties of continuity<br />
6 Section 1.3 4. Use the Intermediate Value Theorem 59–63 odd, All AP ® Practice Problems<br />
7 Section 1.4<br />
1. Use the Squeeze Theorem to find a limit<br />
9–45 odd<br />
2. Find limits involving trigonometric functions<br />
(continued next page)<br />
Chapter 1 • Limits and Continuity<br />
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<strong>Sullivan</strong> AP˙<strong>Sullivan</strong>˙Chapter01 October 8, 2016 17:4<br />
<strong>Sullivan</strong><br />
88 Chapter 1 • Limits and Continuity<br />
33.<br />
2x 2 if x < 1<br />
f (x) =<br />
3x 2 − 1 if x > 1<br />
at c = 1<br />
34. f (x) =<br />
x 2 − 1 if x > −1<br />
x 3 if x < −1<br />
at c =−1<br />
<br />
35. f (x) =<br />
at c = 0<br />
2x + 1 if x > 0<br />
⎧<br />
⎨ x 2 if x < 1<br />
x 2 if x ≤ 0<br />
36. f (x) = 2 if x = 1 at c = 1<br />
⎩<br />
−3x + 2 if x > 1<br />
Calculus AB Pacing Guide (continued)<br />
Applications and Extensions<br />
In Problems 37–40, sketch a graph of a function with the given<br />
properties. Answers will vary.<br />
37. lim f (x) = 3;<br />
x→2<br />
lim f (x) = 3;<br />
x→3− lim f (x) = 1;<br />
x→3 +<br />
f (2) = 3; f (3) = 1<br />
38. lim f (x) = 0;<br />
x→−1 lim f (x) =−2;<br />
x→2− lim f (x) =−2;<br />
x→2 +<br />
f (−1) is not defined; f (2) =−2<br />
39. lim f (x) = 4;<br />
x→1<br />
lim f (x) =−1;<br />
x→0− lim f (x) = 0;<br />
x→0 +<br />
f (0) =−1; f (1) = 2<br />
f (x) = 1;<br />
x→1<br />
In Problems 41–50, use either a graph or a table to investigate<br />
each limit.<br />
|x − 5|<br />
|x − 5|<br />
41. lim<br />
42. lim<br />
43. lim<br />
x→5 + x − 5<br />
x→5 − x − 5<br />
<br />
x→ 12<br />
2x<br />
−<br />
44. lim <br />
x→ 12<br />
2x 45. lim<br />
+ <br />
x→ 23<br />
2x 46. lim<br />
− <br />
x→ 23<br />
2x<br />
+<br />
<br />
47. lim |x|−x 48. lim |x|−x<br />
x→2 + x→2 −<br />
3 3<br />
49. lim x−x 50. lim x−x<br />
x→2 + x→2 −<br />
51. Slope of a Tangent Line For f (x) = 3x 2 :<br />
(a) Find the slope of the secant line containing the points (2, 12)<br />
and (3, 27).<br />
(b) Find the slope of the secant line containing the points (2, 12)<br />
and (x, f (x)), x = 2.<br />
(c) Create a table to investigate the slope of the tangent line to the<br />
graph of f at 2 using the result from (b).<br />
(d) On the same set of axes, graph f , the tangent line to the graph<br />
of f at the point (2, 12), and the secant line from (a).<br />
(a) Find the slope of the secant line containing the points (2, 8)<br />
and (3, 27).<br />
(b) Find the slope of the secant line containing the points (2, 8)<br />
and (x, f (x)), x = 2.<br />
(c) Create a table to investigate the slope of the tangent line to the<br />
graph of f at 2 using the result from (b).<br />
(d) On the same set of axes, graph f , the tangent line to the graph<br />
of f at the point (2, 8), and the secant line from (a).<br />
53. Slope of a Tangent Line For f (x) = 1 2 x2 − 1:<br />
Day Topic <strong>Sullivan</strong>/Miranda Chapter Objectives Suggested Assignment<br />
8 Section 1.4<br />
9 Section 1.5<br />
10 Section 1.5<br />
3. Determine where the trigonometric functions are continuous<br />
4. Determine where an exponential or a logarithmic function is<br />
continuous<br />
1. Investigate infinite limits<br />
2. Find the vertical asymptotes of a graph<br />
3. Investigate limits 40. at lim f (x) = 2; lim f (x) = 0; lim<br />
x→2<br />
infinity<br />
x→−1<br />
4. Find the horizontal asymptotes f (−1) = 1; f (2) of = a 3graph<br />
5. Find the asymptotes of the graph of a rational function<br />
(a) Find the slope m sec of the secant line containing the<br />
points P = (2, f (2)) and Q = (2 + h, f (2 + h)).<br />
(b) Use the result from (a) to complete the following table:<br />
h −0.5 −0.1 −0.001 0.001 0.1 0.5<br />
m sec<br />
(c) Investigate the limit of the slope of the secant line found in (a)<br />
as h → 0.<br />
(d) What is the slope of the tangent line to the graph of f at the<br />
point P = (2, f (2))?<br />
(e) On the same set of axes, graph f and the tangent line to f at<br />
P = (2, f (2)).<br />
54. Slope of a Tangent Line For f (x) = x 2 − 1:<br />
All AP ® Practice Problems<br />
(a) Find the slope m sec of the secant line containing the<br />
points P = (−1, f (−1)) and Q = (−1 + h, f (−1 + h)).<br />
(b) Use the result from (a) to complete the following table:<br />
2–26, 27–41 odd<br />
h −0.1 −0.01 −0.001 −0.0001 0.0001 0.001 0.01 0.1<br />
m sec<br />
43–59 odd, 67–71 odd, All AP ® Practice<br />
Problems<br />
(c) Investigate the limit of the slope of the secant line found<br />
in (a) as h → 0.<br />
(d) What is the slope of the tangent line to the graph of f at the<br />
point P = (−1, f (−1))?<br />
(e) On the same set of axes, graph f and the tangent line to f<br />
at P = (−1, f (−1)).<br />
11 Review Chapter 1 Review Exercises Chapter 1 AP ® Review Problems<br />
12 Test Chapter 1 Test AP ® Practice Exam, Big Idea 1: Limits<br />
Calculus BC Pacing Guide<br />
PAGE<br />
85 55. (a) Investigate lim cos π by using a table and evaluating the<br />
x→0 x<br />
function f (x) = cos π x at<br />
Day Topic <strong>Sullivan</strong>/Miranda Chapter Objectives Suggested Assignment<br />
1 Section 1.1<br />
2 Section 1.2<br />
3 Section 1.2<br />
4 Section 1.3<br />
5 Section 1.4<br />
6 Section 1.5<br />
1. Discuss the slope of the tangent line to the graph<br />
2. Investigate a limit using a table<br />
3. Investigate a limit using a graph<br />
1. Find the limit of a sum, a difference, and a product<br />
2. Find the limit of a power and the limit of a root<br />
3. Find the limit of a polynomial<br />
52. Slope of a Tangent Line For f (x) = x 3 :<br />
4. Find the limit of a quotient<br />
5. Find the limit of an average rate of change<br />
6. Find the limit of a difference quotient<br />
1. Determine whether a function is continuous at a number<br />
2. Determine intervals on which a function is continuous<br />
3. Use properties of continuity<br />
4. Use the Intermediate Value Theorem<br />
1. Use the Squeeze Theorem to find a limit<br />
2. Find limits involving trigonometric functions<br />
3. Determine where the trigonometric functions are continuous<br />
4. Determine where an exponential or a logarithmic function is<br />
continuous<br />
1. Investigate infinite limits<br />
2. Find the vertical asymptotes of a graph<br />
3. Investigate limits at infinity<br />
4. Find the horizontal asymptotes of a graph<br />
5. Find the asymptotes of the graph of a rational function<br />
x =− 1 2 , − 1 4 , − 1 8 , − 1 10 , − 1 12 ,..., 1<br />
12 , 1 10 , 1 8 , 1 4 , 1 2 .<br />
(b) Investigate lim cos π by using a table and evaluating the<br />
x→0 x<br />
function f (x) = cos π x at<br />
9, 11, 13, 15, 17, 19, 31, 37, 39, 43, 47,<br />
51, 55, 59, 63–66, AP ® Practice Problems<br />
3–8<br />
x =−1, − 1 3 , − 1 5 , − 1 7 , − 1 9 ,..., 1 9 , 1 7 , 1 5 , 1 3 , 1.<br />
(c) Compare the results from (a) and (b). What do you conclude<br />
about the limit? Why do you think this happens? What is<br />
your view about using a table to draw a conclusion about<br />
limits?<br />
(d) Use technology to graph f . Begin with the x-window<br />
[−2π, 2π] and the y-window [−1, 1]. If you were finding<br />
lim f (x) using a graph, what would you conclude? Zoom in<br />
x→0<br />
on the graph. Describe what you see. (Hint: Be sure your<br />
calculator is set to the radian mode.)<br />
10, 13, 19, 35, 37, 39, 43, 47, 51, 55, 59<br />
63, 66, 74, 75, 83, 93, 96<br />
All AP ® Practice Problems (especially 5)<br />
56. (a) Investigate lim cos π by using a table and evaluating the<br />
x→0 x2 function f (x) = cos π at x =−0.1, −0.01, −0.001,<br />
x2 −0.0001, 0.0001, 0.001, 0.01, 0.1.<br />
13–18, 24, 29, 45, 51–56, 59, 62, 63,<br />
73, 79<br />
All AP ® Practice Problems<br />
11,15,19, 23, 24, 25, 35, 38, 43, 45, 55,<br />
70<br />
All AP ® Practice Problems<br />
17–26, 30–35, 40, 45, 50, 53, 55, 60, 63,<br />
70, 73, 74, 75, 78, 86<br />
All AP ® Practice Problems<br />
7 Review Chapter 1 Review Exercises Chapter 1 AP ® Review Problems<br />
8 Test Chapter 1 Test AP ® Practice Exam, Big Idea 1: Limits<br />
1-6 Chapter 1 • Limits and Continuity<br />
TE_<strong>Sullivan</strong>_Chapter01_PART 0.indd 5<br />
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<strong>Sullivan</strong> AP˙<strong>Sullivan</strong>˙Chapter01 September October 8, 20, 2016 2016 17:416:7<br />
1 Limits and Continuity<br />
(b) Investigate lim cos π by using a table and evaluating the<br />
x→0 x2 function f (x) = cos π x 2 at<br />
x =− 2 3 , − 2 5 , − 2 7 , − 2 9 ,..., 2 9 , 2 7 , 2 5 , 2 3 .<br />
(c) Compare the results from (a) and (b). What do you conclude<br />
about the limit? Why do you think this happens? What is your<br />
view about using a table to draw a conclusion about limits?<br />
(d) Use technology to graph f . Begin with the x-window<br />
[−2π, 2π] and the y-window [−1, 1]. If you were finding<br />
lim f (x) using a graph, what would you conclude? Zoom in<br />
x→0<br />
on the graph. Describe what you see. (Hint: Be sure your<br />
calculator is set to the radian mode.)<br />
PAGE<br />
x − 8<br />
85 57. (a) Use a table to investigate lim .<br />
x→2 2<br />
(b) How close must x be to 2, so that f (x) is within 0.1 of the<br />
limit?<br />
(c) How close must x be to 2, so that f (x) is within 0.01 of the<br />
limit?<br />
58. (a) Use a table to investigate lim(5 − 2x).<br />
x→2<br />
(b) How close must x be to 2, so that f (x) is within 0.1 of the<br />
limit?<br />
(c) How close must x be to 2, so that f (x) is within 0.01 of the<br />
limit?<br />
59. First-Class Mail As of April<br />
2016, the U.S. Postal Service<br />
charged $0.47 postage for<br />
first-class letters weighing up to<br />
and including 1 ounce, plus a flat<br />
fee of $0.21 for each additional<br />
or partial ounce up to and<br />
including 3.5 ounces. First-class<br />
letter rates do not apply to letters<br />
weighing more than 3.5 ounces.<br />
Source: U.S. Postal Service Notice 123<br />
(a) Find a function C that models the first-class postage charged,<br />
in dollars, for a letter weighing w ounces. Assume w>0.<br />
(b) What is the domain of C?<br />
(c) Graph the function C.<br />
(d) Use the graph to investigate lim C(w) and lim C(w). Do<br />
w→2− w→2 +<br />
1.1 Limits of Functions Using<br />
Numerical these suggest and that Graphical lim C(w) exists?<br />
w→2<br />
(e) Techniques Use the graph to investigate lim C(w).<br />
w→0 +<br />
1.2(f) Limits Use the ofgraph Functions to investigate Using lim<br />
Properties of Limits<br />
C(w).<br />
w→3.5 −<br />
60. 1.3 First-Class Continuity Mail As of April 2016, the U.S. Postal Service<br />
charged $0.94 postage for first-class large envelope weighing up to<br />
1.4 Limits and Continuity of<br />
and including 1 ounce, plus a flat fee of $0.21 for each additional<br />
or partial Trigonometric, ounce up toExponential, and includingand<br />
13 ounces. First-class rates do<br />
notLogarithmic apply to largeFunctions<br />
envelopes weighing more than 13 ounces.<br />
1.5 Source: Infinite U.S. Limits; Postal Service LimitsNotice at Infinity; 123<br />
Asymptotes<br />
(a) Find a function C that models the first-class postage charged,<br />
1.6 The in dollars, ε-δ Definition for a largeofenvelope a Limitweighing w ounces. Assume<br />
Chapter w>0. Review<br />
(b)<br />
Chapter<br />
What is<br />
Project<br />
the domain of C?<br />
Kathryn Sidenstricker /Dreamstime.com<br />
(c) Graph the function C.<br />
(d) Use the graph to investigate lim C(w) and lim C(w). Do<br />
w→1− w→1 +<br />
these suggest that lim C(w) exists?<br />
w→1<br />
(e) Use the graph to investigate lim C(w) and lim C(w).<br />
w→12− w→12 +<br />
Do these suggest that lim C(w) exists?<br />
w→12<br />
(f) Use the graph to investigate lim C(w).<br />
w→0 +<br />
(g) Use the graph to investigate lim C(w).<br />
w→13 −<br />
61. Correlating Student Success to Study Time Professor Smith<br />
claims that a student’s final exam score is a function of the time t<br />
(in hours) that the student studies. He claims that the closer to<br />
seven hours one studies, the closer to 100% the student scores<br />
on the final. He claims that studying significantly less than seven<br />
hours may cause one to be underprepared for the test, while<br />
studying significantly more than seven hours may cause<br />
“burnout.”<br />
(a) Write Professor Smith’s claim symbolically as a limit.<br />
(b) Write Professor Smith’s claim using the ε-δ definition<br />
of limit.<br />
Source: Submitted by the students of Millikin University.<br />
62. The definition of the slope of the tangent line to the graph of<br />
f (x) − f (c)<br />
y = f (x) at the point (c, f (c)) is m tan = lim<br />
.<br />
x→c x − c<br />
Another way to express this slope is to define a new variable<br />
h = x − c. Rewrite the slope of the tangent line m tan using h and c.<br />
63. If f (2) = 6, can you conclude anything about lim f (x)? Explain<br />
x→2<br />
your reasoning.<br />
64. If lim f (x) = 6, can you conclude anything about f (2)? Explain<br />
x→2<br />
your reasoning.<br />
65. The graph of f (x) = x − 3 is a straight line with a point punched<br />
3 − x<br />
out.<br />
(a) What straight line and what point?<br />
(b) Use the graph of f to investigate the one-sided limits of f as<br />
x approaches 3.<br />
AP Photo<br />
(c) Does the graph suggest that lim f (x) exists? If so, what is it?<br />
x→3<br />
Oil Spills and Dispersant Chemicals<br />
66. (a) Use a table to investigate lim(1 + x) 1/x .<br />
x→0<br />
On April 20, 2010, the Deepwater (b) Horizon Use graphing drillingtechnology rig exploded toand graph initiated g(x) = the (1 worst + x) 1/x marine . oil<br />
spill in recent history. Oil gushed (c) from What thedo well (a) for and three (b) suggest monthsabout and released lim(1 + millions x) 1/x ? of gallons<br />
of crude oil into the Gulf of Mexico. One technique used to help clean x→0<br />
up during and after the<br />
CAS<br />
spill was the use of the chemical (d) dispersant Find lim(1 + x) 1/x .<br />
x→0<br />
Corexit. Oil dispersants allow the oil particles to<br />
spread more freely in the water, thus allowing the oil to biodegrade more quickly. Their use is<br />
debated, however, because some of their ingredients are carcinogens. Further, the use of oil<br />
dispersants can increase toxic Challenge hydrocarbon Problems levels affecting sea life. Over time, the pollution<br />
caused by the oil spill and the dispersants will eventually diminish and sea life will return, more<br />
or less, to its previous condition.<br />
For Problems<br />
In the short<br />
67–70,<br />
term,<br />
investigate<br />
however,<br />
each of the following limits.<br />
{<br />
pollution raises serious questions<br />
about the health of the local sea life and the safety of fish 1 and if shellfish x an for integer human consumption.<br />
f (x) =<br />
Explore a hypothetical situation of pollution in a lake in the Chapter 0 1 Project if x isonnot p. 156. an integer<br />
67. lim f (x) 68. lim f (x) 69. lim f (x) 70. lim f (x)<br />
x→2 x→1/2 x→3 x→0<br />
77<br />
PD Chapter 1 Overview<br />
Watch the chapter overview video for<br />
expert advice on teaching the content in<br />
this chapter, anticipating likely problem<br />
spots, and guidance on staying on pace.<br />
Teaching Tip<br />
Limits are foundational to the study of<br />
calculus. This chapter opens with a<br />
practical application of limits by modeling<br />
the level of contamination in Clear Lake.<br />
Students will have an opportunity to<br />
apply what they have learned through<br />
this chapter by completing the Chapter 1<br />
Project, Pollution in Clear Lake, at the end<br />
of this unit of study.<br />
TRM Chapter 1 Bell Ringers<br />
Each day, the students review and<br />
evaluate 5 trigonometric functions and 2<br />
logarithmic functions without the use of<br />
a calculator. Each problem is presented<br />
on a PowerPoint slide, and the problems<br />
transition on a timer, allowing you time to<br />
take attendance or check homework while<br />
your students work the problems. At the<br />
end of the Bell Ringer file is a 20-question<br />
quiz that can be given on the day that you<br />
review the chapter.<br />
WEB SITE<br />
Mathisfun: The Math Is Fun Web site is a<br />
helpful resource for the student who would<br />
benefit from learning about limits in simple<br />
terms with lots of illustrations. A link to this<br />
resource is available on the Additional<br />
Chapter 1 Resources document, available<br />
for download.<br />
Section 1.1 • Assess Your Understanding 89<br />
Where to Find the<br />
u Teacher Resources?<br />
t<br />
All of the Teacher Resource Materials listed in<br />
the blue pages for this chapter and referenced<br />
through the PD and TRM icons may be<br />
found by clicking on the links in the Teacher’s<br />
e-Book (TE-book), logging in to LaunchPad<br />
(password required) highschool.bfwpub.com/<br />
launchpad/apsullivan2e, or opening the<br />
Teacher’s Resource Flash Drive (TRFD).<br />
TRM AP® Calc AB Exam Prep Flashcards<br />
You may want to give your students the AP ® Calc<br />
AB Exam Prep Flashcards now so that they may<br />
use them throughout the year. Have them separate<br />
the Chapter 1 cards by referring to the chapter<br />
number in the bottom corner.<br />
Chapter 1 • Limits and Continuity<br />
77<br />
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<strong>Sullivan</strong><br />
TRM Alternate Examples Section 1.1<br />
You can find the Alternate Examples for<br />
this section in PDF format in the Teacher’s<br />
Resource Materials.<br />
TRM AP® Calc Skill Builders<br />
Section 1.1<br />
You can find the AP ® Calc Skill Builders for<br />
this section in PDF format in the Teacher’s<br />
Resource Materials.<br />
WEB SITE<br />
Many Web sites provide help for teaching<br />
and learning calculus. Some provide<br />
advice and concept development videos<br />
or applets, while others offer step-by-step<br />
support in working calculus problems.<br />
Three very useful sites:<br />
• Khan Academy is a well-known source<br />
of good tutorials, and the College Board<br />
has partnered with Khan to provide<br />
instructional videos on released AP ®<br />
Exam problems. https://www.<br />
khanacademy.org/math/calculus-home/<br />
ap-calc-topic<br />
• Lin McMullin’s Teaching Calculus blog<br />
offers an extensive set of resources for<br />
teachers and students, including<br />
instructional videos for most topics,<br />
suggestions for addressing the MPACs,<br />
and other helpful information.<br />
https://teachingcalculus.com/<br />
• The Wolfram|Alpha site offers help with<br />
solving many types of problems by “doing<br />
dynamic computations based on a vast<br />
collection of built-in data, algorithms, and<br />
methods.” It offers much helpful guidance<br />
to students when they are stuck and<br />
looking for quick specific help.<br />
https://www.wolframalpha.com/<br />
Links to these resources, and others,<br />
are found in the Additional Chapter 1<br />
Resources document, available for download.<br />
88 78 Chapter 1 • Limits and Continuity<br />
T<br />
2x 2 if x < 1<br />
33. f (x) =<br />
3x 2 at c = 1<br />
he concept of a limit 53. is Slope central of a to Tangent calculus. Line To understand For f (x) = calculus, 1<br />
− 1 if x > 1<br />
2 x2 − 1: it is essential to<br />
know what it means for a function to have a limit, and then how to find a limit of a<br />
x 3 function. Chapter 1 explains (a) what Finda the limit slope is, shows m<br />
if x < −1<br />
sec ofhow the secant to findline a limit containing of a function, the and<br />
34. f (x) =<br />
x 2 at c =−1 demonstrates how to prove thatpoints limitsP exist = (2, using f (2)) the anddefinition Q = (2 + of h, limit. f (2 + h)).<br />
− 1 if x > −1<br />
We begin the chapter using (b) Use numerical the result and from graphical (a) to complete approaches the following to explore table: the idea<br />
x 2 if x ≤ 0<br />
35. f (x) =<br />
at c = 0 of a limit. Although these methods seem to work well, there are instances in which they<br />
2x + 1 if x > 0<br />
⎧<br />
fail to identify the correct limit. h −0.5 −0.1 −0.001 0.001 0.1 0.5<br />
⎨ x 2 if x < 1<br />
In Section 1.2, we provide analytic m sec techniques for finding limits. Some of the proofs<br />
36. f (x) = 2 if x = 1 at c = 1<br />
⎩<br />
of these techniques are found in Section 1.6, others in Appendix B. A limit found by<br />
−3x + 2 if x > 1<br />
correctly applying these analytic (c) Investigate techniques theis limit precise; of thethere slopeis ofno thedoubt secantthat lineit found is correct. in (a)<br />
as h → 0.<br />
In Sections 1.3–1.5, we continue to study limits and some ways that they are used.<br />
(d) What is the slope of the tangent line to the graph of f at the<br />
Applications and Extensions<br />
For example, we use limits to define continuity, an important property of a function.<br />
point P = (2, f (2))?<br />
In Problems 37–40, sketch a graph of a function with the Section given 1.6 provides a(e) precise On the definition same set of ofaxes, limit, graph the so-called f and the tangent ε-δ (epsilon-delta)<br />
line to f at<br />
properties. Answers will vary.<br />
definition, which we use to showP = when (2, f a(2)).<br />
limit does, and does not, exist. ■<br />
37. lim f (x) = 3; lim f (x) = 3; lim f (x) = 1;<br />
x→2 x→3− x→3 +<br />
f (2) = 3; f (3) = 1<br />
54. Slope of a Tangent Line For f (x) = x 2 − 1:<br />
1.1 Limits of Functions Using Numerical<br />
38. lim f (x) = 0; lim f (x) =−2; lim f (x) =−2;<br />
(a) Find the slope m sec of the secant line containing the<br />
x→−1 x→2− x→2 + and Graphical Techniques<br />
points P = (−1, f (−1)) and Q = (−1 + h, f (−1 + h)).<br />
f (−1) is not defined; f (2) =−2<br />
39. lim f (x) = 4; lim f (x) =−1; lim f<br />
OBJECTIVES<br />
(x) = 0;<br />
When you(b) finish Usethis the result section, from you (a) to should complete be able the following to: table:<br />
x→1 x→0− x→0 +<br />
AP®<br />
f<br />
EXAM<br />
(0) =−1;<br />
INSIGHT<br />
f (1)<br />
Limits<br />
= 2<br />
is the first 1 Discuss the slope of a tangent line to a graph (p. 78)<br />
h −0.1 −0.01 −0.001 −0.0001 0.0001 0.001 0.01 0.1<br />
Big Idea in the AP® Calculus curriculum.<br />
40. lim f (x) = 2; lim f (x) = 0; lim 2 Investigate a limit using a table (p. 80)<br />
f (x) = 1;<br />
m sec<br />
x→2 x→−1 x→1 3 Investigate a limit using a graph (p. 82)<br />
f (−1) = 1; f (2) = 3<br />
(c) Investigate the limit of the slope of the secant line found<br />
Calculus can be used to solve certain (a) asfundamental h → 0. questions in geometry. Two of these<br />
In Problems 41–50, use either a graph or a table toquestions investigate are:<br />
(d) What is the slope of the tangent line to the graph of f at the<br />
each limit.<br />
point P = (−1, f (−1))?<br />
|x − 5|<br />
|x − 5|<br />
41. lim<br />
42. lim<br />
43.<br />
• Given<br />
lim<br />
a<br />
x→5 + x − 5<br />
x→5 − x − 5<br />
<br />
x→ 12<br />
2x<br />
function f and(e) a point On theP same on its set graph, of axes, what graph is the f and slope the tangent of the line<br />
to f<br />
tangent −<br />
to the graph of f at atP? P = See (−1, Figure f (−1)). 1.<br />
• Given a nonnegative function f whose domain<br />
PAGE<br />
44. lim <br />
x→ 12<br />
2x 45. lim<br />
+ <br />
x→ 23<br />
2x 46. lim<br />
− <br />
x→ 23<br />
2x<br />
+<br />
85 55. (a) Investigate lim cos π is the closed interval [a, b], what is<br />
the area of the region enclosed by the by using a table evaluating the<br />
x→0<br />
graph of x f , the x-axis, and the vertical lines<br />
x = a and x = b? See Figure 2.<br />
function f (x) = cos π<br />
47. lim |x|−x 48. lim |x|−x<br />
x at<br />
x→2 + x→2 −<br />
x =− 1<br />
3 3 2 , − 1 4 , − 1 8 , − 1 10 , − 1 12 ,..., 1<br />
12 , 1 10 , 1 8 , 1 4 , 1 y<br />
y f (x)<br />
y<br />
2 .<br />
49. lim x−x 50. lim x−x<br />
x→2 + x→2 − l T<br />
51. Slope of a Tangent Line For f (x) = 3x 2 (b) Investigate lim cos π y f (x)<br />
by using a table and evaluating the<br />
:<br />
Tangent x→0 x<br />
P<br />
line<br />
(a) Find the slope of the secant line containing the points (2, 12)<br />
function f (x) = cos π x at<br />
and (3, 27).<br />
(b) Find the slope of the secant line containing the points (2, 12)<br />
x =−1, − 1 3 , − 1 5 , − 1 7 , − 1 9 ,..., 1 9 , 1 7 , 1 5 , 1 x<br />
a<br />
3 , 1. b x<br />
and (x, f (x)), x = 2.<br />
(c) Compare the results from (a) and (b). What do you conclude<br />
(c) Create a table to investigate the slope of the DFtangent Figureline 1 to the<br />
Figure 2<br />
about the limit? Why do you think this happens? What is<br />
graph of f at 2 using the result from (b).<br />
your view about using a table to draw a conclusion about<br />
(d) On the same set of axes, graph f , the tangent line to the graph<br />
These questions, traditionally limits? called the tangent problem and the area problem,<br />
of f at the point (2, 12), and the secant line from (a).<br />
were solved by Gottfried Wilhelm von Leibniz and Sir Isaac Newton during the late<br />
(d) Use technology to graph f . Begin with the x-window<br />
52. Slope of a Tangent Line For f (x) = x 3 : seventeenth and early eighteenth [−2π, centuries. 2π] and The thesolutions y-window to[−1, the1]. twoIfseemingly you were finding different<br />
problems are both based on the idea<br />
(a) Find the slope of the secant line containing the points (2, 8)<br />
lim f of (x) a limit. using Their a graph, solutions what would not you onlyconclude? are related Zoom to each in<br />
x→0<br />
other, but are also applicable to many other problems in science and geometry. Here,<br />
and (3, 27).<br />
on the graph. Describe what you see. (Hint: Be sure your<br />
we begin to discuss the tangent problem. The discussion of the area problem begins in<br />
(b) Find the slope of the secant line containing the points (2, 8)<br />
calculator is set to the radian mode.)<br />
Chapter 5.<br />
and (x, f (x)), x = 2.<br />
56. (a) Investigate lim cos π by using a table and evaluating the<br />
(c) Create a table to investigate the slope of the tangent line to the<br />
x→0 x2 graph of f at 2 using the result from (b). 1 Discuss the Slope offunction a Tangent f (x) = Line cos π to at x a=−0.1, Graph −0.01, −0.001,<br />
(d) On the same set of axes, graph f , the tangent line to the graph<br />
x2 NEED TO REVIEW? The slope of a line<br />
Notice that the line <br />
is discussed of finatAppendix the point A.3, (2, 8), p. A-18.<br />
T in Figure 1 just touches the graph of f at the point P. This unique<br />
and the secant line from (a).<br />
−0.0001, 0.0001, 0.001, 0.01, 0.1.<br />
line is the tangent line to the graph of f at P. But how is the tangent line defined?<br />
∑ Mathematical Practices Tip<br />
MPAC 4: Connecting Multiple Representations<br />
Because limits can be presented numerically,<br />
graphically, and algebraically, you will have many<br />
opportunities to reinforce MPAC 4. Students<br />
may find that one method is more useful or<br />
appealing than another for a given function, but<br />
it is important that they have experience with<br />
multiple approaches. Show students how to solve<br />
the same problem by creating a table, examining<br />
a graph, and manipulating the problem using<br />
algebra. By constructing one representational form<br />
from another, you will help to solidify the concept<br />
in the students’ minds as they see that the various<br />
methods all lead to the same result.<br />
Teaching Tip<br />
In the AB pacing guide, two days have been<br />
allocated to Section 1.1 and two days have been<br />
allocated to Section 1.2. If you teach limits with all<br />
three techniques together (tabular, graphical, and<br />
analytical), then you may spend 4 days working<br />
on limits and still adhere to the schedule. The<br />
suggested pacing guide will allow you to cover all<br />
concepts on the AP ® Calculus Exam and still have<br />
approximately 3 weeks for review before the test.<br />
78<br />
Chapter 1 • Limits and Continuity<br />
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Section 1.1 • Limits of Functions Using Section Numerical 1.1 • Assess and Graphical Your Understanding Techniques 89 79<br />
(b) Investigate lim cos π by using a table and evaluating In planethe<br />
geometry, a tangent (c) Graph linethe tofunction a circleC.<br />
is defined as a line having exactly one<br />
x→0 x2 function f (x) = cos π point in common with the circle, as shown in Figure 3. However, this definition does<br />
x 2 at<br />
(d) Use the graph to investigate lim C(w) and lim C(w). Do<br />
not work for graphs in general. For example, in Figure 4, w→1 the − three lines w→1 1 + , 2 , and 3<br />
x =− 2 3 , − 2 5 , − 2 7 , − 2 9 ,..., 2 9 , 2 7 , 2 5 , 2 contain the point P and have exactly these suggest one point thatinlimcommon C(w) exists? with the graph of f , but they<br />
3 .<br />
w→1<br />
P<br />
do not meet the requirement of just touching the graph at P. On the other hand,<br />
(e) Use the graph to investigate lim C(w) and lim C(w). the line<br />
T just touches the graph of f at P, but it intersects thew→12 graph − at other points. w→12<br />
(c) Compare the results from (a) and (b). What do you conclude<br />
+ It is the<br />
slope of the tangent line <br />
about the limit? Why do you think this happens? What is your T that Do distinguishes these suggestithat fromlimallC(w) otherexists?<br />
lines containing P.<br />
w→12<br />
Figure 3 Tangent line to a circle at the<br />
view about using a table to draw a conclusion about So before limits? defining a tangent (f) Use line, thewe graph investigate to investigate its slope, lim which C(w). we denote by m<br />
point P.<br />
tan .<br />
w→0<br />
We begin with the graph of a function f , a point P on its graph, +<br />
(d) Use technology to graph f . Begin with the x-window<br />
(g) Use the graph to investigate lim C(w). and the tangent line T<br />
to f at P, as shown in Figure 5.<br />
w→13<br />
[−2π, 2π] and the y-window [−1, 1]. If you were finding<br />
− y<br />
lim f (x) l using l a graph, what would you conclude? Zoom in<br />
2 1<br />
The tangent line T 61. to Correlating the graph ofStudent f at P Success must contain to Study theTime point P. Professor We denote Smiththe<br />
x→0l y f (x)<br />
3<br />
on the graph. Describe what you see. (Hint:<br />
coordinates<br />
Be sure your<br />
of P by (c, f (c)). claims Since that afinding student’s a final slopexam requires score two is a function points, of and thewe time have t<br />
calculator is set to the radian mode.) only one point on the tangent (in hours) line T that , wethe proceed studentas studies. follows. He claims that the closer to<br />
PAGE<br />
x −l 8 T<br />
85 57. (a) Use a table to investigate lim .<br />
Suppose we choose any seven point hours Q one = studies, (x, f (x)), the closer othertothan 100% P, the onstudent the graph scoresof f ,<br />
x→2 2<br />
on the final. He claims that studying significantly less than seven<br />
Tangent as shown in Figure 6. (Q can be to the left or to the right of P; we chose Q to be to the<br />
(b) How close must x be to 2, soline<br />
that f (x) is within 0.1 of the hours may cause one to be underprepared for the test, while<br />
limit?<br />
right of P.) The line containing studying thesignificantly points P = more (c, f than (c)) seven and Q hours = (x, mayf cause (x)) is called a<br />
P<br />
secant line of the graph of<br />
(c) How close must x be to 2, so that f (x) is within 0.01 of the “burnout.” f . The slope m sec of this secant line is<br />
limit?<br />
(a) Write Professor f (x) Smith’s − f (c)<br />
m claim symbolically as a limit.<br />
sec = (1)<br />
58. (a) Use a table to investigate lim(5 − 2x). x<br />
x→2 (b) Write Professorx Smith’s − c claim using the ε-δ definition<br />
(b) How close must x be to 2, so that f (x) is within 0.1 of the<br />
of limit.<br />
Figure 4<br />
Figure 7 shows three different points Q 1 , Q 2 , and Q 3 on the graph of f that are<br />
limit?<br />
successively closer to pointSource: P, andSubmitted three associated by the students secantof lines Millikin 1 , University.<br />
2 , and 3 . The closer<br />
(c) How close must x be to 2, so that f (x) isthe within point 0.01 Q is ofto the the point 62. P, The thedefinition closer theofsecant the slope lineof isthe to the tangent tangent line line to the T graph . Theof<br />
line T ,<br />
limit?<br />
the limiting position of these secant lines, is the tangent line to the graph<br />
f (x)<br />
of<br />
− f<br />
(c)<br />
at P.<br />
59. First-Class Mail As of April<br />
y = f (x) at the point (c, f (c)) is m tan = lim<br />
.<br />
y<br />
y<br />
y<br />
x→c<br />
2016, the U.S. Postal Service<br />
x − c<br />
y f (x)<br />
y f (x)<br />
charged $0.47 postage for<br />
Another Secant<br />
y f (x)<br />
way to express this slope is to define a new variable l 1 Secant<br />
first-class letters weighing up to<br />
h = x<br />
line<br />
l2l3 lines<br />
Q (x, f (x)) − c. Rewrite the slope of the tangent line m tan Q 1<br />
using h and c.<br />
and including 1 ounce, plus a flat<br />
63. If f (2) = 6, can you conclude anything about lim f (x)? Explain<br />
fee of $0.21 for each additional l x→2<br />
T<br />
your reasoning.<br />
l Q T<br />
2 l T<br />
Q<br />
or partial ounce up to and<br />
3<br />
Tangent<br />
64. If limTangent<br />
f (x) = 6, can you conclude anything about f (2)? Tangent Explain<br />
including P 3.5 (c, f ounces. (c)) First-class<br />
P (c, f (c)) x→2<br />
P (c, f (c))<br />
line<br />
line<br />
line<br />
letter rates do not apply to letters<br />
your reasoning.<br />
weighing more than 3.5 ounces.<br />
65. The graph of f (x) = x − 3 is a straight line with a point punched<br />
Source: U.S. Postal c Service Notice 123 x<br />
c<br />
3 − x<br />
out. x<br />
x<br />
c x x<br />
3 x 2 x 1<br />
x<br />
Figure (a) 5Find m tan<br />
a = function slope of C the thattangent modelsline.<br />
the first-class postage charged,<br />
Figure 6 m sec = slope of a(a) secant What line. straight line and DF what Figure point? 7<br />
in dollars, for a letter weighing w ounces. Assume w>0.<br />
(b) Use the graph of f to investigate the one-sided limits of f as<br />
(b) What is the domain of C?<br />
If the limiting position of xthe approaches secant lines 3. is the tangent line, then the limit of the<br />
(c) Graph the function C.<br />
slopes of the secant lines should (c) Does equal the graph the slope suggest of the thattangent lim f (x) exists? If so, what is it?<br />
(d) Use the graph to investigate lim C(w) and lim C(w). Do<br />
x→3<br />
line. Notice in Figure 7<br />
w→2− thatw→2 as + the point Q moves closer to the point P, the numbers x get closer to c. So,<br />
66. (a) Use a table to investigate lim(1 + x)<br />
these suggest that lim C(w) exists? equation (1) suggests that<br />
1/x .<br />
x→0<br />
w→2<br />
(e) Use the graph to investigate lim C(w).<br />
m (b) Use graphing technology to graph g(x) = (1 + x) 1/x tan = [Slope of the tangent line to f at P]<br />
.<br />
[<br />
]<br />
Tangent line at P w→0 + (c) What f do (x) (a) −and f (c) (b) suggest about lim(1 + x) 1/x ?<br />
(f) Use the graph to investigate lim C(w).<br />
= Limit of<br />
as x gets closer x→0 to c<br />
w→3.5 − x − c<br />
CAS (d) Find lim(1 + x) 1/x .<br />
x→0<br />
60. First-Class Mail As of April 2016, the U.S. InPostal symbols, Service we write<br />
P<br />
f (x) − f (c)<br />
charged $0.94 postage for first-class large envelope weighing up to<br />
m tan = lim<br />
and including 1 ounce, plus a flat fee of $0.21 for each additional<br />
x→c x − c<br />
or partial ounce up to and including 13 ounces. The First-class notation rates limdois read, Challenge “the limit Problems as x approaches c.”<br />
x→c<br />
not apply to large envelopes weighing more than 13 ounces.<br />
The tangent line to Forthe Problems graph of 67–70, a function investigate f at each a point of theP following = (c, f limits. (c)) is the line<br />
Source: U.S. Postal Service Notice 123<br />
{<br />
containing the point P whose slope is<br />
1 if x is an integer<br />
Kathryn Sidenstricker /Dreamstime.com<br />
f (x) =<br />
(a) Find a function C that models the first-class postage charged,<br />
f (x) − f (c)<br />
in dollars, for a large envelope weighing w ounces. Assume<br />
m tan = lim<br />
w>0.<br />
67. lim f (x) x→c 68. limx −f (x) c 69. lim<br />
x→2 x→1/2 x→3<br />
(b) What is the domain ofSecant<br />
C?<br />
provided the limit exists.<br />
Figure 8<br />
lines<br />
WEB SITE<br />
Mathscoop: The Mathscoop Web site has a<br />
secant and tangent line applet that you may want<br />
to demo in class. You can drag one point of the<br />
secant line to the point of tangency to show the<br />
relationship between the two lines. The applet<br />
allows you to click Play to show the animation<br />
to the students. This can help to show how x<br />
approaches c when demonstrating how the slope<br />
of the secant line approaches that of the tangent<br />
line. A link to this resource is available on the<br />
Additional Chapter 1 Resources document,<br />
available for download.<br />
0 if x is not an integer<br />
f (x) 70. lim<br />
x→0 f (x)<br />
As Figure 8 illustrates, this new idea of a tangent line is consistent with the traditional<br />
definition of a tangent line to a circle.<br />
Teaching Tip<br />
Students should be familiar with finding the slope<br />
of a line given two points. When those two points<br />
fall on the graph of a function, we call the line that<br />
connects them the secant line. The word “secant”<br />
comes from the Latin word secare, meaning to<br />
cut. As the two points that create the secant line<br />
become closer and closer together on the curve,<br />
the slope of the secant line approaches the slope<br />
of the tangent line to the point upon which they are<br />
converging.<br />
common error<br />
Students may think that a tangent line<br />
must only touch the function at one point<br />
and cannot cross or touch the function<br />
at any other point. However, the tangent<br />
line is a localized concept. The tangent<br />
line may intersect the function at another<br />
point on the function. See in the figure that<br />
the graph of y = sin x, with a tangent line<br />
drawn at x = π / 3. This line intersects the<br />
function again in quadrant III.<br />
2π<br />
y<br />
1<br />
21<br />
Tangent line<br />
Teaching Tip<br />
Remind the students of the definition of<br />
slope with which they are familiar when<br />
naming the two points (x 1<br />
, y 1<br />
) and (x 2<br />
, y 2<br />
):<br />
m −<br />
=<br />
y y 2 1<br />
x − x<br />
2 1<br />
Ask the students to compare this to the<br />
formula for the slope of a secant line.<br />
fx ( ) − fc ()<br />
msec<br />
=<br />
x−<br />
c<br />
The only difference between these two<br />
formulas is the way the two points are<br />
named. In this case, the points are named<br />
(c,f(c)) and (c,f(x)).<br />
Then, point out that as the two points get<br />
closer and closer together (as x approaches<br />
c), the slope of the secant line approaches<br />
the slope of the tangent line. Hence the<br />
formula:<br />
fx fc<br />
m lim ( ) − ()<br />
x c<br />
.<br />
tan =<br />
x→c<br />
−<br />
Teaching Tip<br />
The word “tangent” comes from the Latin<br />
tangere, meaning to touch. Students may<br />
be familiar with the term “tangent line” from<br />
their geometry course, where they learned<br />
that a line that touches a circle at just one<br />
point is tangent to the circle. Let them know<br />
that a tangent line can be found for any<br />
function, not just circles.<br />
π<br />
x<br />
Section 1.1 • Limits of Functions Using Numerical and Graphical Techniques 79<br />
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<strong>Sullivan</strong><br />
88 80 Chapter 1 •• Limits and Continuity<br />
Teaching Tip<br />
Consider explaining limits in this way to the<br />
students: The limit of a function is the value<br />
that the function approaches as the x value<br />
gets very close to some value of interest.<br />
Practically speaking, the easiest way to find<br />
a limit is to just plug in the number, and if<br />
the result is an indeterminate form such as<br />
0<br />
, this means that you must now do more<br />
0<br />
work.<br />
Teaching Tip<br />
Clarify that the limit of a function as x<br />
approaches some number (let’s call that<br />
number c) is not necessarily the same as<br />
f (c). The limit of a function is the y value,<br />
f (c), that the function gets close to as the<br />
x value gets close to (but not equal to)<br />
the value, c. In the figure below, the value<br />
of f (c) = 3, while the limit of f (x) as x<br />
approaches c is 1.<br />
y<br />
3<br />
2<br />
1<br />
c<br />
(c, 3)<br />
y f (x)<br />
AP® CaLC skill builder<br />
for example 1<br />
Interpreting a Limit Expressed<br />
Symbolically<br />
Express the limit lim 5x<br />
= 10 in words and<br />
x→2<br />
interpret it.<br />
Solution<br />
In words, lim 5x<br />
= 10 is stated: The limit 5x<br />
x→2<br />
as x approaches 2 is equal to 10.<br />
The limit is interpreted as: The value<br />
of the function f(x) = 5x can be made as<br />
close as we please to 10 by choosing x<br />
sufficiently close to but not equal to 2.<br />
On the graph of the function y = 5x<br />
shown below, explain that as you choose<br />
values of x closer and closer to the 2 points<br />
on the graph close in on the value y = 10.<br />
y<br />
16<br />
14<br />
12<br />
10<br />
8<br />
6<br />
4<br />
2<br />
22 22<br />
2<br />
4<br />
6<br />
x<br />
x<br />
2x 2 if x < 1<br />
We have begun to<br />
33. f (x) =<br />
3x 2 at c = 1<br />
53.<br />
answer<br />
Slope<br />
the<br />
of a<br />
tangent<br />
Tangent<br />
problem<br />
Line For<br />
by<br />
f<br />
introducing<br />
(x) = 1<br />
− 1 if x > 1<br />
2 x2 −<br />
the<br />
1:<br />
idea of a limit.<br />
Now we describe the idea of a limit in more detail.<br />
x 3 (a) Find the slope m<br />
if x < −1<br />
sec of the secant line containing the<br />
34. f (x) =<br />
x 2 at c =−1 The Idea of a Limit points P = (2, f (2)) and Q = (2 + h, f (2 + h)).<br />
− 1 if x > −1<br />
We begin by asking a question: (b) Use What the result doesfrom it mean (a) tofor complete a function the following f to have table: a limit L<br />
x 2 if x ≤ 0<br />
35. f (x) =<br />
at c = 0 as x approaches some fixed number c? To answer the question, we need to be more precise<br />
2x + 1 if x > 0<br />
⎧<br />
about f , L, and c. To have a limit h at −0.5 c, the function −0.1 −0.001 f must be defined 0.001 everywhere 0.1 0.5 in<br />
⎨ x 2 if x < 1<br />
an open interval containing themnumber sec c, except possibly at c, and L must be a number.<br />
36. NEED f (x) TO=<br />
REVIEW? 2 If a < if b, x the = 1open<br />
at c = 1 Using these restrictions, we introduce the notation<br />
interval (a, b) ⎩consists −3x + 2of all if numbers x > 1 x<br />
(c) Investigate the limit of the slope of the secant line found in (a)<br />
for which a < x < b. Interval notation is<br />
lim<br />
as h → 0.<br />
f (x) = L<br />
x→c<br />
discussed in Appendix A.1, p. A-5.<br />
(d) What is the slope of the tangent line to the graph of f at the<br />
Applications and Extensions<br />
which is read, “the limit as x point approaches P = (2, cf (2))? of f (x) is equal to the number L.” The<br />
In Problems 37–40, sketch a graph of a function with notation the given lim f (x) = L can(e) be described On the same as<br />
x→c set of axes, graph f and the tangent line to f at<br />
properties. Answers will vary.<br />
37. lim f (x) = 3; lim f (x) = 3; lim<br />
The value f (x) P = can (2, be f (2)). made as close as we please to L,<br />
f (x) = 1;<br />
x→2 x→3− x→3 + for x sufficiently close to c, but not equal to c.<br />
f (2) = 3; f (3) = 1<br />
54. Slope of a Tangent Line For f (x) = x 2 − 1:<br />
Figure 9 shows that as x gets closer to c, the value f (x) gets closer to L. In<br />
38. lim f (x) = 0; lim f (x) =−2; lim Figure f (x) =−2; 9(a), lim f (x) = L(a) and Find f (c) the slope = L, mwhile sec of the in secant Figureline 9(b), containing lim f (x) the = L, but<br />
x→−1 x→2− x→2 + x→c x→c<br />
points P = (−1, f (−1)) and Q = (−1 + h, f (−1 + h)).<br />
f (−1) is not defined; f (2) =−2<br />
f (c) = L. In Figure 9(c) lim f (x) = L, but f is not defined at c.<br />
x→c<br />
39. lim f (x) = 4; lim f (x) =−1; lim<br />
(b) Use the result from (a) to complete the following table:<br />
f (x) = 0;<br />
x→1 x→0− x→0 +<br />
y<br />
y<br />
f (0) =−1; f (1) = 2<br />
(c, f(c))<br />
y<br />
h −0.1 −0.01 −0.001 −0.0001 0.0001 0.001 0.01 0.1<br />
40. lim f (x) = 2; lim f (x) = 0; lim f (x) = 1;<br />
m sec<br />
x→2 x→−1 x→1<br />
L f (−1) = 1; f (2) = 3<br />
L<br />
(c) Investigate the limit L of the slope of the secant line found<br />
(c, L)<br />
in (a) as h → 0.<br />
In Problems 41–50, use either a graph or a table to investigate<br />
(d) What is the slope of the tangent line to the graph of f at the<br />
each limit.<br />
point P = (−1, f (−1))?<br />
|x − 5|<br />
y 5 |xf (x) − 5|<br />
y 5 f (x)<br />
y 5 f (x)<br />
41. lim<br />
42. lim<br />
43. lim<br />
x→5 + x − 5<br />
x→5 − x − 5<br />
<br />
x→ 12<br />
2x<br />
(e) On the same set of axes, graph f and the tangent line to f<br />
−<br />
at P = (−1, f (−1)).<br />
a c b x<br />
a c b x<br />
PAGE<br />
44. lim <br />
x→ 12<br />
2x 45. lim<br />
+ <br />
x→ 23<br />
2x 46. lim<br />
− <br />
x→ 23<br />
2x<br />
+<br />
85 55. (a) Investigate lim cos π a c b<br />
x<br />
by using a table and evaluating the<br />
(a) f is defined at c; f(c) 5 L (b) f is defined at c; f(c) Þ L x→0 x<br />
function f (x) = cos π (c) f is not defined at c<br />
Figure 9 lim f (x) = L<br />
47. lim |x|−x 48. lim |x|−x<br />
x at<br />
x→c<br />
x→2 + x→2 −<br />
x =− 1<br />
3 3 2 , − 1 4 , − 1 8 , − 1 10 , − 1 12 ,..., 1<br />
12 , 1 10 , 1 8 , 1 4 , 1 2 .<br />
49. lim x−x 50. lim x−x<br />
x→2 + x→2 − EXAMPLE 1 Interpreting a Limit Expressed<br />
51. Slope of a Tangent Line For f (x) = 3x 2 (b) Investigate lim cos π Symbolically<br />
by using a table and evaluating the<br />
: Express the limit<br />
x→0 x<br />
(a) Find the slope of the secant line containing the points (2, 12)<br />
function f x(x) = cos π 2 − 1<br />
lim<br />
x at<br />
and (3, 27).<br />
x→−1 x<br />
(b) Find the slope of the secant line containing the points (2, 12)<br />
x =−1, − 1 +<br />
3 , − 1 =−2<br />
1 5 , − 1 7 , − 1 9 ,..., 1 9 , 1 7 , 1 5 , 1 in words and interpret it.<br />
3 , 1.<br />
and (x, f (x)), x = 2.<br />
(c) Compare the results from (a) and (b). What do you conclude<br />
(c) Create a table to investigate the slope of the tangent line to the x 2 − 1<br />
Solution In words, lim about =−2, the limit? is stated, Why do“The you think limitthis as happens? x approaches What is −1 of<br />
graph of f at 2 using the result from (b).<br />
x→−1 x + 1<br />
your view about using a table to draw a conclusion about<br />
(d) On the same set of axes, graph f , the tangent x 2 −line 1 to the graph<br />
limits?<br />
of f at the point (2, 12), and the secant line from (a). is equal to the number −2.” The limit is interpreted “The value of the function<br />
x + 1<br />
(d) Use technology to graph f . Begin with the x-window<br />
52. Slope of a Tangent Line For f (x) = x 3 :<br />
f (x) = x 2 − 1<br />
can be made [−2π, as close 2π] asand wethe please y-window to −2 [−1, by1]. choosing If you were x sufficiently finding<br />
x + 1<br />
(a) Find the slope of the secant line containing the points (2, 8)<br />
lim f (x) using a graph, what would you conclude? Zoom in<br />
x→0<br />
close to, but not equal to, −1.” ■<br />
and (3, 27).<br />
on the graph. Describe what you see. (Hint: Be sure your<br />
(b) Find the slope of the secant line containing the points (2, 8)<br />
calculator is set to theNOW radianWORK mode.) AP® Practice Problem 7.<br />
and (x, f (x)), x = 2.<br />
56. (a) Investigate lim cos π by using a table and evaluating the<br />
(c) Create a table to investigate the slope of the tangent line to the<br />
x→0 x2 graph of f at 2 using the result from (b). 2 Investigate a Limit Using functiona f Table (x) = cos π at x =−0.1, −0.01, −0.001,<br />
(d) On the same set of axes, graph f , the tangent line to the graph<br />
x2 We can use a table to better understand what it means for a function to have a limit as x<br />
of f at the point (2, 8), and the secant line from (a).<br />
−0.0001, 0.0001, 0.001, 0.01, 0.1.<br />
approaches a number c.<br />
AP® Exam Tip<br />
Questions about limits that appear on the<br />
exams often require algebraic manipulation.<br />
Students must become comfortable with<br />
factoring the difference of squares and<br />
trinomials.<br />
Teaching Tip<br />
Consider teaching the students about limits<br />
using graphical, tabular, and algebraic methods<br />
together. For example, first estimate the limit of<br />
a function graphically. Then solidify that estimate<br />
using tabular techniques. Finally, use algebraic<br />
manipulation (when necessary) and substitution to<br />
show that all three methods give the same answer.<br />
∑ Mathematical Practices Tip<br />
Connecting Multiple Representations<br />
To find the limit as x approaches a, we choose x<br />
sufficiently close to a. Ask students to define the<br />
term “sufficiently.” Observe that as we choose<br />
any x as close to a as we want, the values of<br />
f(x) approach a number b. Draw the graph of<br />
a function and have students describe how the<br />
points on the graph of f move along the curve<br />
as x approaches b. Explain that the limit is an<br />
important concept that will be used to define the<br />
continuity of a function.<br />
On the graph that follows, note that the curve<br />
is smooth and unbroken so that the points on<br />
80<br />
Chapter 1 • Limits and Continuity<br />
TE_<strong>Sullivan</strong>_Chapter01_PART 0.indd 9<br />
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Section 1.1 • Limits of Functions Using Section Numerical 1.1 • Assess and Graphical Your Understanding Techniques 89 81<br />
(b) Investigate lim cos π by using a table and evaluating the<br />
x→0 x2 EXAMPLE 2 Investigating (c) Grapha Limit the function Using C. a Table<br />
function f (x) = cos π x 2 at<br />
Investigate lim(2x + 5) using (d) aUse table. the graph to investigate lim C(w) and lim C(w). Do<br />
w→1− w→1 + x→2<br />
x =− 2 3 , − 2 5 , − 2 7 , − 2 9 ,..., 2 9 , 2 7 , 2 5 , 2 these suggest that lim C(w) exists?<br />
3 . Solution We create Table 1 by evaluating f (x) w→1 = 2x +5 at values of x near 2, choosing<br />
numbers x slightly less than(e) 2 and Use numbers the graph to x slightly investigate greater lim than C(w) 2. and lim C(w).<br />
w→12− w→12<br />
(c) Compare the results from (a) and (b). What do you conclude<br />
+ Do these suggest that lim C(w) exists?<br />
about the limit? Why do you think this happens? What is your<br />
w→12<br />
TABLE 1view about using a table to draw a conclusion about limits? (f) Use the graph to investigate lim C(w).<br />
w→0<br />
(d) Use technology to graph numbers f . Begin x slightly with the less x-window than 2<br />
+ (g) Use the graph to investigate numbers limx slightly C(w). greater than 2<br />
−−−−−−−−−−−−−−−−−−−−−→<br />
←−−−−−−−−−−−−−−−−−−−−−−−−<br />
w→13<br />
[−2π, 2π] and the y-window [−1, 1]. If you were finding<br />
−<br />
x 1.99 1.999 1.9999 1.99999 → 2 ← 2.00001 2.0001 2.001 2.01<br />
lim f (x) using a graph, what would you conclude? Zoom in 61. Correlating Student Success to Study Time Professor Smith<br />
f (x) = x→0 2x + 5 8.98 8.998 8.9998 8.99998 f (x)<br />
on the graph. Describe what you see. (Hint: Be sure your<br />
claims approaches that a student’s 9 9.00002 final exam score 9.0002 is a function 9.002 of the time 9.02 t<br />
calculator is set to the radian mode.)<br />
(in hours) that the student studies. He claims that the closer to<br />
PAGE<br />
x − 8 Table 1 suggests that the value sevenof hours f (x) one = studies, 2x + 5the can closer be made to 100% “asthe close student as we scores please”<br />
85 57. (a) Use a table to investigate lim .<br />
x→2 2<br />
on the final. He claims that studying significantly less than seven<br />
to 9 by choosing x “sufficiently close” to 2. This suggests that lim(2x + 5) = 9. ■<br />
(b) How close must x be to 2, so that f (x) is within 0.1 of the hours may cause one to be underprepared x→2 for the test, while<br />
limit?<br />
studying significantly more than seven hours may cause<br />
NOW WORK Problem 9.<br />
(c) How close must x be to 2, so that f (x) is within 0.01 of the “burnout.”<br />
limit?<br />
(a) Write Professor Smith’s claim symbolically as a limit.<br />
In creating Table 1, first we used numbers x close to 2 but less than 2, and then we<br />
58. (a) Use a table to investigate lim(5 − 2x).<br />
x→2 used numbers x close to 2(b) but Write greater Professor than 2. Smith’s When claim x < 2, using we the say, ε-δ “xdefinition<br />
is approaching<br />
(b) How close must x be to 2, so that f (x) is within 0.1 of the<br />
of limit.<br />
2 from the left,” and the number 9 is called the left-hand limit. When x > 2, we say,<br />
limit?<br />
“x is approaching 2 from the Source: right,” Submitted and the bynumber the students 9 is ofcalled Millikinthe University. right-hand limit.<br />
(c) How close must x be to 2, so that f (x) isTogether, within 0.01 these of the are called<br />
62.<br />
the<br />
The<br />
one-sided<br />
definition of<br />
limits<br />
the slope<br />
of f<br />
of<br />
as<br />
the<br />
x approaches<br />
tangent line to<br />
2.<br />
the graph of<br />
limit?<br />
One-sided limits are symbolized as follows. The left-hand limit, written f (x) − f (c)<br />
59. First-Class Mail As of April<br />
y = f (x) at the point (c, f (c)) is m tan = lim<br />
.<br />
x→c<br />
2016, the U.S. Postal Service<br />
lim f (x) = L x − c<br />
left<br />
x→c<br />
charged $0.47 postage for<br />
Another way −<br />
to express this slope is to define a new variable<br />
first-class letters weighing up to<br />
h = x − c. Rewrite the slope of the tangent line m tan using h and c.<br />
and including 1 ounce, plus a flat<br />
is read, “The limit as x approaches c from the left of f (x) equals L<br />
63. If f (2) = 6, can you conclude anything about left .” It means that<br />
lim f (x)? Explain<br />
fee of $0.21 for each additional<br />
the value of f can be made as close as we please to the number L left by x→2 choosing x < c<br />
your reasoning.<br />
or partial ounce up to and<br />
and sufficiently close to c.<br />
64. If lim f (x) = 6, can you conclude anything about f (2)? Explain<br />
including 3.5 ounces. First-class<br />
Similarly, the right-hand x→2 limit, written<br />
letter rates do not apply to letters<br />
your reasoning.<br />
weighing more than 3.5 ounces.<br />
lim<br />
65. The graph of f f (x) (x) = = xL− right 3<br />
x→c +<br />
is a straight line with a point punched<br />
Source: U.S. Postal Service Notice 123<br />
3 − x<br />
out.<br />
is read, “The limit as x approaches c from the right of f (x) equals L right .” It means that<br />
(a) Find a function C that models the first-class<br />
the<br />
postage<br />
value of<br />
charged,<br />
f can be made(a) as close Whatas straight we please line and to the what number point? L right by choosing x > c<br />
in dollars, for a letter weighing w ounces. Assume w>0.<br />
and sufficiently close to c.<br />
(b) What is the domain of C?<br />
(c) Graph the function C.<br />
Kathryn Sidenstricker /Dreamstime.com<br />
EXAMPLE 3 Investigating (c) Does a Limit the graph Using suggest a Table that lim f (x) exists? If so, what is it?<br />
(d) Use the graph to investigate lim C(w) and lim C(w). Do<br />
x→3<br />
w→2− w→2 + e x − 1<br />
Investigate lim 66. using (a) a Use table. a table to investigate lim(1 + x)<br />
these suggest that lim C(w) exists?<br />
1/x .<br />
x→0 x<br />
x→0<br />
w→2<br />
(e) Use the graph to investigate lim C(w).<br />
(b) Use graphing technology to graph g(x) = (1 + x) 1/x .<br />
w→0 + Solution The domain of f (c) (x) What = ex − 1<br />
do (a) and is {x|x (b) suggest = 0}. So, about f lim is defined (1 + x) 1/x everywhere ? in<br />
(f) Use the graph to investigate lim C(w).<br />
x<br />
x→0<br />
w→3.5 − an open interval containing CAS (d) theFind number lim(1 0, + except x) 1/x . for 0.<br />
x→0<br />
60. First-Class Mail As of April 2016, the U.S. Postal Service<br />
e x − 1<br />
We create Table 2, investigating the left-hand limit lim and the right-hand<br />
charged $0.94 postage for first-class large envelope weighing up to<br />
x→0 − x<br />
and including 1 ounce, plus a flat fee of $0.21 for each additional e x − 1<br />
limit lim . First,<br />
or partial ounce up to and including 13 ounces. First-class rates do Challenge we evaluate Problems f at numbers less than 0, but close to zero, and<br />
x→0 + x<br />
not apply to large envelopes weighing more than then13at ounces. numbers greaterFor than Problems 0, but close 67–70, toinvestigate zero. each of the following limits.<br />
Source: U.S. Postal Service Notice 123<br />
{ 1 if x is an integer<br />
TABLE f (x) =<br />
(a) 2Find a function C that models the first-class postage charged,<br />
0 if x is not an integer<br />
in dollars, for a large envelope x approaches weighing0w from ounces. theAssume<br />
left<br />
x approaches 0 from the right<br />
w>0.<br />
−−−−−−−−−−−−−−−−−−−−→ 67. lim f (x) 68. lim f (x) 69. lim f (x) 70. lim f (x)<br />
x→2<br />
←−−−−−−−−−−−−−−−−−−−−−<br />
x→1/2 x→3 x→0<br />
(b) Use the graph of f to investigate the one-sided limits of f as<br />
x approaches 3.<br />
(b) What x is the domain −0.01 of C? −0.001 −0.0001 −0.00001 → 0 ← 0.00001 0.0001 0.001 0.01<br />
f (x) = ex − 1<br />
0.995 0.9995 0.99995 0.999995 f (x) approaches 1 1.000005 1.00005 1.0005 1.005<br />
x<br />
∑ MPAC Tip continued<br />
the graph gradually close in on the limiting value<br />
3<br />
y = as the values of x approach 1.<br />
2<br />
y<br />
3<br />
2<br />
1<br />
1<br />
2<br />
3<br />
x<br />
Teaching Tip<br />
Remember to model proper notation for your<br />
students when working with one-sided limits.<br />
• One-sided limit of f(x) as x approaches a from<br />
the left: lim fx ( )<br />
x→a<br />
−<br />
• One-sided limit of f(x) as x approaches a from<br />
the right: lim fx ( )<br />
x→ a<br />
+<br />
Calculator Tip<br />
Most calculators have table settings<br />
that default to display a starting value<br />
(usually zero) and additional values with a<br />
consistent incremental change thereafter<br />
(like 0, 1, 2, 3…). Table settings can<br />
be changed (usually from Auto to Ask<br />
so students can input any independent<br />
variable they want. The table will then<br />
display the y-values that correspond to<br />
the selected x-values.<br />
TRM Section 1.1: Worksheet 1<br />
This worksheet includes 3 problems for<br />
which a student completes a premade<br />
table to find the limit of a function by using<br />
a table.<br />
AP® CaLC skill builder<br />
for example 3<br />
Investigating a Limit Using a Table<br />
2<br />
x −1<br />
Investigate lim using a table.<br />
x→1<br />
x −1<br />
Solution<br />
2<br />
x −1<br />
The domain of fx ( ) = is {x | x ≠ 1}. So<br />
x −1<br />
f is defined everywhere in an open interval<br />
containing 1, except for 1. We create a<br />
table by evaluating the function at values of<br />
x near 1, choosing values that are slightly<br />
less than 1 and slightly greater than 1.<br />
x<br />
f(x)<br />
0.9 1.9<br />
0.99 1.99<br />
0.999 1.999<br />
1 Not defined<br />
1.001 2.001<br />
1.01 2.01<br />
1.1 2.1<br />
This table suggests that the value of<br />
2<br />
x −1<br />
fx ( ) = gets as close as we please<br />
x −1<br />
to 2 as x gets very close to 1. That is, there<br />
2<br />
x −1<br />
is evidence indicating that lim<br />
− = 2.<br />
x→1<br />
x 1<br />
In Section 1.2, students will verify<br />
2<br />
x −1<br />
lim algebraically by factoring the<br />
x→1<br />
x −1<br />
numerator and reducing the expression<br />
before substituting the value of 1.<br />
Section 1.1 • Limits of Functions Using Numerical and Graphical Techniques<br />
81<br />
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<strong>Sullivan</strong> AP˙<strong>Sullivan</strong>˙Chapter01 October 8, 2016 17:4<br />
<strong>Sullivan</strong><br />
88 82 Chapter 1 • Limits and Continuity<br />
AP® Exam Tip<br />
It is helpful to be familiar with limits of<br />
ax<br />
the form lim sin( ) before taking the<br />
x→0<br />
exam.<br />
x<br />
ax<br />
lim sin( ) = a<br />
x→0<br />
x<br />
Consider working with variations of this<br />
problem as well, such as<br />
ax a<br />
lim sin( ) =<br />
x→0<br />
bx b<br />
These limits can be found using the<br />
Squeeze Theorem, but familiarity will<br />
save time during the exam.<br />
On completion of Section 4.5, the<br />
student will be able to solve problems<br />
of this form using L’Hôpital’s Rule, but<br />
in the meantime, students may prefer<br />
to rely on these two forms rather than<br />
finding these limits using the Squeeze<br />
Theorem.<br />
AP® CaLC skill builder<br />
for example 4<br />
Investigating a Limit Using a Table and<br />
Technology<br />
x<br />
Investigate lim sin(5 ) using a table.<br />
x→0<br />
x<br />
Solution<br />
sin(5 x)<br />
The domain of the function fx ( ) =<br />
x<br />
is {x | x ≠ 0}. So f is defined everywhere<br />
on the open interval containing 0, except<br />
at 0. We create a table by evaluating the<br />
function at values of x near 0, choosing<br />
values that are slightly less than 0 and<br />
slightly greater than 0.<br />
2x 2 if x < 1<br />
33. f (x) =<br />
3x 2 at c = 1<br />
53. Slope<br />
e x of<br />
−<br />
a<br />
1<br />
Tangent Line For<br />
e x −<br />
f (x)<br />
1<br />
Table 2 suggests that lim = 1 and lim = 1.<br />
1 − 1 if x > 1<br />
2 x2 − 1:<br />
x→0<br />
− x<br />
x→0 + x<br />
x 3 if x < −1<br />
e x − 1 (a) Find the slope m sec of the secant line containing the<br />
34. f (x) =<br />
x 2 at c =−1 This suggests lim = 1. points ■<br />
− 1 if x > −1<br />
x→0<br />
P = (2, f (2)) and Q = (2 + h, f (2 + h)).<br />
x<br />
(b) Use the result from (a) to complete the following table:<br />
x 2 if x ≤ 0<br />
NOW WORK Problem 13 and AP® Practice Problem 6.<br />
35. f (x) =<br />
at c = 0<br />
2x + 1 if x > 0<br />
⎧<br />
h −0.5 −0.1 −0.001 0.001 0.1 0.5<br />
⎨ x 2 if x < 1<br />
m sec<br />
36. f (x) = 2 if x = 1 at c = 1 EXAMPLE 4 Investigating a Limit Using a Table and Technology<br />
⎩<br />
−3x + 2 if x > 1<br />
sin x (c) Investigate the limit of the slope of the secant line found in (a)<br />
Investigate lim using a table.<br />
x→0 x<br />
as h → 0.<br />
(d) What is the slope of the tangent line to the graph of f at the<br />
Applications and Extensions<br />
Solution The domain of the function point P = f (x) (2, f = (2))?<br />
sin x is {x | x = 0}. So, f is defined<br />
In Problems 37–40, sketch a graph of a function with the given<br />
x<br />
(e) On the same set of axes, graph f and the tangent line to f at<br />
properties. Answers will vary.<br />
everywhere in an open interval containing 0, except for 0.<br />
P = (2, f (2)).<br />
37. lim f (x) = 3; lim f (x) = 3; lim f (x) = 1;<br />
x→2 x→3− x→3 + We investigate the one-sided limits of sin x as x approaches 0 by using a graphing<br />
f (2) = 3; f (3) = 1<br />
54. Slope of a Tangent<br />
x<br />
Line For f (x) = x 2 − 1:<br />
calculator to set up a table. When making the table, we choose numbers x (in radians)<br />
38. lim f (x) = 0; lim f (x) =−2; lim slightly f (x) =−2; less than 0 and numbers (a) Find slightly the slope greater m sec than of the 0. secant See Figure line containing 10. the<br />
x→−1 x→2− x→2 +<br />
points P = (−1, f (−1)) and Q = (−1 + h, f (−1 + h)).<br />
Figure f (−1) 10<br />
The table in Figure 10 suggests that lim f (x) = 1 and lim f (x) = 1. This<br />
is not defined; f (2) =−2<br />
x→0 −<br />
x→0 +<br />
39. lim f (x) = 4; lim f (x) =−1; lim<br />
(b) Use the result from (a) to complete the following table:<br />
f (x) = 0; sin x<br />
x→1 x→0− x→0 + suggests that lim = 1. ■<br />
NOTE f (x) = sin x is an even function,<br />
f (0) =−1;<br />
x<br />
x→0 x<br />
f (1) = 2<br />
h −0.1 −0.01 −0.001 −0.0001 0.0001 0.001 0.01 0.1<br />
so the Y 1 values in Figure 10 are<br />
40. lim f (x) = 2; lim f (x) = 0; lim f (x) = 1;<br />
m sec<br />
NOW WORK Problem 15 and AP® Practice Problem 8.<br />
symmetric x→2<br />
about x = 0.<br />
x→−1 x→1<br />
f (−1) = 1; f (2) = 3<br />
(c) Investigate the limit of the slope of the secant line found<br />
in (a) as h → 0.<br />
In Problems 41–50, use either a graph or a table to 3investigate<br />
Investigate a Limit (d) Using What isa the Graph slope of the tangent line to the graph of f at the<br />
each limit.<br />
The graph of a function can alsopoint helpP us = investigate (−1, f (−1))? limits. Figure 11 shows the graphs<br />
|x − 5|<br />
|x − 5|<br />
41. lim<br />
42. lim<br />
43. of three lim<br />
x→5 + x − 5<br />
x→5 − x − 5<br />
<br />
different<br />
x→ 12<br />
2x functions (e) f , g, Onand theh. same Observe set of axes, that in graph eachf function, and the tangent as x line getstocloser<br />
f<br />
−<br />
to c, whether from the left or from at P the = (−1, right, f the (−1)). value of the function gets closer to the<br />
number L. This is the key idea of a limit.<br />
PAGE<br />
44. lim <br />
x→ 12<br />
2x 45. lim<br />
+ <br />
x→ 23<br />
2x 46. lim<br />
− <br />
x→ 23<br />
2x<br />
+<br />
85 55. (a) Investigate lim cos π by using a table and evaluating the<br />
Notice in Figure 11(b) that the value x→0 of g atxc does not affect the limit. Notice in<br />
<br />
Figure 11(c) that h is not defined function at c, but f (x) the= value cos π of<br />
47. lim |x|−x 48. lim |x|−x<br />
x at h gets closer to the number L for<br />
x sufficiently close to c. This suggests<br />
x→2 + x→2 −<br />
x =− 1 that<br />
3 3 2 , − 1 the limit<br />
4 , − 1 of<br />
8 , − 1 a function<br />
10 , − 1 as<br />
12 ,..., 1 x approaches<br />
12 , 1 10 , 1 8 , 1 4 , 1 c does<br />
not depend on the value, if it exists, of the function at c.<br />
2 .<br />
49. lim x−x 50. lim x−x<br />
x→2 + x→2 −<br />
51. Slope of a Tangent Line For f (x) = 3x 2 (b) Investigate lim cos π y<br />
by using a table and evaluating the<br />
:<br />
y<br />
y x→0 x<br />
(c, g(c))<br />
(a) Find the slope of the secant y 5 line f (x) containing the points (2, 12)<br />
y function 5 g(x) f (x) = cos π x at<br />
y 5 h(x)<br />
and (3, 27).<br />
y<br />
y<br />
(b) Find the slope of the secant line containing the points (2, 12)<br />
x =−1, − 1 3 , − 1 y<br />
5 , − 1 7 , − 1 9 ,..., 1 9 , 1 7 , 1 5 , 1 (c, L)<br />
3 , 1.<br />
L<br />
L<br />
L<br />
and (x, f (x)), x = 2.<br />
y<br />
y<br />
(c) Compare the results y from (a) and (b). What do you conclude<br />
(c) Create a table to investigate the slope of the tangent line to the<br />
about the limit? Why do you think this happens? What is<br />
graph of f at 2 using the result from (b).<br />
your view about using a table to draw a conclusion about<br />
(d) On the same set of axes, graph f , the tangent line to the graph<br />
limits?<br />
of f at the point (2, 12), and the secant line from (a).<br />
c<br />
x<br />
(d) Use technology to graph f . Begin with the x-window<br />
52. Slope of a Tangent Line For f (x) = x 3 x c x<br />
x<br />
x c x<br />
x<br />
:<br />
[−2π, 2π] and the y-window [−1, 1]. If you were finding<br />
(a) f (c) 5 L (b) g(c) Þ L (c) h(c) is not defined<br />
(a) Find the slope of the secant line containing the points (2, 8)<br />
lim f (x) using a graph, what would you conclude? Zoom in<br />
x→0<br />
DF Figure and 11 (3, 27).<br />
on the graph. Describe what you see. (Hint: Be sure your<br />
(b) Find the slope of the secant line containing the points (2, 8)<br />
calculator is set to the radian mode.)<br />
and (x, f (x)), x = 2.<br />
56. (a) Investigate lim cos π by using a table and evaluating the<br />
(c) Create a table to investigate the slope of the tangent line to the<br />
x→0 x2 graph of f at 2 using the result from (b). EXAMPLE 5 Investigatingfunction a Limit f (x) Using = cos a Graph<br />
NEED TO REVIEW? Piecewise-defined<br />
{ π at x =−0.1, −0.01, −0.001,<br />
(d) On the same set of axes, graph f , the tangent line to the graph<br />
x2 functions are discussed in Section P.1,<br />
3x + 1 if x = 2<br />
p. 8. of f at the point (2, 8), and the secant line<br />
Use<br />
from<br />
a graph<br />
(a).<br />
to investigate lim f<br />
x→2 −0.0001, (x) if f (x) 0.0001, = 0.001, 10 0.01, 0.1. if x = 2 .<br />
This table suggests that the value of<br />
sin(5 x)<br />
fx ( ) = gets closer and closer to<br />
x<br />
5 as x gets very close to 0. That is, there is<br />
sin(5 x)<br />
evidence indicating that limx→0<br />
= 5.<br />
x<br />
Teaching Tip<br />
Students should have become familiar with<br />
graphing constant, linear, and basic quadratic,<br />
cubic, radical, rational, and trigonometric functions<br />
before taking this course. If they are not, consider<br />
how to remediate these concepts, because basic<br />
graphing skills are necessary to complete the<br />
exam on time. There are many videos on the Khan<br />
Academy Web site and others that can guide a<br />
student to a stronger understanding of the basic<br />
functions. Consult the Additional Chapter 1<br />
Resources document for suggested URLs.<br />
TRM Section 1.1: Worksheet 2<br />
This worksheet includes 19 problems that have<br />
students determine limits using graphs.<br />
TRM Section 1.1: Worksheet 3<br />
This worksheet contains 3 problems in which the<br />
student is given a piecewise function. They are<br />
asked to find the one- and two-sided limits for<br />
a particular x-value and then verify the limits by<br />
graphing the function.<br />
82<br />
Chapter 1 • Limits and Continuity<br />
TE_<strong>Sullivan</strong>_Chapter01_PART 0.indd 11<br />
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<strong>Sullivan</strong> AP˙<strong>Sullivan</strong>˙Chapter01 October 8, 2016 17:4<br />
Section 1.1 • Limits of Functions Using Section Numerical 1.1 • Assess and Graphical Your Understanding Techniques 89 83<br />
(b) Investigate lim cos π by using a table and Solution evaluating Thethe<br />
function f is a(c) piecewise-defined Graph the function function. C. Its graph is shown in Figure 12.<br />
x→0 x2 y<br />
function f (x) = cos π Observe that as x approaches<br />
x 2 at<br />
(d)<br />
2<br />
Use<br />
from<br />
the graph<br />
the left,<br />
to investigate<br />
the value<br />
lim<br />
of f<br />
C(w)<br />
is close<br />
and<br />
to<br />
lim<br />
7,<br />
C(w).<br />
and as<br />
Do<br />
x<br />
w→1− w→1 +<br />
10<br />
approaches 2 from the right, the value of f is close to 7. In fact, we can make the value<br />
(2, 10)<br />
x =− 2 3 , − 2 5 , − 2 7 , − 2 9 ,..., 2 9 , 2 7 , 2 5 , 2 of f as close as we please to 7 these by choosing suggest that x sufficiently lim C(w) exists?<br />
3 .<br />
w→1<br />
close to 2 but not equal to 2.<br />
This suggests lim f (x) = 7. ■<br />
(e) Use the graph to investigate lim C(w) and lim x→2<br />
w→12− w→12<br />
(c) Compare the results from (a) and (b). What do you conclude<br />
7<br />
If we use a table to investigate lim f (x), the result is the same. See Table 3.<br />
Do these<br />
about the limit? Why do you think this happens? What is your<br />
x→2 suggest that lim C(w) exists?<br />
w→12<br />
view about using a table to draw a conclusion about limits? (f) Use the graph to investigate lim<br />
TABLE 3<br />
C(w).<br />
w→0 +<br />
(d) Use technology to graph f . Begin with the x-window<br />
(g) Use the graph to investigate lim w→13<br />
[−2π, 2π] and the y-window [−1, 1]. If you were x approaches finding 2 from the left<br />
x approaches − 2 from the right<br />
−−−−−−−−−−−−−−−−−−−−−→<br />
←−−−−−−−−−−−−−−−−−−−−−−<br />
lim f (x) using a graph, what would you conclude? Zoom in 61. Correlating Student Success to Study Time Professor Smith<br />
x→0 x 1.99 1.999 1.9999 1.99999 → 2 ← 2.00001 2.0001 2.001 2.01<br />
on the graph. Describe what you see. (Hint: Be sure your<br />
claims that a student’s final exam score is a function of the time t<br />
f (x) 6.97 6.997 6.9997 6.99997<br />
calculator is set to the radian mode.)<br />
(in hours) that f (x) the approaches student studies. 7 7.00003 He claims 7.0003 that the7.003 closer 7.03 to<br />
PAGE<br />
x − 8<br />
seven hours one studies, the closer to 100% the student scores<br />
85 57. (a) Use a table 1 to investigate 2 3 limx<br />
.<br />
x→2 2<br />
on the final. He claims that studying significantly less than seven<br />
(b) How close must { x be to 2, so that f (x) is within<br />
Figure<br />
0.1 of<br />
12<br />
the<br />
shows that f hours (2) = may 10cause but that one to this bevalue underprepared has no impact for the test, on the while limit as<br />
3x + 1 if x = 2<br />
Figure 12 limit? f (x) =<br />
x approaches 2. In fact, f (2) studying can equal significantly any number, more than and seven it would hours have mayno cause effect on the<br />
10 if x = 2<br />
(c) How close must x be to 2, so that f (x) islimit within as0.01 x approaches of the 2. “burnout.”<br />
limit?<br />
58. (a) Use a table to investigate lim(5 − 2x).<br />
x→2<br />
We make the following(a) observations: Write Professor Smith’s claim symbolically as a limit.<br />
• The limit L of a function (b) y Write = f (x) Professor as x approaches Smith’s claima number using thec ε-δ does definition not depend<br />
(b) How close must x be to 2, so that f (x) is within on the 0.1 of value the of f at c. of limit.<br />
limit?<br />
• The limit L of a function Source: y = Submitted f (x) as xbyapproaches the students aofnumber Millikin University. c is unique; that is,<br />
(c) How close must x be to 2, so that f (x) is within a function 0.01 of the cannot have more than one limit. (A proof of this property is given in<br />
62. The definition of the slope of the tangent line to the graph of<br />
limit?<br />
Appendix B.)<br />
f (x) − f (c)<br />
59. First-Class Mail As of April<br />
• If there is no single number y = f that (x) at the the value point of (c, f f approaches (c)) is m tan = aslim<br />
.<br />
x→c<br />
x gets<br />
2016, the U.S. Postal Service<br />
xclose − c to c, we<br />
say that f has no limit as x approaches c, or more simply, that the limit of f does<br />
charged $0.47 postage for<br />
Another way to express this slope is to define a new variable<br />
not exist at c.<br />
first-class letters weighing up to<br />
h = x − c. Rewrite the slope of the tangent line m tan using h and c.<br />
and including 1 ounce, plus a flat<br />
63. If f (2) = 6, can you conclude anything about lim f (x)? Explain<br />
fee of $0.21 for each additional<br />
x→2<br />
your reasoning. NOW WORK AP® Practice Problem 4.<br />
or partial ounce up to and<br />
64. If lim f (x) = 6, can you conclude anything about f (2)? Explain<br />
including 3.5 ounces. First-class<br />
x→2<br />
letter rates do not apply to letters Examples 6 and 7 that follow your illustrate reasoning. situations in which a limit does not exist.<br />
weighing more than 3.5 ounces.<br />
65. The graph of f (x) = x − 3<br />
CALC<br />
is a straight line with a point punched<br />
Source: U.S. Postal Service Notice 123 EXAMPLE 6 Investigating 3 − x<br />
out. a Limit Using a Graph<br />
CLIP<br />
{<br />
(a) Find a function C that models the first-class postage charged, (a) What straight line and x<br />
what if<br />
point? x < 0<br />
Use a graph to investigate lim f (x) if f (x) =<br />
in dollars, for a letter weighing w ounces. Assume w>0.<br />
x→0 1 if x > 0 .<br />
(b) Use the graph of f to investigate the one-sided limits of f as<br />
(b) What is the y domain of C?<br />
Solution Figure 13 shows the graph x approaches of f . We 3. first investigate the one-sided limits. The<br />
(c) Graph the2<br />
function C.<br />
graph suggests that, as x approaches (c) Does the 0 from graphthe suggest left, that lim f (x) exists? If so, what is it?<br />
(d) Use the graph to investigate lim C(w) and lim C(w). Do<br />
x→3<br />
1<br />
w→2− w→2 + lim<br />
66. (a) Use a table to investigate x→0 lim(1 + x)<br />
these suggest that lim C(w) exists?<br />
.<br />
x→0<br />
w→2<br />
(e) Use the graph to investigate lim and as x approaches 0 from(b) theUse right, graphing technology to graph g(x) = (1 + x) 1/x .<br />
2 1 1 2 3 x<br />
w→0 + (c) What lim do (a) and (b) suggest about lim(1 + x) 1/x ?<br />
1<br />
(f) Use the graph to investigate lim x→0 + x→0<br />
w→3.5 − CAS (d) Find lim(1 + x) 1/x .<br />
2<br />
Since there is no single number that x→0<br />
the values of f approach when x is close to 0, we<br />
60. First-Class Mail As of April 2016, the U.S. conclude Postal Service that lim f (x) does not exist. ■<br />
charged $0.94 postage { for first-class large envelope weighing x→0<br />
x if x < 0<br />
up to<br />
Figure and13 including f (x) = 1 ounce, plus a flat fee of $0.21 for each Tableadditional<br />
4 uses a numerical approach to support the conclusion that lim f (x) does not<br />
1 if x > 0<br />
or partial ounce up to and including 13 ounces. First-class rates do Challenge Problems<br />
x→0<br />
exist.<br />
not apply to large envelopes weighing more than 13 ounces. For Problems 67–70, investigate each of the following limits.<br />
Source: U.S. Postal Service Notice 123<br />
{<br />
TABLE 4<br />
1 if x is an integer<br />
f (x) =<br />
(a) Find a function C that models the first-class postage x approaches charged, 0 from the left<br />
0 if xis approaches not an integer 0 from the right<br />
in dollars, for a large envelope weighing w ounces. −−−−−−−−−−−−−−−−−−−−→<br />
Assume<br />
←−−−−−−−−−−−−−−−−−−−−−<br />
w>0.<br />
x −0.01 −0.00167. −0.0001 lim f (x) x→2<br />
→68. 0 lim f ←(x) x→1/2<br />
69. 0.0001 lim f (x) 0.001 70.<br />
x→3<br />
lim 0.01<br />
(b) What is the domain of C?<br />
f (x) −0.01 −0.001 −0.0001 no single number 1 1 1<br />
Teaching Tip<br />
In Section 1.3, students will learn the formal<br />
definition of continuity. Problems about continuity<br />
on the exams often involve piecewise functions.<br />
When investigating the limits of piecewise functions<br />
graphically, consider checking the limits using<br />
direct substitution into the appropriate portions<br />
of the piecewise function. Doing so will lay the<br />
groundwork for Section 1.3. Here you want to focus<br />
heavily on getting students to write the one-sided<br />
limits correctly. A graphic organizer can be useful<br />
here, and a teacher modeling this correctly when<br />
students first see it will save many mistakes later.<br />
This is the template.<br />
lim ( what function ?) = ( what yvalue ?)<br />
x → 1<br />
+<br />
( x approaches what value ?)<br />
The student can then set up the organizer with<br />
blanks for the parenthetical expressions.<br />
Kathryn Sidenstricker /Dreamstime.com<br />
Examples 5 and 6 lead to the following result.<br />
AP® CaLC skill builder<br />
for example 6<br />
Investigate a Limit Using a Graph<br />
Use a graph to investigate lim fx ( ) if<br />
x→2<br />
⎧⎪<br />
2<br />
fx ( ) =<br />
− x + 4 x ≠ 2<br />
⎨<br />
⎩⎪ 2 x = 2<br />
Solution<br />
The function f is a piecewise-defined function.<br />
The graph is shown. y<br />
24<br />
22<br />
4<br />
2<br />
22<br />
24<br />
2<br />
4<br />
x<br />
The graph suggests that as x approaches 2<br />
from the left,<br />
lim fx ( ) = 0.<br />
x→2<br />
−<br />
and as x approaches 2 from the right,<br />
lim fx ( ) = 0.<br />
x→ 2<br />
+<br />
Because the right- and left-hand limits<br />
approach the same number, the graph<br />
suggests that<br />
lim fx ( ) = 0.<br />
x→2<br />
Note: If the left- and right-hand limits<br />
agree, then the limit exists. It does not<br />
matter that the value f (2) does not equal 0.<br />
AP® CaLC skill builder<br />
for example 6<br />
Analyzing a Limit Using a Graph<br />
Suppose the function f has the graph<br />
shown.<br />
f (x)<br />
a b c d e<br />
Investigate whether or not each of the<br />
following limits exists.<br />
(a) lim fx ( )<br />
(e) lim fx ( )<br />
x→a<br />
− x→ b<br />
+<br />
(b) lim fx ( ) (f) lim fx ( )<br />
x→ a<br />
+ x→b<br />
(c) lim fx ( ) (g) lim fx ( )<br />
x→a<br />
x→d<br />
(d) lim fx ( ) (h) lim fx ( )<br />
x→b<br />
− x→e<br />
Solution<br />
(a) The graph suggests lim fx ( ) exists.<br />
x→a<br />
−<br />
(b) The graph suggests lim fx ( ) exists.<br />
x→ a<br />
+<br />
(c) The graph suggests lim fx ( ) exists.<br />
x→a<br />
(d) The graph suggests lim fx ( ) exists.<br />
x→b<br />
−<br />
(e) The graph suggests lim fx ( ) exists.<br />
x→ b<br />
+<br />
(f) The graph suggests lim fx ( ) does not<br />
x→b<br />
exist because the left-hand and righthand<br />
limits are not equal.<br />
(g) The graph suggests lim fx ( ) exists.<br />
(h) The graph suggests<br />
x→d<br />
x→e<br />
x<br />
lim fx ( ) exists.<br />
x→0 f (x) 83<br />
Section 1.1 • Limits of Functions Using Numerical and Graphical Techniques<br />
TE_<strong>Sullivan</strong>_Chapter01_PART 0.indd 12<br />
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<strong>Sullivan</strong> AP˙<strong>Sullivan</strong>˙Chapter01 October 8, 2016 17:4<br />
<strong>Sullivan</strong> <strong>Sullivan</strong>AP<br />
88 84 Chapter 1 • Limits and Continuity<br />
Alternate Example<br />
Investigating a Limit<br />
x<br />
Investigate lim | − 1| .<br />
x→1<br />
x −1<br />
Solution<br />
We can investigate a limit such as this one<br />
in three ways: graphically, algebraically, and<br />
numerically.<br />
Graphically<br />
The graph suggests that lim fx ( ) does not<br />
x→1<br />
exist because lim fx ( ) ≠ lim fx ( ).<br />
24<br />
Algebraically<br />
− +<br />
x→1 x→1<br />
22<br />
y<br />
4<br />
2<br />
22<br />
24<br />
| x −1|<br />
fx ( ) = is equivalent to the<br />
x −1<br />
piecewise-defined function.<br />
⎧ −( x −1)<br />
=−1<br />
x < 1<br />
⎪<br />
fx ( ) =<br />
x −1<br />
⎨<br />
⎪ ( x −1)<br />
− = 1 x > 1<br />
⎩<br />
⎪ x 1<br />
lim fx ( ) does not exist because<br />
x→1<br />
lim fx ( ) ≠ lim fx ( ).<br />
− +<br />
x→1 x→1<br />
Using a Table<br />
x f(x)<br />
0.9 −1<br />
0.99 −1<br />
0.999 −1<br />
1.001 1<br />
1.01 1<br />
1.1 1<br />
The table suggests lim fx ( ) does not exist<br />
x→1<br />
because lim fx ( ) ≠ lim fx ( ).<br />
− +<br />
x→1 x→1<br />
2<br />
4<br />
x<br />
2x 2 if x < 1<br />
33. f (x) =<br />
3x 2 at c = 1 THEOREM 53. Slope of a Tangent Line For f (x) = 1<br />
− 1 if x > 1<br />
2 x2 − 1:<br />
The limit L of a function y = f (x) as x approaches a number c exists if and only if<br />
x 3 (a) Find the slope m<br />
if x < −1<br />
sec of the secant line containing the<br />
both one-sided limits exist at c and both one-sided limits are equal. That is,<br />
34. f (x) =<br />
x 2 at c =−1<br />
points P = (2, f (2)) and Q = (2 + h, f (2 + h)).<br />
− 1 if x > −1<br />
lim f (x) = L(b) ifUse andthe only result if from lim(a) f to (x) complete = limthe following x 2 x→c table:<br />
x→c<br />
if x ≤ 0<br />
x→c +<br />
35. f (x) =<br />
at c = 0<br />
2x + 1 if x > 0<br />
NOW<br />
⎧<br />
h WORK −0.5Problems −0.1 25, −0.001 31, and AP® 0.001 Practice 0.1 Problem 0.5 5.<br />
⎨ x 2 if x < 1<br />
m sec<br />
36. f (x) = 2 if x = 1 at c = 1 A one-sided limit is used to describe the behavior of functions such as<br />
⎩<br />
−3x + 2 if x > 1<br />
f (x) = √ √ x − 1 near x = (c) 1. Since Investigate the √domain the limit of of ftheis slope {x|xof ≥the 1}, secant the left-hand line foundlimit,<br />
in (a)<br />
lim x − 1 makes no sense. But as h lim → 0. x − 1 = 0 suggests how f behaves near and<br />
x→1 − x→1 +<br />
Applications y and Extensions<br />
to the right of 1. See Figure(d) 14 and What Table is the5. slope They of suggest the tangent limlinef to(x) the= graph 0. of f at the<br />
point P = (2, f (2))? x→1 +<br />
In Problems 2 37–40, sketch a graph of a function with the given<br />
(e) On the same set of axes, graph f and the tangent line to f at<br />
properties. Answers will vary.<br />
TABLE 5<br />
1<br />
P = (2, f (2)).<br />
37. lim f (x) = 3; lim f (x) = 3; lim f (x) = 1;<br />
x approaches 1 from the right<br />
x→2 x→3− x→3 +<br />
−−−−−−−−−−−−−−−−−−−−−→<br />
f (2) = 3; f (3) = 1<br />
54. Slope of a Tangent Line For f (x) = x 2 − 1:<br />
1 2 3 4 5 6 x<br />
x 1.009 1.0009 1.00009 1.000009 1.0000009 → 1<br />
38. lim f (x) = 0; lim f (x) =−2; lim<br />
=−2;<br />
(a) Find the slope m sec of the secant line containing the<br />
Figure x→−1 14 f (x) = √ f (x) = √ x − 1 0.0949 0.03 0.00949 0.003 0.000949 f (x) approaches 0<br />
xx→2 − 1− x→2 +<br />
points P = (−1, f (−1)) and Q = (−1 + h, f (−1 + h)).<br />
f (−1) is not defined; f (2) =−2<br />
39. lim f (x) = 4;<br />
x→1<br />
lim f (x) =−1;<br />
x→0− lim f (x) = 0;<br />
x→0 +<br />
(b) Use the result from (a) NOW to complete WORK the AP® following Practicetable:<br />
Problem 1.<br />
f (0) =−1; f (1) = 2<br />
Using numeric tables and/orhgraphs −0.1gives −0.01us−0.001 an idea−0.0001 of what a0.0001 limit might 0.001 0.01 be. That 0.1<br />
40. lim f (x) = 2;<br />
x→2<br />
lim f (x) = 0;<br />
x→−1 lim is, these methods suggest a limit,<br />
f (x) = 1;<br />
m sec<br />
but there are dangers in using these methods, as the<br />
x→1 following example illustrates.<br />
f (−1) = 1; f (2) = 3<br />
(c) Investigate the limit of the slope of the secant line found<br />
EXAMPLE 7 Investigatingin a(a) Limit as h → 0.<br />
In Problems 41–50, use either a graph or a table to investigate<br />
(d) What is the slope of the tangent line to the graph of f at the<br />
each limit.<br />
Investigate lim sin π<br />
x→0 x . 2 point P = (−1, f (−1))?<br />
|x − 5|<br />
|x − 5|<br />
41. lim<br />
42. lim<br />
43. lim<br />
x→5 + x − 5<br />
x→5 − x − 5<br />
<br />
x→ 12<br />
2x<br />
(e) On the same set of axes, graph f and the tangent line to f<br />
Solution The −<br />
domain of the function at P = (−1, f (x) f = (−1)). sin π is {x|x = 0}.<br />
x<br />
2<br />
PAGE<br />
44. lim <br />
x→ 12<br />
2x 45. lim<br />
+ <br />
x→ 23<br />
2x 46. Suppose lim<br />
− <br />
x→ 23<br />
2x we let<br />
+<br />
85 x approach 55. (a) Investigate zero in thelimfollowing cos π byway:<br />
using a table and evaluating the<br />
x→0 x<br />
function f (x) = cos π TABLE 47. lim6<br />
|x|−x 48. lim |x|−x<br />
x at<br />
x→2 + x→2 −<br />
x approaches 0 from the left<br />
x =− 1<br />
3 3 2 , − 1 4 , − 1 8 , − 1 x<br />
10 , approaches − 1 12 ,..., 1<br />
0 from<br />
12 , 1 10 , the 1 8 , 1 right<br />
4 , 1 2 .<br />
49. lim x−x 50. lim<br />
−−−−−−−−−−−−−−−−−−−−→<br />
x−x<br />
←−−−−−−−−−−−−−−−−−−−−−<br />
x→2 + x→2<br />
x − 1<br />
− 1 − 1 − 1<br />
51. Slope of a Tangent Line For f (x) = 3x 2 (b) Investigate lim cos π 1 1 1 1<br />
→ 0 ←<br />
10 100<br />
by using a table and evaluating the<br />
: 1000 10,000<br />
x→0 x10,000<br />
1000 100 10<br />
f (x)(a) = Find sin π the slope of0the secant line 0 containing0the points (2, 12) 0 f (x) approaches function f (x) 0 = cos π x at 0 0 0 0<br />
x 2<br />
and (3, 27).<br />
(b) Find the slope of the secant line containing the points (2, 12)<br />
x =−1, − 1 3 , − 1 5 , − 1 7 , − 1 9 ,..., 1 9 , 1 7 , 1 5 , 1 Table 6 suggests that lim sin π 3 , 1.<br />
and (x, f (x)), x = 2.<br />
x→0 x = 0. 2<br />
(c) Compare the results from (a) and (b). What do you conclude<br />
(c) Create a table to investigate the slope of the tangent<br />
Now suppose<br />
line to the<br />
we let x approach zero as follows:<br />
about the limit? Why do you think this happens? What is<br />
graph of f at 2 using the result from (b).<br />
TABLE your view about using a table to draw a conclusion about<br />
(d) 7On the same set of axes, graph f , the tangent line to the graph<br />
limits?<br />
of f at the point (2,<br />
x<br />
12),<br />
approaches<br />
and the secant<br />
0 from<br />
line<br />
the<br />
from<br />
left<br />
(a).<br />
x approaches 0 from the right<br />
−−−−−−−−−−−−−−−−−−−−→<br />
(d) Use technology to graph f . Begin with the x-window<br />
52. Slope of a Tangent Line For f (x) = x 3 ←−−−−−−−−−−−−−−−−−−−−−<br />
:<br />
[−2π, 2π] and the y-window [−1, 1]. If you were finding<br />
x − 2 − 2 − 2 − 2 − 2 2 2 2 2 2<br />
→ 0<br />
(a) Find the slope of the secant line containing the points (2, 8)<br />
lim<br />
←<br />
3 5 7 9 11<br />
f (x) using a11<br />
graph, what 9 would you 7 conclude? 5 Zoom3in<br />
x→0<br />
f (x) = sin and (3, 27). 0.707 0.707 0.707 0.707 0.707 f (x) approaches on the 0.707 graph. Describe 0.707 what 0.707 you see. 0.707 (Hint: Be 0.707 sure your 0.707<br />
(b) Findx the slope of the secant line containing the points (2, 8)<br />
calculator is set to the radian mode.)<br />
and (x, f (x)), x = 2.<br />
56. (a) Investigate lim cos π by using a table and evaluating the<br />
(c) Create a table to investigate the slope of the tangent line to the<br />
x→0 x2 Table 7 suggests that lim sin π √<br />
2<br />
graph of f at 2 using the result from (b).<br />
function f (x) = cos π x→0 x = 2 2 ≈ 0.707. at x =−0.1, −0.01, −0.001,<br />
(d) On the same set of axes, graph f , the tangent line to the graph<br />
x2 In fact, by carefully selecting x, we can make f appear to approach any number in<br />
of f at the point (2, 8), and the secant line from (a).<br />
−0.0001, 0.0001, 0.001, 0.01, 0.1.<br />
the interval [−1, 1].<br />
Teaching Tip<br />
Have students create their own piecewise<br />
functions on graph paper. Ask them to write 3 or<br />
4 limit questions based upon their (well-labeled)<br />
graph. Point out to students that being able to<br />
write out and explain piecewise functions so<br />
that another student can follow means that they<br />
have to be very clear. Reinforce good habits and<br />
good explanations. Then they can trade with one<br />
another, find the limits, and discuss the solutions<br />
with each other. Khan Academy offers explanatory<br />
videos, as well as a practice video, on piecewise<br />
functions.<br />
WEB SITE<br />
Graphfree: The Graphfree site is an easy tool<br />
for making piecewise graphs to create additional<br />
examples. Make sure you change the plot type to<br />
Piecewise. A link to this resource is available on<br />
the Additional Chapter 1 Resources document,<br />
available for download.<br />
R<br />
N<br />
D<br />
84<br />
Chapter 1 • Limits and Continuity<br />
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17:4<br />
Section 1.1 • Limits of Functions Using Numerical Section 1.1 and• Assess Graphical Your Techniques Understanding 85 89<br />
(b) Investigate lim cos π by using a table and evaluating x→0 x2 Now look at the graphs of (c) f (x) Graph = sin the function π shown C. in Figure 15. In Figure 15(a),<br />
function f (x) = cos π x<br />
2<br />
x 2 at<br />
(d) Use the graph to investigate lim C(w) and lim C(w). Do<br />
the choice of lim sin π = 0 seems reasonable. But in Figure w→115(b), − it appears w→1 + that<br />
x→0 x 2<br />
x =− 2 3 , − 2 5 , − 2 7 , − 2 9 ,..., 2 9 , 2 7 , 2 5 , 2 these suggest that lim C(w) exists?<br />
3 .<br />
w→1<br />
lim sin π<br />
x→0 x =−1 . Figure 15(c) (e) illustrates Use the graph that the to investigate graph of flimoscillates C(w) and rapidly lim as C(w). x<br />
2 2<br />
w→12− w→12<br />
(c) Compare the results from (a) and (b). approaches What do you 0. This conclude suggests that the + Do value these ofsuggest f doesthat not approach lim C(w) aexists?<br />
single number, and<br />
about the limit? Why do you think this happens? What is your<br />
w→12<br />
view about using a table to draw a conclusion that lim sin π about limits? does not exist. (f) Use the graph to investigate lim C(w).<br />
x→0 x<br />
2<br />
w→0 +<br />
(d) Use technology to graph f . Begin with the x-window<br />
(g) Use the graph to investigate lim C(w).<br />
w→13<br />
[−2π, 2π] and the y-window [−1, 1]. If you were finding<br />
−<br />
lim f (x) using a graph, what would you conclude? Zoom in<br />
x→0<br />
on the graph. Describe what you see. (Hint: Be sure your<br />
calculator is set to the radian mode.)<br />
PAGE<br />
x − 8<br />
85 57. (a) Use a table to investigate lim .<br />
x→2 2<br />
(b) How close must x be to 2, so that f (x) is within 0.1 of the<br />
limit?<br />
(c) How close must x be to 2, so that f (x) is within 0.01 of the<br />
limit?<br />
58. (a) Use a table to investigate lim<br />
Figure<br />
(5 − 2x).<br />
15<br />
x→2<br />
(b) How close must x be to 2, so that f (x) is within 0.1 of the<br />
limit?<br />
(c) How close must x be to 2, so that f (x) is within 0.01 of the<br />
limit?<br />
59. First-Class Mail As of April<br />
2016, the U.S. Postal Service<br />
charged $0.47 postage for<br />
EXAMPLE 8<br />
first-class letters weighing up to<br />
and including 1 ounce, plus a flat<br />
fee of $0.21 for each additional<br />
or partial ounce up to and<br />
including 3.5 ounces. First-class<br />
letter rates do not apply to letters<br />
RECALL<br />
weighing<br />
On the number<br />
more than<br />
line,<br />
3.5<br />
the<br />
ounces.<br />
distance between two points with<br />
coordinates Source: a andU.S. b is Postal |a − b|. Service Notice 123<br />
(a) Find a function C that models the first-class postage charged,<br />
in dollars, for a letter weighing w ounces. Assume w>0.<br />
NEED TO REVIEW? Inequalities<br />
(b) What is the domain of C?<br />
involving absolute values are discussed<br />
in Appendix<br />
(c)<br />
A.1,<br />
Graph<br />
p. A-7.<br />
the function C.<br />
(d) Use the graph to investigate lim C(w) and lim C(w). Do<br />
w→2− w→2 +<br />
these suggest that lim C(w) exists?<br />
w→2<br />
(e) Use the graph to investigate lim C(w).<br />
w→0 +<br />
y<br />
(f) Use the graph to investigate lim C(w).<br />
w→3.5 −<br />
10<br />
60. 9 First-Class Mail As of April 2016, the U.S. Postal Service<br />
8 charged $0.94 postage for first-class large envelope weighing up to<br />
and including 1 ounce, plus a flat fee of $0.21 for each additional<br />
6 or partial ounce up to and including 13 ounces. First-class rates do<br />
not apply to large envelopes weighing more than 13 ounces.<br />
4<br />
Source: U.S. Postal Service Notice 123<br />
Source: Submitted by the students of Millikin University.<br />
So, how do we find a limit 62. with The definition certainty? ofThe the slope answer of the liestangent in giving line atovery the graph precise of<br />
definition of limit. The next example helps explain the definition. f (x) − f (c)<br />
y = f (x) at the point (c, f (c)) is m tan = lim<br />
.<br />
x→c x − c<br />
Analyzing a Limit Another way to express this slope is to define a new variable<br />
In Example 2, we claimed that lim<br />
h =<br />
(2x<br />
x −<br />
+<br />
c.<br />
5)<br />
Rewrite<br />
= 9.<br />
the slope of the tangent line m tan using h and c.<br />
63. x→2 If f (2) = 6, can you conclude anything about lim f (x)? Explain<br />
x→2<br />
(a) How close must x be to 2, so your that reasoning. f (x) = 2x + 5 is within 0.1 of 9?<br />
(b) How close must x be to64. 2, so If that lim<br />
6, can you conclude anything about f (2)? Explain<br />
x→2<br />
f (x) = 2x + 5 is within 0.05 of 9?<br />
your reasoning.<br />
Solution (a) The function f (x) = 2x + 5 is within<br />
65. The graph of f (x) = x − 0.1 3 of9, if the distance between<br />
f (x) and 9 is less than 0.1 unit. That is, if | f (x) − 9| is a straight line with a point punched<br />
3 −≤0.1.<br />
x<br />
out.<br />
|(2x + 5) − 9| ≤0.1<br />
(a) What straight line and what point?<br />
|2x − 4| ≤0.1<br />
(b) Use the graph of f to investigate the one-sided limits of f as<br />
|2(xx− approaches 2)| ≤0.13.<br />
(c) Does the graph suggest that lim f (x) exists? If so, what is it?<br />
|x − 2| ≤ 0.1<br />
x→3<br />
2 = 0.05<br />
66. (a) −0.05 Use a table ≤ xto− investigate 2 ≤ 0.05 lim(1 + x) 1/x .<br />
x→0<br />
(b) Use 1.95 graphing ≤ x ≤technology 2.05 to graph g(x) = (1 + x) 1/x .<br />
(c) What do (a) and (b) suggest about lim(1 + x) 1/x ?<br />
x→0<br />
So, if 1.95 ≤ x ≤ 2.05, then f (x) will be within 0.1of9.<br />
CAS (d) Find lim(1 + x) 1/x .<br />
(b) The function f (x) = 2x + 5 is within x→0 0.05 of 9 if | f (x) − 9| ≤0.05. That is,<br />
|(2x + 5) − 9| ≤0.05<br />
Challenge |2x −Problems<br />
4| ≤0.05<br />
For Problems 67–70, investigate each of the following limits.<br />
|x − 2| ≤ 0.05 = { 0.025<br />
2 1 if x is an integer<br />
2<br />
f (x) =<br />
(a) Find a function C that models the first-class So, postage if 1.975charged,<br />
≤ x ≤ 2.025, then f (x) will be within0 0.05ifof x is 9. not ■ an integer<br />
in dollars, for a large envelope weighing w ounces. Assume<br />
w>0. 1 2 x<br />
67. lim f (x) 68. lim f (x) 69. lim f (x) 70. lim f (x)<br />
x→2 x→1/2 x→3 x→0<br />
Notice that the closer we require f to be to the limit 9, the narrower the interval for<br />
(b) What is the domain of C?<br />
DF Figure 16 f (x) = 2x + 5<br />
x becomes. See Figure 16.<br />
Kathryn Sidenstricker /Dreamstime.com<br />
61. Correlating Student Success to Study Time Professor Smith<br />
claims that a student’s final exam score is a function of the time t<br />
(in hours) that the student studies. He claims that the closer to<br />
seven hours one studies, the closer to 100% the student scores<br />
on the final. He claims that studying significantly less than seven<br />
hours may cause one to be underprepared for the test, while<br />
studying significantly more than seven hours may cause<br />
“burnout.”<br />
(a) 24p ≤ x ≤ 4p (b) 2p ≤ x ≤ p (c) 21 ≤ x ≤ 1<br />
(a) Write Professor Smith’s claim symbolically as a limit.<br />
■<br />
(b) Write Professor Smith’s claim using the ε-δ definition<br />
of limit.<br />
NOW WORK Problem 55.<br />
NOW WORK Problem 57.<br />
Optional/Enrichment<br />
If you are pressed for time, you might<br />
skip Example 8, because problems like<br />
this usually do not appear on the exam.<br />
However, if you have time, this example<br />
may deepen students’ conceptual<br />
understanding of limits.<br />
Teaching Tip<br />
Sometimes students come to calculus<br />
class without having developed good<br />
homework habits. Emphasize the message<br />
that practice is necessary. Point out the<br />
benefits of daily practice. If you observe<br />
students who are not completing their<br />
homework, it is critical to speak with them<br />
as soon as you can.<br />
Section 1.1 • Limits of Functions Using Numerical and Graphical Techniques<br />
85<br />
TE_<strong>Sullivan</strong>_Chapter01_PART 0.indd 14<br />
11/01/17 9:52 am
<strong>Sullivan</strong> AP˙<strong>Sullivan</strong>˙Chapter01 October 8, 2016 17:4<br />
<strong>Sullivan</strong><br />
88 86 Chapter 1 • Limits and Continuity<br />
Must-Do Problems for<br />
Exam Readiness<br />
AB: 1–6, 17–28, 31, 35, 37, 38, 41, 42,<br />
AP ® Practice Problems<br />
BC: 11, 13, 17, 19, 31, 37, 39, 51, 59,<br />
63–65, AP ® Practice Problems 3–8<br />
TRM Full Solutions to Section<br />
1.1 Problems and AP® Practice<br />
Problems<br />
Answers to Section 1.1<br />
Problems<br />
1. (c)<br />
2. True.<br />
3. False.<br />
4. False.<br />
5. False.<br />
6. False.<br />
7. lim 2x<br />
= 2. For table values see TSM.<br />
x→1<br />
8. lim ( x + 3) = 5. For table values see<br />
x→2<br />
TSM.<br />
9. lim ( x<br />
2 + 2) = 2. For table values see<br />
x→0<br />
TSM.<br />
10. lim ( x<br />
2 − 2) =−1. For table values<br />
x→−1<br />
see TSM.<br />
2<br />
x − 9<br />
11. lim =−6. For table values<br />
x→−3<br />
x + 3<br />
see TSM.<br />
3<br />
x + 1<br />
12. lim<br />
+ = 3. For table values see<br />
x→−1<br />
x 1<br />
TSM.<br />
2x 2 if x < 1<br />
The discussion in Example<br />
33. f (x) =<br />
3x 2 at c = 1<br />
53. Slope8offorms a Tangent the basis Line ofFor thef definition (x) = 1 of<br />
− 1 if x > 1<br />
2 x2 − 1: a limit. We state<br />
the definition here, but postpone the details until Section 1.6. It is customary to use the<br />
x 3 Greek letters ε (epsilon) and(a) δ (delta) Find the in slope the definition, m<br />
if x < −1<br />
sec of thesosecant we call lineitcontaining the ε-δ definition the of a<br />
34. f (x) =<br />
x 2 at c =−1 limit.<br />
points P = (2, f (2)) and Q = (2 + h, f (2 + h)).<br />
− 1 if x > −1<br />
(b) Use the result from (a) to complete the following table:<br />
x 2 if x ≤ 0<br />
35. f (x) =<br />
at c = 0 DEFINITION ε-δ Definition of a Limit<br />
2x + 1 if x > 0<br />
⎧<br />
Let f be a function defined everywhere h −0.5 in an open −0.1interval −0.001 containing 0.001 c, except 0.1possibly<br />
0.5<br />
⎨ x 2 if x < 1<br />
at c. Then the limit of the function m sec f as x approaches c is the number L, written<br />
36. f (x) = 2 if x = 1 at c = 1<br />
⎩<br />
−3x + 2 if x > 1<br />
(c) Investigate lim f the (x) limit = L of the slope of the secant line found in (a)<br />
x→c<br />
as h → 0.<br />
Applications and Extensions<br />
if, given any number ε>0, (d) there What is is a number the slopeδ>0 of the tangent so that line to the graph of f at the<br />
point P = (2, f (2))?<br />
In Problems 37–40, sketch a graph of a function with the given whenever(e) 0 < On|x the −same c|
<strong>Sullivan</strong> AP˙<strong>Sullivan</strong>˙Chapter01 October 8, 2016 17:4<br />
Section 1.1 • Assess Your Understanding 89 87<br />
In Problems (b) Investigate 13–16, use limtechnology cos π by using to complete a tablethe andtable evaluating and the In Problems (c) Graph 21–28, the function use the graph C. to investigate lim f (x). If the limit<br />
x→0 x2 x→c<br />
investigate the limit.<br />
function f (x) = cos π does<br />
x 2 at<br />
(d)<br />
not exist,<br />
Use the<br />
explain<br />
graph<br />
why.<br />
to investigate lim C(w) and lim C(w). Do<br />
w→1− w→1 +<br />
PAGE<br />
2 − 2e x<br />
82 13. lim x→0 x<br />
x =− 2 3 , − 2 5 , − 2 7 , − 2 9 ,..., 2 9 , 2 7 , 2 5 , 2 3 .<br />
21. these suggest that lim C(w) w→1 22. exists?<br />
y (e) Use the graph to investigate<br />
y<br />
x approaches 0 x approaches 0<br />
lim C(w) and lim C(w).<br />
w→12− w→12<br />
(c) Compare the resultsfrom the (a) left and (b). What from do youthe conclude right<br />
+<br />
about the limit? Why<br />
−−−−−−−−−−→<br />
do you think this happens?<br />
←−−−−−−−−−−<br />
Do these suggest that lim C(w) exists?<br />
What is your 3<br />
w→12 y f (x)<br />
viewx about using −0.2 a table −0.1to −0.01 draw a conclusion → 0 ← about 0.01limits?<br />
0.1 0.2<br />
y f (x)<br />
2<br />
(f) Use the graph to investigate lim C(w).<br />
w→0 +<br />
(d) f (x) Use = 2 technology − 2<br />
2ex<br />
1<br />
to graph f . Begin with the x-window<br />
1 (g) Use the graph to investigate lim<br />
x<br />
C(w).<br />
w→13<br />
[−2π, 2π] and the y-window [−1, 1]. If you were finding<br />
− (c, 1)<br />
(c, 1)<br />
lim ln xf (x) using a graph, what would you conclude? Zoom in<br />
c<br />
x<br />
61. Correlating c Student Success x to Study Time Professor Smith<br />
14. lim x→0<br />
x→1 on x −the 1 graph. Describe what you see. (Hint: Be sure your<br />
claims that a student’s final exam score is a function of the time t<br />
calculator is set to the radian mode.)<br />
(in hours) that the student studies. He claims that the closer to<br />
x approaches 1 x approaches 1<br />
PAGE<br />
x − 8<br />
23. seven hours one studies, the closer 24. to 100% the student scores<br />
85 57. (a) Use a table to investigate from the lim left . from the right y<br />
x→2 2<br />
on the final. He claims that studying y<br />
−−−−−−−−−−→ ←−−−−−−−−−−<br />
significantly less than seven<br />
(b) How x close must 0.9 x0.99 be to0.999 2, so that → f 1(x) ←is within 1.001 0.11.01 of the1.1<br />
hours may cause one to be underprepared for the test, while<br />
limit?<br />
studying significantly more than seven hours may cause<br />
(c) f (x) How =<br />
ln close x<br />
3 (c, 3) 3<br />
must x be to 2, so that f (x) is within 0.01 of the 2 “burnout.”<br />
2<br />
x − 1<br />
limit?<br />
y f (x)<br />
y f (x)<br />
1 (a) Write Professor Smith’s claim1<br />
symbolically as a limit.<br />
58. (a) Use a table to investigate lim(5 − 2x).<br />
PAGE<br />
1 − cos x<br />
(c, 1)<br />
x→2 (b) Write Professor Smith’s claim using the ε-δ definition<br />
82 15. lim , where x is measured in radians<br />
(b) x→0<br />
How<br />
x<br />
close must x be to 2, so that f (x) is within 0.1 of the<br />
ofclimit.<br />
x<br />
c<br />
x<br />
limit? x approaches 0 x approaches 0<br />
Source: Submitted by the students of Millikin University.<br />
(c) How close must x from be to 2, thesoleft<br />
that f (x) is within from the 0.01right<br />
of the<br />
PAGE<br />
limit? −−−−−−−−−−→ ←−−−−−−−−−− 84 25. 62. The definition of the slope of the26.<br />
tangent line to the graph of<br />
59. First-Class<br />
x (in radians)<br />
Mail<br />
−0.2<br />
As of<br />
−0.1<br />
April<br />
−0.01 → 0 ← 0.01 0.1 0.2<br />
y f (x) − f (c)<br />
y = f (x) at the point (c, f (c)) is m tan = lim<br />
.<br />
x→c<br />
2016, the U.S. Postal Service<br />
x − c<br />
f (x) = 1 − cos x<br />
y<br />
2<br />
charged $0.47 x postage for<br />
Another way y f to(x)<br />
express this slope 3is to define a new variable<br />
1<br />
first-class letters weighing up to<br />
h = x − c. Rewrite the slope of the tangent (c, 2) line my tan using f (x)<br />
sin x<br />
h and c.<br />
2<br />
16. and lim including , 1where ounce, xplus is measured a flat in radians<br />
x→0 1 + tan x<br />
63. If f (2) c = 6, can you conclude x anything about lim f (x)? Explain<br />
fee of $0.21 for each additional<br />
1<br />
x→2<br />
your reasoning.<br />
or partial ounce up to and x approaches 0 x approaches 0 2<br />
64. If lim f (x) = 6, can you conclude anything about f (2)? Explain<br />
including 3.5 ounces. First-class<br />
c<br />
x<br />
from the left from the right<br />
x→2<br />
letter rates do not apply−−−−−−−−−−→<br />
to letters<br />
←−−−−−−−−−− your reasoning.<br />
weighing x (in radians) more than−0.2 3.5 ounces. −0.1 −0.01 → 0 ← 0.01 0.1 0.2<br />
65. The graph of f (x) = x − 3 is a straight line with a point punched<br />
Source: U.S. Postal Service Notice 123<br />
3 − x<br />
f (x) =<br />
sin x<br />
27. 28.<br />
y out.<br />
y<br />
1 + tan x<br />
(a) Find a function C that models the first-class postage charged, (a) What straight line and what point?<br />
in dollars, for a letter weighing w ounces. Assume w>0. 3<br />
3<br />
In Problems 17–20, use the graph to investigate<br />
(b) Use the graph of f to investigate the one-sided limits of f as<br />
(a) lim<br />
(b) What<br />
f (x),<br />
is<br />
(b)<br />
the<br />
lim<br />
domain<br />
f (x),<br />
of<br />
(c)<br />
C?<br />
lim<br />
x approaches 3.<br />
(c) Graph the function C.<br />
f (x).<br />
(c, 2)<br />
y f (x)<br />
2<br />
y f (x)<br />
2 (c, 2)<br />
x→2− x→2 + x→2<br />
1 (c) Does the graph suggest that lim1<br />
f (x) exists? If so, what is it?<br />
(d) Use the graph to investigate lim C(w) and lim C(w). Do<br />
x→3<br />
17. 18.<br />
w→2− w→2 +<br />
y<br />
y<br />
66. (a) Use c a table to investigate x lim(1 + x)<br />
these suggest that lim C(w) exists?<br />
1/x . c<br />
x<br />
x→0<br />
4<br />
w→2<br />
(e) Use the graph to investigate lim 4 C(w).<br />
(b) Use graphing technology to graph g(x) = (1 + x) 1/x .<br />
w→0 + (c) What do (a) and (b) suggest about lim(1 + x) 1/x ?<br />
(f) Use the graph to investigate lim C(w).<br />
x→0<br />
2 y f (x)<br />
w→3.5 2 − y f (x)<br />
CAS In Problems (d) Find29–36, lim(1 use + x) a 1/x graph . to investigate lim f (x) at the<br />
(2, 2)<br />
x→c x→0<br />
60. First-Class Mail As of April 2016, the U.S. Postal Service number c. <br />
charged $0.94 postage for first-class large envelope weighing up to<br />
2x + 5 if x ≤ 2<br />
2 4 x<br />
2 4 x 29. f (x) =<br />
at c = 2<br />
and including 1 ounce, plus a flat fee of $0.21 for each additional<br />
4x + 1 if x > 2<br />
or partial ounce up to and including 13 ounces. First-class rates do Challenge Problems 2x + 1 if x ≤ 0<br />
19. not apply to large envelopes weighing 20. more than 13 ounces. 30. For Problems f (x) = 67–70, investigate eachatof c = the0<br />
2x if x > 0 following limits.<br />
y<br />
Source: U.S. Postal Service Notice 123<br />
y<br />
⎧ { 1 if x is an integer<br />
6<br />
8<br />
⎨ 3x −f (x) 1 = if x < 1<br />
(a) (2, Find 6) a function C that models the first-class postage y charged, f (x) PAGE<br />
0 if x is not an integer<br />
in dollars, y for f a(x)<br />
large envelope6<br />
84 31. f (x) = 4 if x = 1 at c = 1<br />
weighing w ounces. Assume<br />
⎩<br />
4x if x > 1<br />
3 w>0.<br />
4<br />
67. lim f (x) 68. lim f (x) 69. lim f (x) 70. lim f (x)<br />
x→2 ⎧ x→1/2 x→3 x→0<br />
(b) What is the domain of C? 2<br />
⎨ x + 2 if x < 2<br />
32. f (x) = 4 if x = 2 at c = 2<br />
⎩<br />
2 4 x<br />
2 4 x<br />
x 2 if x > 2<br />
Kathryn Sidenstricker /Dreamstime.com<br />
− e<br />
13. lim 2 2 x<br />
=−2. For table values<br />
x→0<br />
x<br />
see TSM.<br />
x<br />
14. lim ln − =<br />
→ x 1 1 . For table values see<br />
x 1<br />
TSM.<br />
− x<br />
15. lim 1 cos = 0. For table values<br />
x→0<br />
x<br />
see TSM.<br />
x<br />
16. lim<br />
sin = 0. For table values<br />
x→0<br />
1+<br />
tanx<br />
see TSM.<br />
17. (a) 2<br />
(b) 2<br />
(c) 2<br />
18. (a) 4<br />
(b) 4<br />
(c) 4<br />
19. (a) 3<br />
(b) 6<br />
(c) Limit does not exist.<br />
20. (a) 4<br />
(b) 2<br />
(c) Limit does not exist.<br />
21. 1<br />
22. 1<br />
23. 1<br />
24. Limit does not exist.<br />
25. Limit does not exist, because the two<br />
one-sided limits are not equal.<br />
26. Limit does not exist.<br />
27. Limit does not exist, because the two<br />
one-sided limits are not equal.<br />
28. Limit does not exist.<br />
29. 9<br />
30. Limit does not exist.<br />
31. Limit does not exist.<br />
32. 4<br />
Section 1.1 • Assess Your Understanding<br />
87<br />
TE_<strong>Sullivan</strong>_Chapter01_PART 0.indd 16<br />
11/01/17 9:52 am
<strong>Sullivan</strong> AP˙<strong>Sullivan</strong>˙Chapter01 October 8, 2016 17:4<br />
<strong>Sullivan</strong><br />
88 Chapter 1 • Limits and Continuity<br />
33. 2<br />
34. Limit does not exist.<br />
35. Limit does not exist.<br />
36. Limit does not exist.<br />
37. Answers will vary. Sample answer:<br />
y<br />
4<br />
3<br />
2<br />
1<br />
1 2 3 4<br />
38. Answers will vary. Sample answer:<br />
y<br />
2<br />
1<br />
22 21 1 2 3 x<br />
21<br />
22<br />
39. Answers will vary. Sample answer:<br />
y<br />
21<br />
4<br />
3<br />
2<br />
1<br />
21<br />
1 2<br />
40. Answers will vary. Sample answer:<br />
y<br />
3<br />
2<br />
1<br />
22 21 1 2 3 4 x<br />
21<br />
41. 1 42. −1<br />
43. 0 44. 1<br />
45. 1 46. 1<br />
47. 0 48. 0<br />
49. 0 50. −1<br />
51. (a) m sec<br />
= 15<br />
(b) msec = 3( x+<br />
2)<br />
(c) lim m = 12. For sample table<br />
x→2<br />
see TSM.<br />
(d)<br />
y<br />
30<br />
25<br />
20<br />
15<br />
10<br />
5<br />
sec<br />
(2, 12)<br />
(3, 27)<br />
1 2 3<br />
x<br />
x<br />
Tangent line<br />
y 12x 12<br />
Secant line<br />
y 15x 18<br />
52. (a) msec<br />
= 19<br />
2<br />
(b) msec<br />
= x + 2x<br />
+ 4<br />
(c) lim msec<br />
= 12. For sample table<br />
x→2<br />
see TSM.<br />
x<br />
33.<br />
2x 2 if x < 1<br />
f (x) =<br />
3x 2 − 1 if x > 1<br />
at c = 1<br />
34.<br />
x 3 if x < −1<br />
f (x) =<br />
x 2 − 1<br />
if x > −1<br />
at c =−1<br />
35.<br />
x 2 if x ≤ 0<br />
f (x) =<br />
at c = 0<br />
2x + 1 if x > 0<br />
⎧<br />
⎨<br />
x 2 if x < 1<br />
36. f (x) =<br />
2 if x = 1<br />
at c = 1<br />
⎩<br />
−3x + 2 if x > 1<br />
Applications and Extensions<br />
In Problems 37–40, sketch a graph of a function with the given<br />
properties. Answers will vary.<br />
37. lim<br />
f (x) = 3;<br />
x→2 lim<br />
f (x) = 3;<br />
x→3 − lim<br />
f (x) = 1;<br />
x→3 + f (2) = 3; f (3) = 1<br />
38. lim<br />
f (x) = 0;<br />
x→−1 lim<br />
f (x) =−2;<br />
x→2 − lim<br />
f (x) =−2;<br />
x→2 + f (−1) is not defined; f (2) =−2<br />
39. lim<br />
f (x) = 4;<br />
x→1 lim<br />
f (x) =−1;<br />
x→0 − lim<br />
f (x) = 0;<br />
x→0 + f (0) =−1; f (1) = 2<br />
40. lim<br />
f (x) = 2;<br />
x→2 lim<br />
f (x) = 0;<br />
x→−1 lim<br />
f (x) = 1;<br />
x→1 f (−1) = 1; f (2) = 3<br />
In Problems 41–50, use either a graph or a table to investigate<br />
each limit.<br />
|x − 5|<br />
|x − 5|<br />
41. lim<br />
42. lim<br />
43. lim<br />
2x<br />
x→5 + x − 5<br />
x→5 − x − 5<br />
x→ 12<br />
2x<br />
−<br />
44. lim 2x 2x 2x<br />
x→ 12<br />
2x 45. lim<br />
+ <br />
x→ 23<br />
2x 46. lim<br />
− <br />
x→ 23<br />
2x<br />
+<br />
47. lim<br />
|x|−x 48. lim<br />
|x|−x<br />
x→2 + x→2 −<br />
3 3 49. lim x−x x−x 50. lim x−x x−x<br />
x→2 + x→2 − 51. Slope of a Tangent Line For f (x) = 3x 2 :<br />
(a) Find the slope of the secant line containing the points (2, 12)<br />
and (3, 27).<br />
(b) Find the slope of the secant line containing the points (2, 12)<br />
and (x, f (x)), x = = 2.<br />
(c) Create a table to investigate the slope of the tangent line to the<br />
graph of f at 2 using the result from (b).<br />
(d) On the same set of axes, graph f , the tangent line to the graph<br />
of f at the point (2, 12), and the secant line from (a).<br />
52. Slope of a Tangent Line For f (x) = x 3 :<br />
(d)<br />
(a) Find the slope of the secant line containing the points (2, 8)<br />
and (3, 27).<br />
(b) Find the slope of the secant line containing the points (2, 8)<br />
and (x, f (x)), x = = 2.<br />
(c) Create a table to investigate the slope of the tangent line to the<br />
graph of f at 2 using the result from (b).<br />
(d) On the same set of axes, graph f , the tangent line to the graph<br />
of f at the point (2, 8), and the secant line from (a).<br />
y<br />
40<br />
30<br />
(3, 27)<br />
Secant line<br />
20 y 5 19x 2 30<br />
10<br />
(2, 8)<br />
1 2 3<br />
Tangent line<br />
y 5 12x 2 16<br />
1<br />
53. (a) msec = 2+<br />
h for h ≠ 0<br />
2<br />
(b) For table see TSM.<br />
(c) lim m = 2<br />
h→0<br />
sec<br />
(d) m = 2<br />
tan<br />
x<br />
53. Slope of a Tangent Line For f (x) = 1 2 x2 − 1:<br />
(a) Find the slope m sec of the secant line containing the<br />
points P = (2, f (2)) and Q = (2 + h, f (2 + h)).<br />
(b) Use the result from (a) to complete the following table:<br />
h −0.5 −0.1 −0.001 0.001 0.1 0.5<br />
m sec<br />
(c) Investigate the limit of the slope of the secant line found in (a)<br />
as h → 0.<br />
(d) What is the slope of the tangent line to the graph of f at the<br />
point P = (2, f (2))?<br />
(e) On the same set of axes, graph f and the tangent line to f at<br />
P = (2, f (2)).<br />
54. Slope of a Tangent Line For f (x) = x 2 − 1:<br />
(a) Find the slope m sec of the secant line containing the<br />
points P = (−1, f (−1)) and Q = (−1 + h, f (−1 + h)).<br />
(b) Use the result from (a) to complete the following table:<br />
h −0.1 −0.01 −0.001 −0.0001 0.0001 0.001 0.01 0.1<br />
m sec<br />
(c) Investigate the limit of the slope of the secant line found<br />
in (a) as h → 0.<br />
(d) What is the slope of the tangent line to the graph of f at the<br />
point P = (−1, f (−1))?<br />
(e) On the same set of axes, graph f and the tangent line to f<br />
at P = (−1, f (−1)).<br />
PAGE<br />
85 55. (a) Investigate lim cos π by using a table and evaluating the<br />
x→0 x function f (x) = cos π x at<br />
x =− 1 2 , − 1 4 , − 1 8 , − 1 10 , − 1 12 ,..., 1<br />
12 , 1 10 , 1 8 , 1 4 , 1 2 .<br />
(b) Investigate lim<br />
cos π by using a table and evaluating the<br />
x→0 x function f (x) = cos π x at<br />
x =−1, − 1 3 , − 1 5 , − 1 7 , − 1 9 ,..., 1 9 , 1 7 , 1 5 , 1 3 , 1.<br />
(c) Compare the results from (a) and (b). What do you conclude<br />
about the limit? Why do you think this happens? What is<br />
your view about using a table to draw a conclusion about<br />
limits?<br />
(d) Use technology to graph f . Begin with the x-window<br />
[−2π, 2π] and the y-window [−1, 1]. If you were finding<br />
lim<br />
f (x) using a graph, what would you conclude? Zoom in<br />
x→0 on the graph. Describe what you see. (Hint: Be sure your<br />
calculator is set to the radian mode.)<br />
56. (a) Investigate lim<br />
cos π by using a table and evaluating the<br />
x→0 x 2 function f (x) = cos π at x =−0.1, −0.01, −0.001,<br />
x 2 −0.0001, 0.0001, 0.001, 0.01, 0.1.<br />
(e)<br />
3<br />
2<br />
1<br />
1<br />
y<br />
1 2 3<br />
y 2x 3<br />
54. (a) msec =− 2+<br />
h for h ≠ 0<br />
(b) For table see TSM.<br />
(c) lim msec<br />
=−2<br />
h→0<br />
(d) m =−2<br />
(e)<br />
tan<br />
y 5 22x 2 2<br />
22 21<br />
y<br />
3<br />
2<br />
1<br />
21<br />
x<br />
1<br />
x<br />
Answers continue on p. 89<br />
88<br />
Chapter 1 • Limits and Continuity<br />
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Section 1.1 • Assess Your Understanding 89<br />
(b) Investigate lim<br />
cos π by using a table and evaluating the<br />
x→0 x 2 function f (x) = cos π x 2 at<br />
x =− 2 3 , − 2 5 , − 2 7 , − 2 9 ,..., 2 9 , 2 7 , 2 5 , 2 3 .<br />
(c) Compare the results from (a) and (b). What do you conclude<br />
about the limit? Why do you think this happens? What is your<br />
view about using a table to draw a conclusion about limits?<br />
(d) Use technology to graph f . Begin with the x-window<br />
[−2π, 2π] and the y-window [−1, 1]. If you were finding<br />
lim<br />
f (x) using a graph, what would you conclude? Zoom in<br />
x→0 on the graph. Describe what you see. (Hint: Be sure your<br />
calculator is set to the radian mode.)<br />
PAGE<br />
x − 8<br />
85 57. (a) Use a table to investigate lim<br />
.<br />
x→2 2<br />
(b) How close must x be to 2, so that f (x) is within 0.1 of the<br />
limit?<br />
(c) How close must x be to 2, so that f (x) is within 0.01 of the<br />
limit?<br />
58. (a) Use a table to investigate lim<br />
(5 − 2x).<br />
x→2 (b) How close must x be to 2, so that f (x) is within 0.1 of the<br />
limit?<br />
(c) How close must x be to 2, so that f (x) is within 0.01 of the<br />
limit?<br />
59. First-Class Mail As of April<br />
2016, the U.S. Postal Service<br />
charged $0.47 postage for<br />
first-class letters weighing up to<br />
and including 1 ounce, plus a flat<br />
fee of $0.21 for each additional<br />
or partial ounce up to and<br />
including 3.5 ounces. First-class<br />
letter rates do not apply to letters<br />
weighing more than 3.5 ounces.<br />
Source: U.S. Postal Service Notice 123<br />
(a) Find a function C that models the first-class postage charged,<br />
in dollars, for a letter weighing w ounces. Assume w>0.<br />
(b) What is the domain of C?<br />
(c) Graph the function C.<br />
(d) Use the graph to investigate lim<br />
C(w) and lim<br />
C(w). Do<br />
w→2 − w→2 + these suggest that lim<br />
C(w) exists?<br />
w→2 (e) Use the graph to investigate lim<br />
C(w).<br />
w→0 + (f) Use the graph to investigate<br />
lim<br />
C(w).<br />
w→3.5 − 60. First-Class Mail As of April 2016, the U.S. Postal Service<br />
charged $0.94 postage for first-class large envelope weighing up to<br />
and including 1 ounce, plus a flat fee of $0.21 for each additional<br />
or partial ounce up to and including 13 ounces. First-class rates do<br />
not apply to large envelopes weighing more than 13 ounces.<br />
Source: U.S. Postal Service Notice 123<br />
(a) Find a function C that models the first-class postage charged,<br />
in dollars, for a large envelope weighing w ounces. Assume<br />
w>0.<br />
(b) What is the domain of C?<br />
55. (a) For table see TSM. Values in table<br />
π<br />
suggest lim cos = 1.<br />
x→0<br />
x<br />
(b) For table see TSM. Values in table<br />
π<br />
suggest lim cos =− 1.<br />
x→0<br />
x<br />
(c) Limit does not exist (see TSM).<br />
(d)<br />
y<br />
1<br />
0.5<br />
23 22 21 1 2 3 x<br />
20.5<br />
21<br />
y<br />
1<br />
0.5<br />
Kathryn Sidenstricker /Dreamstime.com<br />
(c) Graph the function C.<br />
(d) Use the graph to investigate lim<br />
C(w) and lim<br />
C(w). Do<br />
w→1 − w→1 + these suggest that lim<br />
C(w) exists?<br />
w→1 (e) Use the graph to investigate<br />
lim<br />
C(w) and<br />
lim<br />
C(w).<br />
w→12 − w→12 + Do these suggest that lim<br />
C(w) exists?<br />
w→12 (f) Use the graph to investigate lim<br />
C(w).<br />
w→0 + (g) Use the graph to investigate<br />
lim<br />
C(w).<br />
w→13 −<br />
61. Correlating Student Success to Study Time Professor Smith<br />
claims that a student’s final exam score is a function of the time t<br />
(in hours) that the student studies. He claims that the closer to<br />
seven hours one studies, the closer to 100% the student scores<br />
on the final. He claims that studying significantly less than seven<br />
hours may cause one to be underprepared for the test, while<br />
studying significantly more than seven hours may cause<br />
“burnout.”<br />
(a) Write Professor Smith’s claim symbolically as a limit.<br />
(b) Write Professor Smith’s claim using the ε-δ definition<br />
of limit.<br />
Source: Submitted by the students of Millikin University.<br />
62. The definition of the slope of the tangent line to the graph of<br />
f (x) − f (c)<br />
y = f (x) at the point (c, f (c)) is m tan = lim<br />
.<br />
x→c x − c<br />
Another way to express this slope is to define a new variable<br />
h = x − c. Rewrite the slope of the tangent line m tan using h and c.<br />
63. If f (2) = 6, can you conclude anything about lim<br />
f (x)? Explain<br />
x→2 your reasoning.<br />
64. If lim<br />
f (x) = 6, can you conclude anything about f (2)? Explain<br />
x→2 your reasoning.<br />
65. The graph of f (x) = x − 3<br />
is a straight line with a point punched<br />
3 − x out.<br />
(a) What straight line and what point?<br />
(b) Use the graph of f to investigate the one-sided limits of f as<br />
x approaches 3.<br />
(c) Does the graph suggest that lim<br />
f (x) exists? If so, what is it?<br />
x→3 66. (a) Use a table to investigate lim<br />
(1 + x) 1/x .<br />
x→0 (b) Use graphing technology to graph g(x) = (1 + x) 1/x .<br />
(c) What do (a) and (b) suggest about lim<br />
(1 + x) 1/x ?<br />
x→0 CAS<br />
(d) Find lim<br />
(1 + x) 1/x .<br />
x→0<br />
Challenge Problems<br />
For Problems 67–70, investigate each of the following limits.<br />
{ 1 if x is an integer<br />
f (x) =<br />
0 if x is not an integer<br />
67. lim<br />
f (x) 68. lim<br />
f (x) 69. lim<br />
f (x) 70. lim<br />
f (x)<br />
x→2 x→1/2 x→3 x→0<br />
56. (a) For table see TSM. Values in the table<br />
π<br />
suggest lim cos =<br />
→ x<br />
1 .<br />
x 0<br />
2<br />
(b) For table see TSM. Values in table<br />
π 2<br />
suggest. lim cos = .<br />
x→0<br />
2<br />
x 2<br />
(c) Limit does not exist (see TSM).<br />
(d)<br />
y<br />
1<br />
0.5<br />
24 23 22 21 1 2 3 4 x<br />
20.5<br />
21<br />
y<br />
1<br />
0.5<br />
21 20.5 0.5 1 x<br />
20.5<br />
21<br />
x − 8<br />
57. (a) lim =− 3. For sample table<br />
x→2<br />
2<br />
see TSM.<br />
(b) 1.8≤ x ≤2.2<br />
(c) 1.98 ≤ x ≤2.02<br />
58. (a) lim (5− 2 x) = 1. For sample<br />
x→ 2<br />
table see TSM.<br />
(b) 1.95 ≤ x ≤2.05<br />
(c) 1.995 ≤ x ≤2.005<br />
⎧0.47 if 0< w ≤1<br />
⎪<br />
⎪0.68 if 1< w ≤2<br />
59. (a) Cw ( ) = ⎨<br />
⎪0.89 if 2 < w ≤3<br />
⎪<br />
⎩⎪<br />
1.10 if 3 < w ≤3.5<br />
(b) { w|0< w ≤3.5}<br />
(c)<br />
C(w)<br />
1.2<br />
1.0<br />
0.8<br />
0.6<br />
0.4<br />
0.2<br />
1 2 3 3.5 w<br />
(d) lim Cw ( ) = 0.68, lim Cw ( ) = 0.89,<br />
w→2<br />
− +<br />
w→2 w→2<br />
lim Cw ( )does not exist.<br />
(e)<br />
(f)<br />
60. (a)<br />
⎧<br />
⎪<br />
⎪<br />
⎪<br />
⎪<br />
⎪<br />
⎪<br />
⎪<br />
Cw ( ) = ⎨<br />
⎪<br />
⎪<br />
⎪<br />
⎪<br />
⎪<br />
⎪<br />
⎪<br />
⎪<br />
⎩<br />
lim Cw ( ) = 0.47<br />
w → 0<br />
+<br />
lim Cw ( ) = 1.10<br />
w →3.5<br />
−<br />
0.94 if 0 < w ≤1<br />
1.15 if 1< w ≤2<br />
1.36 if 2 < w ≤3<br />
1.57 if 3 < w ≤4<br />
1.78 if 4 < w ≤5<br />
1.99 if 5 < w ≤6<br />
2.20 if 6 < w ≤7<br />
2.41 if 7 < w ≤8<br />
2.62 if 8 < w ≤9<br />
2.83 if 9 < w ≤10<br />
3.04 if 10< w ≤11<br />
3.25 if 11< w ≤12<br />
3.46 if 12< w ≤13<br />
20.1 20.05 0.05 0.1 x<br />
20.5<br />
21 Answers continue on p. 90<br />
Section 1.1 • Assess Your Understanding<br />
89<br />
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<strong>Sullivan</strong><br />
90 Chapter 1 • Limits and Continuity<br />
(b) { w|0< w ≤13}<br />
(c)<br />
C(w)<br />
4<br />
3<br />
2<br />
1<br />
1 2 3 4 5 6 7 8 9 10 11 12 13 w<br />
(d) lim Cw ( ) = 0.94; lim Cw ( ) = 1.15;<br />
w →1<br />
− w → 1<br />
+<br />
lim Cw ( ) does not exist.<br />
w →1<br />
(e) lim Cw ( ) = 3.25;<br />
w →12<br />
−<br />
lim Cw ( ) = 3.46; lim Cw ( ) does<br />
+<br />
w → 12 w →12<br />
not exist.<br />
(f) lim Cw ( ) = 0.94<br />
w → 0<br />
+<br />
(g) lim Cw ( ) = 3.46<br />
−<br />
w →13<br />
61. (a) lim St () = 100<br />
t→7<br />
(b) Given any ε > 0, there is a<br />
number δ > 0 such that whenever<br />
0 < | t − 7| < δ.<br />
| St ( − 100)|
<strong>Sullivan</strong> AP˙<strong>Sullivan</strong>˙Chapter01 October 8, 2016 17:4<br />
Section 1.2 • Limits of Functions Using Properties of Limits 91<br />
y<br />
A<br />
f (x) 5 A<br />
THEOREM The Limit of a Constant<br />
If f (x) = A, where A is a constant, then for any real number c,<br />
lim f (x) = lim A = A (1)<br />
x→c x→c<br />
The theorem is proved in Section 1.6. See Figure 17 and Table 8 for graphical and<br />
numerical support of lim<br />
x→c<br />
A = A.<br />
Teaching Tip<br />
Depending on the function given, a<br />
graphical approach may be helpful.<br />
Consider verifying the limits using graphical<br />
and analytical (that is, algebraic) methods.<br />
Figure 17 For x close to c, the value<br />
of f remains at A; lim x→c<br />
A = A.<br />
y<br />
c<br />
c<br />
c<br />
f (x) 5 x<br />
Figure 18 For x close to c, the<br />
value of f is just as close to c; lim x→c<br />
x = c.<br />
x<br />
x<br />
TABLE 8<br />
x approaches c from the left<br />
−−−−−−−−−−−−−−−−−−−−→<br />
x approaches c from the right<br />
←−−−−−−−−−−−−−−−−−−−−−<br />
x c− 0.01 c − 0.001 c − 0.0001 → c ← c + 0.0001 c + 0.001 c + 0.01<br />
f (x) = A A A A f(x) remains at A A A A<br />
For example,<br />
lim<br />
x→5 2 = 2<br />
lim 1<br />
3 = 1 3<br />
x→ √ 2<br />
THEOREM The Limit of the Identity Function<br />
If f (x) = x, then for any number c,<br />
lim(−π) =−π<br />
x→5<br />
lim f (x) = lim x = c (2)<br />
x→c x→c<br />
This theorem is proved in Section 1.6. See Figure 18 and Table 9 for graphical and<br />
numerical support of lim<br />
x→c<br />
x = c.<br />
TABLE 9<br />
x approaches c from the left<br />
−−−−−−−−−−−−−−−−−−−−→<br />
x approaches c from the right<br />
←−−−−−−−−−−−−−−−−−−−−−<br />
x c− 0.01 c − 0.001 c − 0.0001 → c ← c + 0.0001 c + 0.001 c + 0.01<br />
f (x) = x c− 0.01 c − 0.001 c − 0.0001 f (x) approaches cc+ 0.0001 c + 0.001 c + 0.01<br />
For example,<br />
lim<br />
x→−5 x =−5<br />
lim<br />
x→ √ x = √ 3<br />
3<br />
lim x = 0<br />
x→0<br />
1 Find the Limit of a Sum, a Difference, and a Product<br />
Many functions are combinations of sums, differences, and products of a constant<br />
function and the identity function. The following properties can be used to find the<br />
limit of such functions.<br />
Teaching Tip<br />
Students do not have to be able to state<br />
the theorems presented in this section.<br />
However, they must be able to apply them.<br />
Consider presenting these general rules:<br />
• When finding the limit of a function, first<br />
try to substitute the value of interest. If<br />
that results in a finite value, or something<br />
in the undefined form N for a nonzero N<br />
0<br />
(in which case the limit doesn’t exist), then<br />
you’re done.<br />
• If the substitution results in an<br />
indeterminate expression (like 0 0 or ∞ ∞ ),<br />
try to simplify the expression. If the<br />
expression is rational, try to factor the<br />
numerator and denominator and simplify.<br />
If the rational expression contains a<br />
binomial and one of the terms is a radical,<br />
try multiplying by the conjugate.<br />
• Remember, the ultimate goal is to<br />
determine what value the function is<br />
approaching as x gets very close to the<br />
given value of interest.<br />
IN WORDS The limit of the sum of two<br />
functions equals the sum of their limits.<br />
THEOREM Limit of a Sum<br />
If f and g are functions for which lim f (x) and lim g(x) both exist,<br />
x→c x→c<br />
then lim[ f (x) + g(x)] exists and<br />
x→c<br />
A proof is given in Appendix B.<br />
lim[ f (x) + g(x)] = lim f (x) + lim g(x)<br />
x→c x→c x→c<br />
TRM Section 1.2: Worksheet 1<br />
This worksheet contains 6 questions for<br />
finding limits by direct substitution. Graphs<br />
are provided for answer verification.<br />
Teaching Tip<br />
As you get into the meat of this chapter, it may<br />
be time to refer to the pacing guide and adjust<br />
pacing. There are two schools of thought to<br />
consider: (1) teaching to mastery level initially,<br />
leaving only 3 weeks to review for the end-of-term<br />
exam, or (2) moving more quickly through the<br />
material, which then will leave 4 to 5 weeks for<br />
review. The rationale for the second approach<br />
is that many of the topics that students found<br />
difficult at the beginning of the fall semester are<br />
ones with which they are very comfortable by the<br />
midterm in December. Here in Chapter 1, they still<br />
don’t yet have the background and experience<br />
to ask the good questions needed to push their<br />
understanding of the concepts because they are<br />
still focused on the nuts and bolts of getting the<br />
correct answer.<br />
Section 1.2 • Limits of Functions Using Properties of Limits<br />
91<br />
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<strong>Sullivan</strong><br />
88 92 Chapter 1 • Limits and Continuity<br />
common error<br />
Remind students that when they use<br />
substitution to find the limit, they are not<br />
evaluating the function at the x-value<br />
of interest. Rather, they are trying to<br />
determine the value of the function when<br />
we approach the given value of interest.<br />
For continuous functions, the function value<br />
and the limit close to the value of interest<br />
will be the same.<br />
2x 2 if x < 1<br />
33. f (x) =<br />
3x 2 at c = 1 EXAMPLE 1 Finding 53. the Slope Limit of a Tangent of a Sum Line For f (x) = 1<br />
− 1 if x > 1<br />
2 x2 − 1:<br />
Find lim (x + 4).<br />
x 3 x→−3 (a) Find the slope m<br />
if x < −1<br />
sec of the secant line containing the<br />
34. f (x) =<br />
x 2 at c =−1<br />
points P = (2, f (2)) and Q = (2 + h, f (2 + h)).<br />
− 1 if x > −1<br />
Solution F(x) = x + 4 is the sum of two functions f (x) = x and g(x) = 4.<br />
(b) Use the result from (a) to complete the following table:<br />
x 2 if x ≤ 0<br />
From the limits given in (1) and (2), we have<br />
35. f (x) =<br />
at c = 0<br />
2x + 1 if x > 0<br />
⎧<br />
lim f (x) = lim x =−3 h −0.5and−0.1 lim −0.001 g(x) = lim 0.001 4 = 0.1 4 0.5<br />
⎨ x 2 x→−3 x→−3 x→−3 x→−3<br />
if x < 1<br />
m sec<br />
36. f (x) = 2 if x = 1 at c = 1 Then, using the Limit of a Sum, we have<br />
⎩<br />
−3x + 2 if x > 1<br />
lim (x + (c) 4) Investigate = lim x the + limit of 4 the =−3 slope + of 4 the = secant 1 line found in (a)<br />
x→−3 as h x→−3 → 0. x→−3<br />
■<br />
(d) What is the slope of the tangent line to the graph of f at the<br />
Applications and Extensions<br />
THEOREM Limit of a Difference point P = (2, f (2))?<br />
In Problems 37–40, sketch a graph of a function with If the f and given g are functions for(e) which On the limsame f (x) setand of axes, lim g(x) graphboth f andexist,<br />
the tangent line to f at<br />
properties. Answers will vary.<br />
x→c x→c<br />
then lim[ f (x) − g(x)] exists and P = (2, f (2)).<br />
37. lim f (x) = 3; lim f (x) = 3; lim f (x) = 1; x→c<br />
x→2 x→3− x→3 +<br />
IN WORDS f (2) = 3; The limit f (3) of = the 1 difference<br />
lim 54. [ f Slope (x) −ofg(x)] a Tangent = limLine f (x) For − lim f (x) g(x) = x 2 − 1:<br />
x→c x→c x→c<br />
of two functions equals<br />
38. f (x) = 0; lim<br />
the difference of<br />
their limits.<br />
f (x) =−2; lim f (x) =−2;<br />
(a) Find the slope m sec of the secant line containing the<br />
x→−1 x→2− x→2 +<br />
points P = (−1, f (−1)) and Q = (−1 + h, f (−1 + h)).<br />
f (−1) is not defined; f (2) =−2 EXAMPLE 2 Finding the Limit of a Difference<br />
39. lim f (x) = 4; lim f (x) =−1; lim<br />
(b) Use the result from (a) to complete the following table:<br />
f (x) = 0;<br />
x→1<br />
Find lim(6 − x).<br />
x→0− x→0 + x→4<br />
f (0) =−1; f (1) = 2<br />
h −0.1 −0.01 −0.001 −0.0001 0.0001 0.001 0.01 0.1<br />
Solution F(x) = 6 − x is the difference of two functions f (x) = 6 and g(x) = x.<br />
40. lim f (x) = 2; lim f (x) = 0; lim f (x) = 1;<br />
m sec<br />
x→2 x→−1 x→1<br />
lim f (x) = lim 6 = 6 and lim g(x) = lim<br />
f (−1) = 1; f (2) = 3<br />
x = 4<br />
x→4 x→4 (c) Investigate the limit of x→4 the slope ofx→4 the secant line found<br />
in (a) as h → 0.<br />
In Problems 41–50, use either a graph or a table to Then, investigate using the Limit of a Difference, we have<br />
(d) What is the slope of the tangent line to the graph of f at the<br />
each limit.<br />
lim(6 − point x) = P lim= |x − 5|<br />
|x − 5|<br />
6 (−1, − lim f (−1))? x = 6 − 4 = 2<br />
x→4 x→4 x→4<br />
■<br />
41. lim<br />
42. lim<br />
43. lim<br />
x→5 + x − 5<br />
x→5 − x − 5<br />
<br />
x→ 12<br />
2x<br />
(e) On the same set of axes, graph f and the tangent line to f<br />
−<br />
at P = (−1, f (−1)).<br />
THEOREM Limit of a Product<br />
PAGE<br />
44. lim <br />
x→ 12<br />
2x 45. lim<br />
+ <br />
x→ 23<br />
2x 46. If f and lim<br />
− <br />
x→ 23<br />
<br />
g are 2xfunctions +<br />
85 55. for (a) which Investigate lim f (x) lim and cos π limbyg(x) usingboth a table exist, and evaluating the<br />
x→c x→0 x x→c<br />
then lim[ f (x) · g(x)] exists and<br />
function f (x) = cos π x→c<br />
47. IN WORDS lim |x|−x The limit of 48. the product lim |x|−x of<br />
x at<br />
x→2 + x→2 − lim[ f (x) · g(x)] x =− 1 =<br />
3 3 2 , − 1 lim<br />
4 , − 1 f (x)<br />
8 , − 1 · lim<br />
two functions equals the product of their<br />
10 , − 1g(x)<br />
x→c x→c x→c<br />
12 ,..., 1<br />
12 , 1 10 , 1 8 , 1 4 , 1 limits.<br />
2 .<br />
49. lim x−x 50. lim x−x<br />
x→2 + x→2 − A proof is given in Appendix B.<br />
51. Slope of a Tangent Line For f (x) = 3x 2 (b) Investigate lim cos π by using a table and evaluating the<br />
:<br />
x→0 x<br />
EXAMPLE 3 Finding the Limit<br />
(a) Find the slope of the secant line containing the points (2, 12)<br />
function off a(x) Product = cos π x at<br />
and (3, 27).<br />
Find:<br />
(b) Find the slope of the secant line containing the points (2, 12)<br />
x =−1, − 1 3 , − 1 5 , − 1 7 , − 1 9 ,..., 1 9 , 1 7 , 1 5 , 1 3 , 1.<br />
and (x, f (x)), x = 2.<br />
(a) lim x 2 (b) lim (−4x)<br />
x→3 x→−5<br />
(c) Compare the results from (a) and (b). What do you conclude<br />
(c) Create a table to investigate the slope of the tangent line to the<br />
about the limit? Why do you think this happens? What is<br />
graph of f at 2 using the result from (b). Solution (a) F(x) = x 2 is the product of two functions, f (x) = x and g(x) = x.<br />
your view about using a table to draw a conclusion about<br />
(d) On the same set of axes, graph f , the tangent<br />
Then,<br />
line<br />
using<br />
to the<br />
the<br />
graph<br />
Limit of a Product, we have<br />
limits?<br />
of f at the point (2, 12), and the secant line from (a).<br />
lim<br />
(d) x 2 = lim x · lim x = (3)(3) = 9<br />
Use technology to graph f . Begin with the x-window<br />
52. Slope of a Tangent Line For f (x) = x 3 x→3 x→3 x→3<br />
:<br />
[−2π, 2π] and the y-window [−1, 1]. If you were finding<br />
(b) F(x) =−4x is the product of two functions, f (x) =−4 and g(x) = x. Then,<br />
(a) Find the slope of the secant line containing the points (2, 8)<br />
lim f (x) using a graph, what would you conclude? Zoom in<br />
using the Limit of a Product, wex→0 have<br />
and (3, 27).<br />
on the graph. Describe what you see. (Hint: Be sure your<br />
(b) Find the slope of the secant line containing the points (2, 8) lim (−4x) = calculator lim (−4) is· set lim to the x = radian (−4)(−5) mode.) = 20<br />
x→−5 x→−5 x→−5<br />
■<br />
and (x, f (x)), x = 2.<br />
56. (a) Investigate lim cos π by using a table and evaluating the<br />
(c) Create a table to investigate the slope of the tangent<br />
A corollary<br />
line to ∗ the<br />
of the Limit of a Product x→0 Theorem x2 is the special case when f (x) = k<br />
graph of f at 2 using the result from (b). is a constant function.<br />
function f (x) = cos π at x =−0.1, −0.01, −0.001,<br />
(d) On the same set of axes, graph f , the tangent line to the graph<br />
x2 of f at the point (2, 8), and the secant line ∗ Afrom corollary (a). is a theorem that follows<br />
−0.0001,<br />
directly<br />
0.0001,<br />
from a<br />
0.001,<br />
previously<br />
0.01,<br />
proved<br />
0.1.<br />
theorem.<br />
92<br />
Chapter 1 • Limits and Continuity<br />
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Section 1.2 • Limits Section of Functions 1.1 • Assess UsingYour Properties Understanding of Limits 89 93<br />
(b) Investigate lim cos π by using a table and evaluating the<br />
x→0 x2 COROLLARY Limit of a Constant (c) GraphTimes the function a Function C.<br />
function f (x) = cos π x 2 at<br />
If g is a function for which(d) lim Use g(x) the exists graphand to investigate if k is any real lim number, C(w) and then lim limC(w). [kg(x)] Do<br />
x→c w→1− w→1 x→c +<br />
exists and<br />
x =− 2 3 , − 2 5 , − 2 7 , − 2 9 ,..., 2 9 , 2 7 , 2 5 , 2 these suggest that lim C(w) exists?<br />
IN WORDS The limit of a constant<br />
3 .<br />
w→1<br />
times a function equals the constant<br />
(e) lim Use[kg(x)] the graph= tok investigate lim g(x) lim C(w) and<br />
x→c x→c lim C(w).<br />
w→12− w→12<br />
times (c) theCompare limit of the function. results from (a) and (b). What do you conclude<br />
+ Do these suggest that lim C(w) exists?<br />
about the limit? Why do you think this happens? You areWhat askedisto your prove this corollary in Problem 103. w→12<br />
view about using a table to draw a conclusion about limits? (f) Use the graph to investigate lim<br />
Limit properties often are used in combination.<br />
C(w).<br />
w→0 +<br />
(d) Use technology to graph f . Begin with the x-window<br />
(g) Use the graph to investigate lim C(w).<br />
w→13<br />
[−2π, 2π] and the y-window [−1, 1]. If you were finding<br />
−<br />
lim f (x) using a graph, what would you conclude? Zoom in<br />
x→0<br />
on the graph. Describe what you see. (Hint: Find: Be sure your<br />
calculator is set to the radian mode.)<br />
PAGE<br />
x − 8 (a) lim<br />
85 57. (a) Use a table to investigate lim .<br />
x→2 2<br />
(b) How close must x be to 2, so that f (x) isSolution within 0.1(a)<br />
of the<br />
limit?<br />
lim<br />
(c) How close must x be to 2, so that f (x) is within 0.01 of the<br />
limit?<br />
58. (a) Use a table to investigate lim(5 − 2x).<br />
x→2<br />
(b) How close must x be to 2, so that f (x) is within 0.1 of the<br />
limit?<br />
NOTE (c) The Howlimit close properties must x beare toalso 2, sotrue<br />
that f (x) is within 0.01 of the<br />
(b) We use properties of limits Source: to find Submitted the one-sided by the students limit. of Millikin University.<br />
for one-sided limit? limits.<br />
62. The definition of the slope of [ the tangent ][ line to the graph ] of<br />
lim<br />
f (x) − f (c)<br />
59. First-Class Mail As of April<br />
x→2 +[4x(2<br />
− x)] = 4 lim<br />
y = x→2 f (x) +[x(2 at the − point x)] (c, = 4 lim<br />
f (c)) is m x lim<br />
x→2 + tan = x→2 lim +(2 − x) .<br />
2016, the U.S. Postal Service<br />
[<br />
]<br />
x→c x − c<br />
charged $0.47 postage for<br />
Another = 4 · 2 way limto 2 express − lim this x slope = 4 is · 2 to· define (2 − 2) a new = 0variable<br />
■<br />
x→2 + x→2 +<br />
first-class letters weighing up to<br />
h = x − c. Rewrite the slope of the tangent line m tan using h and c.<br />
and including 1 ounce, plus a flat<br />
63. If f (2) = 6, can you conclude anything NOW about WORK lim f (x)? Problem Explain 13.<br />
fee of $0.21 for each additional<br />
x→2<br />
your reasoning.<br />
or partial ounce up to and<br />
including 3.5 ounces. First-class<br />
To find the limit of 64. piecewise-defined If lim f (x) = 6, can x→2<br />
functions you conclude at numbers anythingwhere about f the (2)? defining Explain<br />
letter rates do not apply to letters equation changes requires the your use reasoning. of one-sided limits.<br />
RECALL weighing The more limit Lthan of a3.5 function ounces.<br />
65. The graph of f (x) = x − 3 is a straight line with a point punched<br />
y = Source: f (x) asU.S. x approaches Postal Service a Notice 123<br />
3 − x<br />
EXAMPLE 5 Finding aout.<br />
Limit for a Piecewise-defined Function<br />
number c exists if and only if<br />
lim<br />
(a)<br />
f (x)<br />
Find<br />
=<br />
a<br />
lim<br />
function C that models the first-class postage charged,<br />
in dollars, for f (x) a letter = L. Find lim f (x), if it exists. (a) What straight line and what point?<br />
x→c− x→2<br />
x→c + weighing w ounces. Assume w>0.<br />
{<br />
(b) Use the graph of f to investigate the one-sided limits of f as<br />
(b) What is the domain of C?<br />
3x + 1 if x < 2<br />
f (x) = x approaches 3.<br />
(c) Graph the function C.<br />
(c) Does<br />
2x(x<br />
the graph<br />
− 1)<br />
suggest<br />
if<br />
that<br />
x<br />
lim<br />
≥ 2<br />
f (x) exists? If so, what is it?<br />
(d) Use the graph to investigate lim C(w) and lim C(w). Do<br />
x→3<br />
w→2− w→2 +<br />
Solution Since the rule66. for(a) f changes Use a table at 2, towe investigate need tolim<br />
find(1 the + x) one-sided<br />
these suggest that lim C(w) exists?<br />
1/x . limits of f as<br />
x→0<br />
w→2 x approaches 2.<br />
(e) Use the graph to investigate lim C(w).<br />
(b) Use graphing technology to graph g(x) = (1 + x) 1/x .<br />
w→0 + For x < 2, we use the left-hand<br />
(c) What<br />
limit.<br />
do<br />
Also,<br />
(a) and<br />
because<br />
(b) suggest<br />
x <<br />
about<br />
2, f (x)<br />
lim<br />
=<br />
(1<br />
3x<br />
+ x)<br />
+ 1/x 1.<br />
?<br />
y<br />
(f) Use the graph to investigate lim C(w).<br />
x→0<br />
20<br />
w→3.5 − lim f (x) = lim CAS (d) Find lim(1 x) 1/x .<br />
x→2 −<br />
x→2−(3x + 1) = lim<br />
x→2 x→0 −(3x)<br />
+ lim 1 = 3 lim x + 1 = 3(2) + 1 = 7<br />
x→2 − x→2<br />
60. First-Class Mail 16<br />
− As of April 2016, the U.S. Postal Service<br />
charged $0.94 12 postage for first-class large envelope<br />
For x<br />
weighing<br />
≥ 2, we<br />
up<br />
use<br />
to<br />
the right-hand limit. Also, because x ≥ 2, f (x) = 2x(x − 1).<br />
and including 1 ounce, plus a flat fee of $0.21 for each additional<br />
8<br />
or partial ounce up to and including 13 ounces. First-class rates do Challenge Problems<br />
not apply to large 4 envelopes weighing more than 13 ounces. lim f (x) = lim<br />
x→2 + Forx→2 Problems +[2x(x<br />
− 1)] = lim<br />
67–70, investigate x→2 +(2x)<br />
· lim<br />
eachx→2 of the +(x<br />
− 1)<br />
(2, 4)<br />
following limits.<br />
Source: U.S. Postal Service Notice 123<br />
[ { ] 1 if x is an integer<br />
4 2<br />
2 4 x<br />
= 2 lim f (x) =<br />
(a) Find a function C that models the first-class postage charged,<br />
x · lim x − lim 1 = 2 · 2 [2 − 1] = 4<br />
x→2 + x→2 + x→2 0 + if x is not an integer<br />
in dollars, for a large envelope weighing w ounces. Assume<br />
w>0.<br />
Since lim f (x) = 7 = 67. lim limf (x) f (x) = 4, 68. lim lim f (x) f does (x) not 69. exist. lim ■f (x) 70. lim<br />
{ f (x)<br />
x→2 −<br />
x→2 x→2 +<br />
x→2x→1/2 x→3 x→0<br />
(b) What is the domain 3x + 1 of C? if x < 2<br />
Figure 19 f (x) =<br />
2x(x − 1) if x ≥ 2 See Figure 19.<br />
NOW WORK Problem 73 and AP® Practice Problems 1 and 5.<br />
Kathryn Sidenstricker /Dreamstime.com<br />
EXAMPLE 4<br />
Finding a Limit<br />
61. Correlating Student Success to Study Time Professor Smith<br />
claims that a student’s final exam score is a function of the time t<br />
(in hours) that the student studies. He claims that the closer to<br />
[2x(x + 4)] seven (b) hours limone studies, the closer to 100% the student scores<br />
x→1 x→2 +[4x(2<br />
− x)]<br />
on the final. He claims that studying significantly less than seven<br />
hours may cause one to be underprepared for the test, while<br />
studying significantly more than seven hours may cause<br />
lim“burnout.”<br />
lim<br />
Limit of a Product<br />
[ ][ ]<br />
[(2x)(x + 4)] = (2x) (x + 4)<br />
x→1 x→1 x→1<br />
[ ] [<br />
]<br />
= 2 ·(a) limWrite x · Professor lim x + Smith’s lim 4 claimLimit symbolically of a Constant as a limit. Times a<br />
(b) x→1 x→1 x→1 Write Professor Smith’s claim<br />
Function,<br />
using the<br />
Limit<br />
ε-δ definition<br />
of a Sum<br />
= (2 · 1) · of (1 limit. + 4) = 10 Use (2) and (1), and simplify.<br />
AP® Exam Tip<br />
If you like to use released AP ® Calculus<br />
multiple-choice questions, be sure to<br />
review each question carefully, as many<br />
questions about limits require more<br />
knowledge than students will have<br />
at this point in the term. Once they<br />
complete this chapter, students will be<br />
able to find limits by direct substitution<br />
or with some algebraic manipulation<br />
followed by substitution.<br />
AP® Exam Tip<br />
Starting with the 2017 exam, L’Hôpital’s<br />
Rule is on the Calculus AB exam.<br />
L’Hôpital’s Rule is a powerful technique<br />
that can be used to find limits of<br />
expressions of the form 0 0 or ∞ ∞ . This<br />
topic will be studied in Section 4.5.<br />
Alternate Example<br />
Finding a Limit for a Piecewise-Defined<br />
Function<br />
⎧ x<br />
Given<br />
⎪ | + 1|<br />
fx ( ) =<br />
x ≤ 0<br />
⎨<br />
2<br />
⎩⎪ 2−<br />
2x<br />
x > 0<br />
.<br />
Find<br />
lim fx ( ) if it exists.<br />
x→0<br />
Solution<br />
Because the rule for f changes at 0, we find<br />
the one-sided limits of f as x approaches<br />
0. For values of x near 0 and less than 0,<br />
f (x) = |x + 1| = x + 1 and<br />
lim fx ( ) = lim x + 1=<br />
1<br />
−<br />
−<br />
x→0 x→0<br />
For values of x near 0 and greater than<br />
zero, f (x) = 2 – 2x 2 and<br />
2<br />
lim fx ( ) = lim 2− 2x<br />
= lim 2− 2(0) = 2<br />
+ + +<br />
x→0 x→0<br />
x→0<br />
Because<br />
lim fx ( )<br />
x→0<br />
lim fx ( ) ≠ lim fx ( ),<br />
x→0<br />
− x→ 0<br />
+<br />
does not exist.<br />
does not exist.<br />
Section 1.2 • Limits of Functions Using Properties of Limits<br />
93<br />
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<strong>Sullivan</strong><br />
88 94 Chapter 1 • Limits and Continuity<br />
Teaching Tip<br />
Most of the limit problems in this section<br />
can be solved using direct substitution.<br />
Consider teaching this section using direct<br />
substitution as the go-to technique. When<br />
substitution does not work, try algebraic<br />
manipulation, a graphical approach, or a<br />
tabular approach. It is not necessary to<br />
teach each of the theorems presented in<br />
this section one at a time in class because<br />
students will not be expected to write a<br />
proof step by step on the AP ® Exam.<br />
AP® Calc Skill Builder<br />
for Example 7<br />
Finding the Limit of a Power<br />
2 2<br />
Investigate lim ( x − 2x+<br />
1)<br />
x→a<br />
Solution<br />
We can rewrite x 2 − 2x + 1 = (x − 1) 2 , which<br />
leads to<br />
lim ( x − 2x+ 1) = lim (( x −1) )<br />
2 2 2 2<br />
x→a x→a<br />
= lim ( x− 1) = ( a−1)<br />
x→a<br />
4 4<br />
{ 2x 2 if x < 1<br />
33. f (x) =<br />
3x 2 at c = 1<br />
53. Slope of a Tangent 0 if Line t < For c<br />
The Heaviside function, u f (x) = 1<br />
− 1 if x > 1<br />
c (t) =<br />
, is a step<br />
2 x2 function − 1: that is used<br />
1 if t ≥ c<br />
x 3 if x < −1<br />
in electrical engineering to(a) model Findathe switch. slope mThe sec ofswitch the secant is off lineif containing t < c, and the it is on<br />
34. f (x) =<br />
x 2 at c =−1<br />
− 1 if x > −1<br />
if t ≥ c.<br />
points P = (2, f (2)) and Q = (2 + h, f (2 + h)).<br />
(b) Use the result from (a) to complete the following table:<br />
x 2 if x ≤ 0<br />
35. f (x) =<br />
at c = 0<br />
2x + 1 if x > 0<br />
EXAMPLE 6 Finding a Limit<br />
⎧<br />
h of the −0.5Heaviside −0.1 −0.001 Function 0.001 0.1 0.5<br />
{<br />
⎨ x 2 if x < 1<br />
m 0sec<br />
if t < 0<br />
Find lim u<br />
36. f (x) = 2 if x = 1 at c = 1<br />
0 (t), where u 0 (t) =<br />
t→0 1 if t ≥ 0<br />
⎩<br />
−3x + 2 if x > 1<br />
(c) Investigate the limit of the slope of the secant line found in (a)<br />
ORIGINS Oliver Heaviside<br />
Solution Since the Heaviside function changes rules at t = 0, we find the one-sided<br />
as h → 0.<br />
(1850--1925) was a self-taught<br />
limits as t approaches 0.<br />
(d) What is the slope of the tangent line to the graph of f at the<br />
Applications mathematician and Extensions<br />
electrical engineer. For t < 0, lim<br />
point P = (2, f (2))?<br />
InHe Problems developed 37–40, a branch sketchofamathematics<br />
u 0(t) = lim 0 = 0 and for t ≥ 0, lim u 0(t) = lim 1 = 1<br />
graph a function with the given t→0 − t→0− t→0<br />
(e) On the same set of axes, graph f and + t→0<br />
the tangent line + to f at<br />
properties. called operational Answers will calculus vary. in which<br />
P = (2, f (2)).<br />
37.<br />
differential<br />
lim f (x)<br />
equations<br />
= 3;<br />
are<br />
lim<br />
solved<br />
f (x)<br />
by<br />
Since the one-sided limits as t approaches 0 are not equal, lim u 0 (t) does not<br />
= 3; lim f (x) = 1;<br />
t→0<br />
converting x→2 them to algebraic x→3− equations. x→3 + exist. ■<br />
Heaviside f (2) = applied 3; f (3) vector = 1calculus to<br />
54. Slope of a Tangent Line For f (x) = x 2 − 1:<br />
NOW WORK Problem 81.<br />
electrical engineering and,<br />
38. lim f (x) = 0; lim perhaps most<br />
f (x) =−2; lim f (x) =−2;<br />
(a) Find the slope m sec of the secant line containing the<br />
significantly, x→−1 he simplified x→2Maxwell's<br />
− x→2 +<br />
points P = (−1, f (−1)) and Q = (−1 + h, f (−1 + h)).<br />
equations f (−1) toisthe notform defined; used by f (2) electrical =−2<br />
engineers to this day. In 1902 Heaviside<br />
39. lim f (x) = 4; lim f (x) =−1; lim<br />
(b) Use the result from (a) to complete the following table:<br />
claimed<br />
f (x) = 0;<br />
x→1 there is a layer x→0 surrounding<br />
− x→0 +<br />
Earthf (0) from=−1; which radio f (1) signals = 2 bounce,<br />
h −0.1 −0.01 −0.001 −0.0001 0.0001 0.001 0.01 0.1<br />
allowing the signals to travel around the 2 Find the Limit of a Power and the Limit of a Root<br />
40. lim f (x) = 2; lim f (x) = 0; lim f (x) = 1;<br />
m sec<br />
Earth. x→2 Heaviside's claim x→−1 was proved truex→1 Using the Limit of a Product, if lim f (x) exists, then<br />
in 1923.<br />
f (−1)<br />
The<br />
=<br />
layer,<br />
1;<br />
contained<br />
f (2) = 3<br />
in the<br />
x→c<br />
[ ]<br />
(c) Investigate the limit of the slope of the secant line found 2<br />
ionosphere, is named the Heaviside<br />
lim<br />
layer. The function we discuss here is<br />
[ f x→c (x)]2 = lim[ f (x)<br />
in<br />
·<br />
(a)<br />
f (x)]<br />
as h<br />
=<br />
→<br />
lim<br />
0.<br />
f (x) · lim f (x) = lim f (x)<br />
x→c x→c x→c x→c<br />
In Problems 41–50, use either a graph or a table to investigate<br />
one of his minor contributions to<br />
(d) What is the slope of the tangent line to the graph of f at the<br />
each limit.<br />
Repeated use of this property produces<br />
point P =<br />
the<br />
(−1,<br />
next<br />
f (−1))?<br />
corollary.<br />
mathematics and electrical engineering.<br />
|x − 5|<br />
|x − 5|<br />
41. lim<br />
42. lim<br />
43. lim<br />
x→5 + x − 5<br />
x→5 − x − 5<br />
<br />
x→ 12<br />
2x<br />
(e) On the same set of axes, graph f and the tangent line to f<br />
COROLLARY −<br />
Limit of a Power at P = (−1, f (−1)).<br />
If lim f (x) exists and if n is a positive integer, then<br />
PAGE<br />
44. lim <br />
x→ 12<br />
2x 45. lim<br />
+ <br />
x→ 23<br />
2x 46. lim<br />
− <br />
x→ 23<br />
2x<br />
+<br />
85 55. (a) Investigate lim cos π x→c by using a table and evaluating the<br />
x→0<br />
[ x<br />
function f (x) = cos π ] n<br />
lim [ f<br />
47. lim |x|−x 48. lim |x|−x<br />
x at<br />
x→c (x)]n = lim f (x)<br />
x→c<br />
x→2 + x→2 −<br />
x =− 1<br />
3 3 2 , − 1 4 , − 1 8 , − 1 10 , − 1 12 ,..., 1<br />
12 , 1 10 , 1 8 , 1 4 , 1 2 .<br />
49. lim x−x 50. lim x−x<br />
x→2 + x→2 −<br />
51. Slope of a Tangent Line For f (x) = 3x 2 (b) Investigate lim cos π by using a table and evaluating the<br />
:<br />
x→0 x<br />
EXAMPLE 7 Finding the Limit of a Power<br />
(a) Find the slope of the secant line containing the points (2, 12)<br />
function f (x) = cos π Find:<br />
x at<br />
and (3, 27).<br />
(b) Find the slope of the secant line containing the points (2, 12)<br />
x =−1, − 1 3 , − 1 5 , − 1 7 , − 1 9 ,..., 1 9 , 1 7 , 1 5 , 1 (a) lim x 5 (b) lim(2x − 3) 3 (c) lim x n n a positive 3 , 1. integer<br />
and (x, f (x)), x = 2.<br />
x→2 x→1 x→c<br />
(c) Compare the results from (a) and (b). What do you conclude<br />
(c) Create a table to investigate the slope of the tangent line to the<br />
( ) 5<br />
about the limit? Why do you think this happens? What is<br />
graph of f at 2 using the result from (b).<br />
Solution (a) lim x 5 = lim x = 2 5 = 32<br />
x→2 x→2<br />
your view about using a table to draw a conclusion about<br />
(d) On the same set of axes, graph f , the tangent line to the graph<br />
[<br />
] 3 [<br />
] 3<br />
limits?<br />
of f at the point (2, 12), and the secant line<br />
(b)<br />
from<br />
lim(2x (a).<br />
− 3) 3 = lim (2x − 3) = lim(2x) − lim 3 = (2 − 3) 3 =−1<br />
x→1 x→1 x→1 x→1<br />
(d) Use technology to graph f . Begin with the x-window<br />
52. Slope of a Tangent Line For f (x) = x 3 :<br />
[ ] n<br />
(c) lim x n = lim x = c<br />
n ■<br />
[−2π, 2π] and the y-window [−1, 1]. If you were finding<br />
x→c x→c<br />
(a) Find the slope of the secant line containing the points (2, 8)<br />
lim f (x) using a graph, what would you conclude? Zoom in<br />
x→0<br />
and (3, 27).<br />
on the graph. Describe what you see. (Hint: Be sure your<br />
The result from Example 7(c) is worth remembering since it is used frequently:<br />
(b) Find the slope of the secant line containing the points (2, 8)<br />
calculator is set to the radian mode.)<br />
and (x, f (x)), x = 2.<br />
56. (a) Investigate lim lim cos π by using a table and evaluating the<br />
(c) Create a table to investigate the slope of the tangent line to the<br />
x n x→0<br />
= c n<br />
x2 x→c<br />
graph of f at 2 using the result from (b).<br />
function f (x) = cos π at x =−0.1, −0.01, −0.001,<br />
where c is a number and n is a positive integer.<br />
(d) On the same set of axes, graph f , the tangent line to the graph<br />
x2 of f at the point (2, 8), and the secant line from (a).<br />
−0.0001, 0.0001, 0.001, 0.01, 0.1. NOW WORK Problem 15.<br />
Science and Society / Science and Society<br />
94<br />
Chapter 1 • Limits and Continuity<br />
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<strong>Sullivan</strong> AP˙<strong>Sullivan</strong>˙Chapter01 October 8, 2016 17:4<br />
Section 1.2 • Limits Section of Functions 1.1 • Assess UsingYour Properties Understanding of Limits 89 95<br />
(b) Investigate lim cos π by using a table and THEOREM evaluating the Limit of a Root (c) Graph the function C.<br />
x→0 x2 function f (x) = cos π x 2 at<br />
If lim f (x) exists and if n (d) ≥ 2Use is antheinteger, graph tothen<br />
investigate lim C(w) and lim C(w). Do<br />
x→c w→1− w→1 +<br />
x =− 2 3 , − 2 5 , − 2 7 , − 2 9 ,..., 2 9 , 2 7 , 2 5 , 2 these √suggest that √ lim C(w) exists?<br />
3 .<br />
n<br />
lim f (x) = n w→1 lim f (x)<br />
(e) x→c Use the graph to investigate x→c lim C(w) and lim C(w).<br />
w→12− w→12<br />
(c) Compare the results from (a) and (b). What do you conclude<br />
+ provided f (x) >0 if n is even. Do these suggest that lim C(w) exists?<br />
about the limit? Why do you think this happens? What is your<br />
w→12<br />
view about using a table to draw a conclusion about limits? (f) Use the graph to investigate lim C(w).<br />
w→0 +<br />
(d) Use technology to graph f . Begin with the EXAMPLE x-window 8 Finding the (g) Limit Use theof grapha Root to investigate lim C(w).<br />
w→13<br />
[−2π, 2π] and the y-window [−1, 1]. If you were finding<br />
−<br />
lim f (x) using a graph, what would you conclude? Zoom in<br />
x→0<br />
on the graph. Describe what you see. (Hint:<br />
Solution<br />
Be sure your<br />
calculator is set to the radian mode.)<br />
PAGE<br />
x − 8<br />
3<br />
85 57. (a) Use a table to investigate lim .<br />
lim<br />
x→2<br />
x→4<br />
2<br />
(b) How close must x be to 2, so that f (x) is within 0.1 of the<br />
limit?<br />
(c) How close must x be to 2, so that f (x) is within 0.01 of the<br />
limit?<br />
58. (a) Use a table to investigate lim(5 − 2x).<br />
x→2<br />
(b) How close must x be to 2, so that f (x) is within 0.1 of the<br />
limit?<br />
(c) How close must x be to 2, so that f (x) is within 0.01 of the<br />
limit?<br />
59. First-Class Mail As of April<br />
2016, the U.S. Postal Service<br />
charged $0.47 postage for<br />
first-class letters weighing up to<br />
and including 1 ounce, plus a flat integers m and n, then<br />
fee of $0.21 for each additional<br />
or partial ounce up to and<br />
including 3.5 ounces. First-class<br />
letter rates do not apply to letters<br />
yweighing more than 3.5 ounces.<br />
30Source: U.S. Postal Service Notice 123 EXAMPLE 9<br />
Source: Submitted by the students of Millikin University.<br />
The Limit of a Power<br />
62.<br />
and<br />
The<br />
the<br />
definition<br />
Limit<br />
of<br />
of<br />
the<br />
a Root<br />
slope<br />
are<br />
of the<br />
used<br />
tangent<br />
together<br />
line to<br />
to<br />
the<br />
find<br />
graph<br />
the limit<br />
of<br />
of<br />
a function with a rational exponent.<br />
f (x) − f (c)<br />
y = f (x) at the point (c, f (c)) is m tan = lim<br />
.<br />
x→c x − c<br />
THEOREM Limit of a Fractional<br />
Another way<br />
Power<br />
to express<br />
[ f (x)]<br />
this m/n<br />
slope is to define a new variable<br />
If f is a function for which h = limx −f c. (x) Rewrite existsthe and slope if [ off (x)] the tangent m/n is line defined m tan using for positive h and c.<br />
x→c<br />
63. If f (2) = 6, can you conclude anything about lim f (x)? Explain<br />
x→2<br />
your reasoning. [ ] m/n<br />
64. If lim lim [ f f = 6, can you conclude anything about f (2)? Explain<br />
x→c (x)]m/n = lim f (x)<br />
x→2 x→c<br />
your reasoning.<br />
65. The graph of f (x) = x − 3 is a straight line with a point punched<br />
Finding the 3 − x<br />
out. Limit of a Fractional Power [f(x)] m/n<br />
27<br />
(8, 27)<br />
(a) Find a function C that models the first-class Findpostage lim(x charged, + 1) 3/2 .<br />
x→8 (a) What straight line and what point?<br />
20 in dollars, for a letter weighing w ounces. Assume w>0.<br />
(b) Use the graph of f to investigate the one-sided limits of f as<br />
(b) What is the domain of C?<br />
Solution Let f (x) = x + 1. Near 8, x + 1 > 0, so (x + 1) 3/2 is defined. Then<br />
x approaches 3.<br />
[ ]<br />
(c) Graph the function C.<br />
3/2<br />
10<br />
lim<br />
(d) Use the graph to investigate lim C(w) and lim<br />
[ C(w). f (c) Does the graph suggest that lim f (x) exists? If so, what is it?<br />
Do<br />
x→3<br />
x→8 (x)]3/2 = lim(x + 1) 3/2 = lim (x + 1) = [8 + 1] 3/2 = 9 3/2 = 27<br />
x→8 ↑[<br />
x→8<br />
w→2− w→2 + m/n<br />
66. lim [(a) f ( x)] Use m/n a=<br />
table limtof ( investigate x)]<br />
lim(1 + x)<br />
these suggest that lim C(w) exists?<br />
1/x .<br />
■<br />
x→c<br />
x→c<br />
x→0<br />
5 w→2 8 10 x<br />
(e) Use the graph to investigate lim C(w).<br />
(b) Use graphing technology to graph g(x) = (1 + x) 1/x .<br />
w→0 + See Figure 20.<br />
Figure 20 f (x) = (x + 1) 3/2 (c) What do (a) and (b) suggest about lim(1 + x) 1/x ?<br />
(f) Use the graph to investigate lim C(w).<br />
x→0<br />
w→3.5 − CAS (d) Find NOW lim(1 WORK + x) 1/x Problem . 23 and AP® Practice Problem 8.<br />
x→0<br />
60. First-Class Mail As of April 2016, the U.S. Postal Service<br />
charged $0.94 postage for first-class large envelope 3 Find weighing theup Limit to of a Polynomial<br />
and including 1 ounce, plus a flat fee of $0.21Sometimes for each additional lim f (x) can be found by substituting c for x in f (x). For example,<br />
or partial ounce up to and including 13 ounces. First-class x→c<br />
rates do Challenge Problems<br />
not apply to large envelopes weighing more than 13 ounces. Forlim<br />
Problems (5x 2 ) = 67–70, 5 liminvestigate x 2 = 5 · 2each 2 = of 20the following limits.<br />
x→2 x→2<br />
Source: U.S. Postal Service Notice 123<br />
{<br />
Since lim x n = c n if n is a positive integer, 1 if x is an integer<br />
f (x) we = can use the Limit of a Constant Times a<br />
(a) Find a function C that models the first-class postage x→c charged,<br />
0 if x is not an integer<br />
in dollars, for a large envelope weighingFunction w ounces. to Assume obtain a formula for the limit of a monomial f (x) = ax n .<br />
w>0.<br />
67. lim f (x) 68. lim f (x) 69. lim f (x) 70. lim f (x)<br />
x→2 lim x→1/2 x→3 x→0<br />
(b) What is the domain of C?<br />
x→c (axn ) = ac n<br />
Kathryn Sidenstricker /Dreamstime.com<br />
Find lim<br />
√ 3<br />
x 2 + 11.<br />
x→4<br />
where a is any number.<br />
61. Correlating Student Success to Study Time Professor Smith<br />
claims that a student’s final exam score is a function of the time t<br />
(in hours) that the student studies. He claims that the closer to<br />
√ √ √<br />
x<br />
2<br />
+<br />
seven<br />
11 =<br />
hours 3 lim<br />
one<br />
(x 2 studies,<br />
+ 11)<br />
the<br />
=<br />
closer 3 lim<br />
to<br />
x 2 100%<br />
+ lim<br />
the<br />
11<br />
student scores<br />
on the ↑ final. x→4 He claims that ↑ studying x→4 significantly x→4 less than seven<br />
Limit hours of amay Root cause one Limit to be ofunderprepared a Sum for the test, while<br />
studying significantly more than seven hours may cause<br />
“burnout.” = 3√ 4 2 + 11 = 3√ 27 = 3<br />
↑<br />
(a) Write Professor Smith’s claim symbolically as a limit.<br />
lim x 2 = c 2<br />
■<br />
x→c (b) Write Professor Smith’s claim using the ε-δ definition<br />
NOWofWORK limit. Problem 19 and AP® Practice Problems 6 and 7.<br />
Section 1.2 • Limits of Functions Using Properties of Limits<br />
95<br />
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<strong>Sullivan</strong><br />
88 96 Chapter 1 • Limits and Continuity<br />
2x 2 if x < 1<br />
Since a polynomial<br />
33. f (x) =<br />
3x 2 at c = 1<br />
53. is the Slope sum of aofTangent monomials Line and Forthe f (x) limit = 1 of a<br />
− 1 if x > 1<br />
2 x2 −sum 1: is the sum of<br />
the limits, we have the following result.<br />
x 3 (a) Find the slope m<br />
if x < −1<br />
sec of the secant line containing the<br />
34. f (x) =<br />
x 2 at c =−1<br />
points P = (2, f (2)) and Q = (2 + h, f (2 + h)).<br />
− 1 if x > −1<br />
THEOREM Limit of a Polynomial Function<br />
(b) Use the result from (a) to complete the following table:<br />
x 2 if x ≤ 0<br />
If P is a polynomial function, then<br />
35. f (x) =<br />
at c = 0<br />
2x + 1 if x > 0<br />
⎧<br />
h lim−0.5 P(x) = −0.1 P(c)<br />
IN WORDS To find the limit of a<br />
−0.001 0.001 0.1 0.5<br />
⎨ x 2 x→c<br />
polynomial as x approaches if c, x evaluate < 1<br />
m sec<br />
36. the polynomial f (x) = at c. 2 if x = 1 at c = 1 for any number c.<br />
⎩<br />
−3x + 2 if x > 1<br />
(c) Investigate the limit of the slope of the secant line found in (a)<br />
Proof If P is a polynomial function, as h →that 0. is, if<br />
(d) What is the slope of the tangent line to the graph of f at the<br />
Applications and Extensions<br />
P(x) = a<br />
point n x n + a<br />
P = n−1 x n−1 +···+a<br />
(2, f (2))? 1 x + a 0<br />
In Problems 37–40, sketch a graph of a function with the given<br />
properties. Answers will vary.<br />
where n is a nonnegative integer, (e) Onthen<br />
same set of axes, graph f and the tangent line to f at<br />
37. lim f (x) = 3; lim f (x) = 3; lim<br />
( P = (2, f (2)).<br />
f (x) = 1; lim P(x) = lim an x n + a<br />
x→2 x→3− x→3 + n−1 x n−1 )<br />
+···+a 1 x + a 0<br />
x→c x→c<br />
(<br />
f (2) = 3; f (3) = 1<br />
54. Slope of a Tangent Line For f (x) = x 2 − 1:<br />
= lim an x n) (<br />
+ lim an−1 x n−1) +···+ lim(a 1 x) + lim a 0<br />
x→c x→c x→c x→c<br />
38. lim f (x) = 0; lim f (x) =−2; lim f (x) =−2;<br />
(a) Find the slope m sec of the secant line containing the<br />
x→−1 x→2− x→2 +<br />
= a n c n + a n−1 c n−1 points +···+a P = (−1, 1 c + f (−1)) a 0 and Q = (−1 Limit + of h, af Monomial (−1 + h)).<br />
f (−1) is not defined; f (2) =−2<br />
= P(c)<br />
39. lim f (x) = 4; lim f (x) =−1; lim<br />
(b) Use the result from (a) to complete the following table: ■<br />
f (x) = 0;<br />
x→1 x→0− x→0 +<br />
f (0) =−1; f (1) = 2<br />
h −0.1 −0.01 −0.001 −0.0001 0.0001 0.001 0.01 0.1<br />
40. lim f (x) = 2; lim f (x) = 0; lim EXAMPLE 10 Finding the<br />
f (x) = 1;<br />
mLimit sec<br />
of a Polynomial<br />
x→2 x→−1 x→1<br />
Find the limit of each polynomial:<br />
f (−1) = 1; f (2) = 3<br />
(c) Investigate the limit of the slope of the secant line found<br />
(a) lim(4x 2 − x + 2) = 4(3) 2 in −(a) 3 + as2 h= →35<br />
x→3<br />
0.<br />
In Problems 41–50, use either a graph or a table to investigate<br />
(d) What is the slope of the tangent line to the graph of f at the<br />
each limit.<br />
(b) lim (7x 5 + 4x 3 − 2x 2 ) = 7(−1) 5 + 4(−1) 3 − 2(−1) 2 =−13<br />
x→−1 point P = (−1, f (−1))?<br />
|x − 5|<br />
|x − 5|<br />
41. lim<br />
42. lim<br />
43. (c) lim lim<br />
x→5 + x − 5<br />
x→5 − x − 5<br />
(10x<br />
x→ 12<br />
2x 6 (e) On the same set of axes, graph f and the tangent line to f<br />
−<br />
− 4x 5 − 8x + 5) = 10(0) 6 − 4(0) 5 − 8(0) + 5 = 5<br />
■<br />
x→0 at P = (−1, f (−1)).<br />
44. lim <br />
x→ 12<br />
2x 45. lim<br />
+ <br />
x→ 23<br />
2x 46. lim<br />
− <br />
x→ 23<br />
2x<br />
+<br />
<br />
47. lim |x|−x 48. lim |x|−x<br />
x→2 + x→2 −<br />
3 3<br />
49. lim x−x 50. lim x−x<br />
x→2 + x→2 − the following result.<br />
51. Slope of a Tangent Line For f (x) = 3x 2 :<br />
(a) Find the slope of the secant line containing the points (2, 12)<br />
and (3, 27).<br />
(b) Find the slope of the secant line containing the points (2, 12)<br />
and (x, f (x)), x = 2.<br />
exists and<br />
(c) Create a table to investigate the slope of the tangent line to the<br />
graph of f at 2 using the result from (b).<br />
IN WORDS<br />
(d) On the<br />
The<br />
same<br />
limit<br />
set<br />
of<br />
of<br />
the<br />
axes,<br />
quotient<br />
graph<br />
of<br />
f , the tangent line to the graph<br />
two functions<br />
of f at<br />
equals<br />
the point<br />
the quotient<br />
(2, 12), and<br />
of<br />
the secant line from (a).<br />
their limits, provided that the limit of the<br />
52. denominator Slope of aisTangent not zero. Line For f (x) = x 3 : provided lim g(x) = 0.<br />
x→c<br />
(a) Find the slope of the secant line containing the points (2, 8)<br />
NEED TO<br />
and<br />
REVIEW?<br />
(3, 27).<br />
Rational functions<br />
are discussed (b) Find the in Section slope of P.2, thep. secant 22. line containing the points (2, 8)<br />
and (x, f (x)), x = 2.<br />
(c) Create a table to investigate the slope of the tangent line to the<br />
graph of f at 2 using the result from (b).<br />
(d) On the same set of axes, graph f , the tangent line to the graph<br />
of f at the point (2, 8), and the secant line from (a).<br />
PAGE<br />
85 55. (a) Investigate lim cos π NOW WORK Problem 29.<br />
by using a table and evaluating the<br />
x→0 x<br />
function f (x) = cos π x at<br />
4 Find the Limit of a Quotient<br />
x =− 1 2 , − 1 4 , − 1 8 , − 1 10 , − 1 12 ,..., 1<br />
12 , 1 10 , 1 8 , 1 4 , 1 2 .<br />
To find the limit of a rational function, which is the quotient of two polynomials, we use<br />
THEOREM Limit of a Quotient<br />
If f and g are functions for which lim<br />
(b) Investigate lim cos π by using a table and evaluating the<br />
x→0 x<br />
function f (x) = cos π x at<br />
[ f (x)<br />
x =−1, f −(x) 1<br />
3 , − and 1<br />
5 , − lim 1 7 , g(x) − 1<br />
9 ,..., both 1 exist,<br />
9 , 1 7 , 1 then<br />
5 , 1 x→c x→c<br />
3 , 1. lim<br />
x→c g(x)<br />
(c) Compare the results from (a) and (b). What do you conclude<br />
about [ the limit? ] f (x) Why lim do f (x) you think this happens? What is<br />
x→c<br />
lim your view about = using a table to draw a conclusion about<br />
x→c g(x) lim<br />
limits?<br />
g(x)<br />
x→c<br />
(d) Use technology to graph f . Begin with the x-window<br />
[−2π, 2π] and the y-window [−1, 1]. If you were finding<br />
lim f (x) using a graph, what would you conclude? Zoom in<br />
x→0<br />
COROLLARY Limit of a Rational the Function graph. Describe what you see. (Hint: Be sure your<br />
calculator is set to the radian mode.)<br />
If the number c is in the domain of a rational function<br />
56. (a) Investigate lim cos π R(x) = p(x)<br />
by using a table and evaluating the<br />
x→0 x2 q(x) , then<br />
function lim R(x) f = cos R(c) π at x =−0.1, −0.01, −0.001, (3)<br />
x→c x2 −0.0001, 0.0001, 0.001, 0.01, 0.1.<br />
You are asked to prove this corollary in Problem 104.<br />
]<br />
96<br />
Chapter 1 • Limits and Continuity<br />
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Section 1.2 • Limits Section of Functions 1.1 • Assess UsingYour Properties Understanding of Limits 89 97<br />
(b) Investigate lim cos π by using a table and evaluating the<br />
x→0 x2 EXAMPLE 11 Finding(c) theGraph Limit theofunction a Rational C. Function<br />
function f (x) = cos π x 2 at<br />
Find:<br />
(d) Use the graph to investigate lim C(w) and lim C(w). Do<br />
w→1− w→1 +<br />
x =− 2 3 , − 2 5 , − 2 7 , − 2 9 ,..., 2 9 , 2 7 , 2 5 , 2 3x 3 − 2x + 1 these suggest that 2x + lim<br />
3 .<br />
w→1<br />
4 C(w) exists?<br />
(a) lim<br />
(b) lim<br />
x→1 4x 2 + 5 (e) Use thex→−2<br />
graph3x to 2 investigate − 1 lim C(w) and lim C(w).<br />
w→12− w→12<br />
(c) Compare the results from (a) and (b). What do you conclude<br />
+<br />
about the limit? Why do you think this happens? Solution What (a) isSince your 1 is in the domain<br />
Do these<br />
of<br />
suggest<br />
the rational<br />
that lim<br />
w→12 function<br />
C(w) exists?<br />
R(x) = 3x 3 − 2x + 1<br />
,<br />
view about using a table to draw a conclusion about limits? (f) Use the graph to investigate lim C(w). 4x 2 + 5<br />
w→0<br />
lim<br />
+<br />
(d) Use technology to graph f . Begin with the x-window<br />
(g) R(x) Use = R(1) = 3 − 2 + 1 = 2<br />
x→1 ↑the graph to investigate 4 + 5 lim9<br />
C(w).<br />
w→13<br />
[−2π, 2π] and the y-window [−1, 1]. If you were finding<br />
− Use (3)<br />
lim f (x) using a graph, what would you conclude? Zoom in 61. Correlating Student Success to Study Time Professor Smith<br />
x→0<br />
on the graph. Describe what you see. (Hint: (b) Be Since sure −2 youris in the domain claims of the thatrational a student’s function final exam H(x) score = is 2x a function + 4 of the time t<br />
calculator is set to the radian mode.)<br />
(in hours) that the student studies. He claims 3x 2 −that 1 ,<br />
the closer to<br />
PAGE<br />
x − 8<br />
seven hours one studies, the closer to 100% the student scores<br />
85 57. (a) Use a table to investigate lim .<br />
lim<br />
x→2 2<br />
on H(x) the = H(−2) = −4 + 4<br />
x→−2 ↑final. He claims12that − studying 1 = 0<br />
11 significantly = 0 less than seven<br />
■<br />
(b) How close must x be to 2, so that f (x) is within 0.1 of the hours Use may (3) cause one to be underprepared for the test, while<br />
limit?<br />
studying significantly more than seven hours may cause<br />
(c) How close must x be to 2, so that f (x) is within 0.01 of the “burnout.”<br />
NOW WORK Problem 33.<br />
limit?<br />
(a) Write Professor Smith’s claim symbolically as a limit.<br />
58. (a) Use a table to investigate lim(5 − 2x).<br />
x→2<br />
EXAMPLE 12 Finding(b) theWrite Limit Professor a Quotient Smith’s claim using the ε-δ definition<br />
√<br />
(b) How close must x be to 2, so that f (x) is within 0.1 of the<br />
of limit.<br />
3x 2<br />
+ 1<br />
limit?<br />
Find lim .<br />
x→4 x − 1 Source: Submitted by the students of Millikin University.<br />
(c) How close must x be to 2, so that f (x) is within 0.01 of the<br />
62. The definition of the slope of the tangent line to graph of<br />
limit?<br />
Solution We seek the limit of the quotient of two functions. Since the limit of the<br />
f (x) − f (c)<br />
59. First-Class Mail As of April<br />
denominator lim(x − 1) = 0, we use the Limit of a Quotient.<br />
x→4 y = f (x) at the point (c, f (c)) is m tan = lim<br />
.<br />
x→c<br />
2016, the U.S. Postal Service<br />
x − c<br />
√<br />
√ √<br />
charged $0.47 postage for<br />
3x 2<br />
+ 1<br />
lim 3x Another 2<br />
+ 1way to express lim (3x 2 √<br />
this + slope 1)<br />
√<br />
is to define a new variable<br />
x→4<br />
x→4 3 · 42 + 1 49<br />
first-class letters weighing up to<br />
lim = h = x − c. Rewrite the slope of the tangent line m tan using h and c.<br />
x→4 x − 1 lim<br />
and including 1 ounce, plus a flat<br />
63. (x If − f (2) 1) =<br />
lim<br />
= 6, can you (x conclude − 1) =<br />
= = 7 4 − 1 3 3<br />
↑ x→4 ↑ x→4 anything about lim f (x)? Explain<br />
fee of $0.21 for each additional<br />
Limit of a Quotient<br />
x→2<br />
your<br />
Limit<br />
reasoning.<br />
of a Root<br />
■<br />
or partial ounce up to and<br />
64. If lim f (x) = 6, can you conclude anything NOW WORK about f (2)? Problem Explain 31.<br />
including 3.5 ounces. First-class<br />
x→2<br />
letter rates do not apply to letters<br />
Based on these examples, your you reasoning. might be tempted<br />
weighing more than 3.5 ounces.<br />
65. The graph of f (x) = x − 3 to conclude that finding a limit as x<br />
approaches c is simply a matter of substituting the number is a straight c intoline thewith function. a pointThe punched next<br />
Source: U.S. Postal Service Notice 123 few examples show that substitution<br />
3 − x<br />
out. cannot always be used and other strategies need to<br />
be employed.<br />
(a) Find a function C that models the first-class postage charged, (a) What straight line and what point?<br />
in dollars, for a letter weighing w ounces. Assume The w>0. limit of a rational function can be found using substitution, provided the<br />
(b) Use the graph of f to investigate the one-sided limits of f as<br />
(b) What is the domain of C?<br />
number c being approached is in the domain of the rational function. The next example<br />
x approaches 3.<br />
(c) Graph the function C.<br />
shows a strategy that can be tried when c is not in the domain.<br />
(c) Does the graph suggest that lim f (x) exists? If so, what is it?<br />
(d) Use the graph to investigate lim C(w) and lim C(w). Do<br />
x→3<br />
w→2− w→2 +<br />
66. (a) Use a table to investigate lim(1 + x)<br />
these suggest that lim C(w) exists? EXAMPLE 13 Finding the Limit of a Rational 1/x .<br />
x→0<br />
Function<br />
w→2<br />
(e) Use the graph to investigate lim<br />
x 2 + 5x + 6 (b) Use graphing technology to graph g(x) = (1 + x) 1/x .<br />
.<br />
x 2 − 4 (c) What do (a) and (b) suggest about lim<br />
Kathryn Sidenstricker /Dreamstime.com<br />
C(w).<br />
w→0 + Find lim<br />
(f) Use the graph to investigate lim C(w).<br />
x→−2<br />
(1 + x) 1/x ?<br />
x→0<br />
w→3.5 − Solution Since −2 is not CAS in(d) theFind domain lim(1 of+ the x) 1/x rational . function, substitution cannot be<br />
60. First-Class Mail As of April 2016, the U.S. used. Postal But Service this does not mean that the x→0<br />
limit does not exist! Factoring the numerator and<br />
charged $0.94 postage for first-class large envelope the denominator, weighing up to we find<br />
and including 1 ounce, plus a flat fee of $0.21 for each additional<br />
or partial ounce up to and including 13 ounces. First-class rates do Challenge x 2 + 5x Problems + 6 (x + 2)(x + 3)<br />
=<br />
not apply to large envelopes weighing more than 13 ounces.<br />
x<br />
For Problems 2 − 4 (x + 2)(x − 2)<br />
67–70, investigate each of the following limits.<br />
Source: U.S. Postal Service Notice 123 Since x = −2, and we are interested in the limit<br />
{<br />
as 1 x approaches if x is an integer −2, the factor x + 2<br />
f (x) =<br />
(a) Find a function C that models the first-class canpostage be divided charged, out. Then<br />
0 if x is not an integer<br />
in dollars, for a large envelope weighing w ounces. Assume x 2 + 5x + 6 (x + 2)(x + 3)<br />
w>0.<br />
lim<br />
67. = lim<br />
f (x) 68. lim f (x) 69. lim f (x) 70. lim f (x)<br />
x→−2 x 2 − 4 x→−2 x→2 (x + 2)(x −x→1/2 2) = lim x + 3<br />
x→−2 x − 2 = −2 + 3<br />
↑<br />
↑<br />
x→3<br />
↑ −2 − 2 =−1 x→0 4<br />
(b) What is the domain of C?<br />
Factor<br />
x = −2<br />
Divide out ( x + 2)<br />
Use the Limit of a<br />
Rational Function<br />
NOW WORK Problem 35 and AP® Practice Problem 2.<br />
■<br />
AP® Calc Skill Builder<br />
for Example 11<br />
Finding the Limit of a Rational Function<br />
3<br />
x − 9x<br />
Find lim<br />
→ x − 3 .<br />
x 3<br />
Solution<br />
3<br />
x 9x<br />
x x<br />
lim lim ( 2<br />
−<br />
−<br />
=<br />
9)<br />
x→3<br />
x − 3 x→3<br />
x − 3<br />
x( x− 3)( x + 3)<br />
= lim<br />
x→3<br />
x − 3<br />
= lim x( x+ 3) = 3(3+ 3) = 18<br />
x→3<br />
To reinforce notational fluency, remind<br />
students that the limit notation must be<br />
included in each step of the factorization.<br />
Also, encourage them to proceed gradually,<br />
step by step, when factoring expressions.<br />
For your visual learners, use the graph of<br />
the function to explain this limit.<br />
24<br />
22<br />
y<br />
18<br />
15<br />
12<br />
9<br />
6<br />
3<br />
Alternate Example<br />
Finding the Limit of a Rational Function<br />
x − x<br />
Find lim 9 3/2 7/2<br />
.<br />
→ + 5/2 3/2<br />
x 0 x − 3x<br />
Solution<br />
Factor numerator and denominator:<br />
x − x<br />
lim 9<br />
→0<br />
x − 3x<br />
x<br />
3/2 7/2<br />
=<br />
lim<br />
+ 5/2 3/2 +<br />
x→0<br />
x<br />
x<br />
2<br />
4<br />
x<br />
(9 − x )<br />
( x−<br />
3)<br />
3/2 2<br />
3/2<br />
TRM Section 1.2: Worksheet 2<br />
In this two-page worksheet, graphs are provided<br />
for 6 limit questions that each require some<br />
form of algebraic manipulation. An additional 6<br />
questions cover the properties of limits given a<br />
table of values.<br />
TRM Section 1.2: Worksheet 3<br />
This worksheet contains 8 limit questions that can<br />
be solved using techniques learned in this section.<br />
Graphs are not provided for answer verification.<br />
=<br />
−x<br />
lim<br />
x<br />
+<br />
x→0<br />
( x −9)<br />
( x−<br />
3)<br />
( x− 3)( x+<br />
3)<br />
=−lim +<br />
x→0<br />
( x − 3)<br />
=− lim ( x + 3) =−3<br />
+<br />
x→0<br />
3/2 2<br />
3/2<br />
Section 1.2 • Limits of Functions Using Properties of Limits<br />
97<br />
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<strong>Sullivan</strong><br />
88 98 Chapter 1 • Limits and Continuity<br />
Alternate Example<br />
Finding the Limit of a Quotient<br />
2<br />
x − 81<br />
Find lim .<br />
x→9<br />
x − 3<br />
Solution<br />
First, we rationalize the denominator<br />
to eliminate the radical term in the<br />
denominator:<br />
2<br />
x 81 x<br />
x<br />
lim lim ( 2<br />
81) 3<br />
x 9<br />
x 9<br />
x<br />
−− 3<br />
= ( x<br />
−− 3)<br />
⋅ +<br />
→<br />
→<br />
x + 3<br />
x x<br />
lim ( 2<br />
− 81) + 3<br />
=<br />
x→9<br />
x − 9<br />
x x x<br />
lim ( − 9)( + 9) + 3<br />
=<br />
x→9<br />
x − 9<br />
= lim ( x + 9) x + 3<br />
x→9<br />
( )<br />
( )<br />
( )<br />
( )<br />
( )<br />
( )<br />
= (9+ 9) 9+ 3 = 108<br />
AP® Calc Skill Builder<br />
for Example 15<br />
Finding the Limit of an Average Rate of<br />
Change<br />
Find the average rate of change of<br />
fx ( ) = x from x = 1 to x. Use this result<br />
to find the limit as x approaches 1 of the<br />
average rate of change of f from 1 to x.<br />
Solution<br />
Since f (1) = 1, the average rate<br />
of change of f from 1 to x is<br />
∆y<br />
∆ = fx ( ) − f (1) x −1<br />
= . The limit of<br />
x x −1<br />
x −1 the average rate of change from 1 to x is<br />
x −1<br />
x −1<br />
=<br />
− − ⋅ x + 1<br />
lim lim<br />
x 1 x→<br />
x 1 x + 1<br />
x −1<br />
= lim<br />
x→1( x− 1)( x + 1)<br />
= lim 1 + = 1<br />
x→1<br />
x 1 2<br />
x→1 1<br />
Explain that the formula for the average<br />
rate of change is fundamental to differential<br />
calculus. Observe that the limit of this<br />
−<br />
formula, lim fx ( ) fc ()<br />
→ x−<br />
c<br />
, gives the<br />
x c<br />
instantaneous rate of change, that is, the<br />
rate of change at a particular moment.<br />
Teaching Tip<br />
When you teach the students to find the<br />
average rate of change of a function<br />
between two points, choose problems that<br />
allow you to clearly illustrate the principle<br />
graphically, as well.<br />
2x 2 if x < 1<br />
The Limit of a Quotient<br />
33. f (x) =<br />
3x 2 at c = 1<br />
53. Slope property of a can Tangent onlyLine be usedFor when f (x) the = limit 1<br />
− 1 if x > 1<br />
2 x2 −of1:<br />
the denominator<br />
of the function is not zero. The next example illustrates a strategy to try if radicals are<br />
x 3 present.<br />
(a) Find the slope m<br />
if x < −1<br />
sec of the secant line containing the<br />
34. f (x) =<br />
x 2 at c =−1<br />
points P = (2, f (2)) and Q = (2 + h, f (2 + h)).<br />
− 1 if x > −1<br />
x 2 EXAMPLE 14 Finding(b) theUse Limit the result of a from Quotient (a) to complete the following table:<br />
if x ≤ 0<br />
35. f (x) =<br />
at c = 0<br />
√ √<br />
2x + 1 if x > 0<br />
x − 5<br />
⎧<br />
Find lim<br />
h −0.5 −0.1 −0.001 0.001 0.1 0.5<br />
⎨ x 2 if x < 1<br />
x→5 x − 5 .<br />
m sec √ √<br />
36. f (x) = 2 if x = 1 at c = 1<br />
x − 5<br />
⎩<br />
Solution The domain of h(x) = is {x|x ≥ 0, x = 5}. Since the limit of the<br />
−3x + 2 if x > 1<br />
(c) Investigate x − 5the limit of the slope of the secant line found in (a)<br />
denominator is<br />
as h → 0.<br />
(d) What is the slope of the tangent line to the graph of f at the<br />
Applications and Extensions<br />
lim g(x) = lim(x − 5) = 0<br />
x→5 point P (2, x→5 f (2))?<br />
In Problems 37–40, sketch a graph of a function with the given<br />
(e) On the same set of axes, graph f and the tangent line to f at<br />
properties. Answers will vary.<br />
we cannot use the Limit of a Quotient<br />
P = (2, f<br />
property.<br />
(2)).<br />
A different strategy is necessary. We<br />
37. lim f (x) = 3; lim f (x) = 3; lim f (x) rationalize = 1; the numerator of the quotient.<br />
x→2 x→3− x→3 +<br />
√ √<br />
f (2) = 3; f (3) = 1<br />
x − 5 54. Slope of a Tangent Line For f (x) = x 2 − 1:<br />
= (√ x − √ 5)<br />
· (√ x + √ 5)<br />
38. lim f (x) = 0; lim f (x) =−2; lim f (x) =−2;<br />
x − 5 (x − 5) (a) Find ( √ the x + √ slope 5) = x − 5<br />
m sec<br />
(x of the − 5)( √ secant x line + √ containing 5) = 1<br />
√ √<br />
↑ x the + 5<br />
x→−1 x→2− x→2 +<br />
points P = (−1, f (−1)) and Q = (−1 + x = h, 5f (−1 + h)).<br />
f (−1) is not defined; f (2) =−2<br />
39. lim f (x) = 4; lim f (x) =−1; lim Do you see why rationalizing (b) the Usenumerator the result from works? (a) to Itcomplete causes the term following x − 5table:<br />
to appear<br />
f (x) = 0;<br />
x→1 x→0− x→0 + in the numerator, and since x = 5, the factor x − 5 can be divided out. Then<br />
f (0) =−1; f (1) = 2<br />
√ √ h −0.1 −0.01 −0.001 −0.0001 0.0001 0.001 0.01 0.1<br />
x − 5<br />
1<br />
lim 1<br />
40. lim f (x) = 2; lim f (x) = 0; lim<br />
x→5<br />
f (x) = lim1;<br />
= lim √ √m sec =<br />
x→2 x→−1 x→1 x→5 x − 5 x→5 x + 5 lim (√ x + √ 5) = 1<br />
√ √ = 1<br />
√<br />
5<br />
5 + 5 2 √ 5 = ■<br />
NOTE When finding a limit, remember<br />
10<br />
f (−1) = 1; f (2) = 3<br />
↑ x→5<br />
to include ''lim'' at each step until you<br />
x→c Use the (c) Limit Investigate of a Quotient the limit of the slope of the secant line found<br />
let x → c.<br />
in (a) as h → 0.<br />
In Problems 41–50, use either a graph or a table to investigate<br />
NOW WORK Problem 41 and AP® Practice Problem 4.<br />
each limit.<br />
|x − 5|<br />
41. lim<br />
x→5 + x − 5<br />
|x − 5|<br />
42. lim<br />
x→5 − x − 5<br />
43. lim <br />
x→ 12<br />
2x<br />
−<br />
(d) What is the slope of the tangent line to the graph of f at the<br />
point P = (−1, f (−1))?<br />
(e) On the same set of axes, graph f and the tangent line to f<br />
at P = (−1, f (−1)).<br />
PAGE<br />
44. lim <br />
x→ 12<br />
2x 45. lim<br />
+ <br />
x→ 23<br />
2x 46. lim<br />
− <br />
x→ 23<br />
2x<br />
+<br />
85 55. (a) Investigate lim cos π by using a table and evaluating the<br />
5 Find the Limit of an Average x→0 Ratexof Change<br />
The next two examples illustrate function limits that f (x) we = cos encounter π<br />
47. lim |x|−x 48. lim |x|−x<br />
x at in Chapter 2.<br />
NEED TO x→2 + REVIEW? Average rate x→2 − of<br />
In Section P.1, we defined average x =− 1 rate<br />
3 3 2 , − 1 of<br />
4 , − 1 change:<br />
8 , − 1 If a<br />
10 , − 1 and b,<br />
12 ,..., 1 where<br />
12 , 1 10 , 1 a =<br />
8 , 1 b,<br />
4 , 1 are in<br />
change is discussed in Section P.1, p. 12. the domain of a function y = f (x), the average rate of change of f from a to b2 is .<br />
49. lim x−x 50. lim x−x<br />
x→2 + x→2 −<br />
51. Slope of a Tangent Line For f (x) = 3x 2 (b) y Investigate f (b) −limf (a) cos π by using a table and evaluating the<br />
:<br />
= x→0 x<br />
(a) Find the slope of the secant line containing the points (2, 12)<br />
function f (x) = cos π a = b<br />
x b − a<br />
x at<br />
and (3, 27).<br />
(b) Find the slope of the secant line containing the points (2, 12)<br />
x =−1, − 1 3 , − 1 5 , − 1 7 , − 1 9 ,..., 1 9 , 1 7 , 1 5 , 1 EXAMPLE 15 Finding the Limit of an Average Rate of Change 3 , 1.<br />
and (x, f (x)), x = 2.<br />
(a) Find the average rate of(c) change Compare of f the (x) results = x 2 from + 3x(a) from and 2(b). toWhat x, x do = 2. you conclude<br />
(c) Create a table to investigate the slope of the tangent line to the<br />
graph of f at 2 using the result from (b).(b) Find the limit as x approaches about 2 the of the limit? average Why do rate youofthink change this of happens? f (x) = What x 2 + is 3x<br />
your view about using a table to draw a conclusion about<br />
(d) On the same set of axes, graph f , the tangent line fromto2the tograph<br />
x.<br />
limits?<br />
of f at the point (2, 12), and the secant line from (a).<br />
Solution (a) The average<br />
(d)<br />
rate<br />
Use<br />
of change<br />
technology<br />
of f<br />
to<br />
from<br />
graph<br />
2<br />
f<br />
to<br />
. Begin<br />
x is<br />
with the x-window<br />
52. Slope of a Tangent Line For f (x) = x 3 :<br />
[−2π, 2π] and the y-window [−1, 1]. If you were finding<br />
y f (x) − f (2)<br />
(a) Find the slope of the secant line containing the= points (2, 8) = (x 2 + 3x) lim − f (x) [2 2 using + 3(2)] a graph, what would you conclude? Zoom in<br />
x→0 = x 2 + 3x − 10 (x + 5)(x − 2)<br />
=<br />
and (3, 27).<br />
x x − 2<br />
on the<br />
x −<br />
graph.<br />
2<br />
Describe what you<br />
x −<br />
see.<br />
2<br />
(Hint: Be sure<br />
x −<br />
your<br />
2<br />
(b) Find the slope of the secant line containing (b) the The points limit (2, of 8) the average rate calculator of change is set is to the radian mode.)<br />
and (x, f (x)), x = 2.<br />
56. (a) Investigate lim cos π by using a table and evaluating the<br />
(c) Create a table to investigate the slope of the tangent line to thef (x) − f (2) (x x→0 + 5)(x x− 2 2)<br />
lim<br />
= lim<br />
graph of f at 2 using the result from (b).<br />
function f (x) = cos π = lim(x + 5) = 7<br />
x→2 x − 2 x→2 x − 2 at x x→2 ■<br />
=−0.1, −0.01, −0.001,<br />
(d) On the same set of axes, graph f , the tangent line to the graph<br />
x2 of f at the point (2, 8), and the secant line from (a).<br />
−0.0001, NOW WORK 0.0001, Problem 0.001, 0.01, 63 0.1. and AP® Practice Problem 3.<br />
∑ Mathematical Practices Tip<br />
MPAC 4: Connecting Multiple Representations<br />
Draw the following curve and secant line. Ask<br />
students to use the illustration to justify why the<br />
average rate of change of a function between two<br />
points is the slope of the secant line that passes<br />
through the points. Remind them that slope =<br />
y2−<br />
y1. Then have them demonstrate what<br />
x2−<br />
x1<br />
happens to the secant lines through the points as<br />
the left-hand point gradually approaches the righthand<br />
point (which is fixed). Ask them to use this<br />
result to show that the limit of the average value<br />
between two points is the slope of the tangent line<br />
at the fixed point.<br />
f(x)<br />
3<br />
2<br />
1<br />
2 4 6 8 10 x<br />
Use the tangent line applet on the Mathscoop Web<br />
site to illustrate this process.<br />
98<br />
Chapter 1 • Limits and Continuity<br />
TE_<strong>Sullivan</strong>_Chapter01_PART 0.indd 27<br />
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<strong>Sullivan</strong> AP˙<strong>Sullivan</strong>˙Chapter01 October September 8, 2016 20, 2016 17:414:45<br />
Section 1.1 1.2 • Assess Your Understanding 89 99<br />
(b) Investigate lim cos π by using a table and evaluating the<br />
(c) Graph the function C.<br />
x→0 x2 function f (x) = cos π x 2 at<br />
In Section P.1, we defined the (d) difference Use the graph quotient to investigate of a function lim C(w) f at xand as lim C(w). Do<br />
w→1− w→1 +<br />
f<br />
x =− 2 3 , − 2 5 , − 2 7 , − 2 9 ,..., 2 9 , 2 7 , 2 5 , 2 these (x + suggest h) − f that (x) lim<br />
3 .<br />
w→1<br />
h C(w) = 0 exists?<br />
h<br />
(e) Use the graph to investigate lim C(w) and lim C(w).<br />
w→12− CALC<br />
w→12<br />
(c) Compare the results from (a) and (b). What do you conclude<br />
+ EXAMPLE 16 Finding theDo Limit theseof suggest a Difference that lim C(w) Quotient exists?<br />
about the limit? Why do you think this CLIPhappens? What is your<br />
w→12<br />
view about using a table to draw a conclusion about limits? (f) Use the graph to investigate lim<br />
(a) For f (x) = 2x 2 f (x C(w). + h) − f (x)<br />
− 3x + 1, find the difference quotient w→0 + , h = 0.<br />
(d) Use technology to graph f . Begin with the x-window<br />
(g) Use the graph to investigate lim C(w). h<br />
w→13<br />
[−2π, 2π] and the y-window [−1, 1]. If<br />
(b)<br />
you were<br />
Find<br />
finding<br />
the limit as h approaches 0 of the difference quotient − of f (x) = 2x 2 − 3x + 1.<br />
lim f (x) using a graph, what would you conclude? Zoom in 61. Correlating Student Success to Study Time Professor Smith<br />
x→0 Solution (a) To find the difference quotient of f, we begin with f (x + h).<br />
on the graph. Describe what you see. (Hint: Be sure your<br />
claims that a student’s final exam score is a function of the time t<br />
calculator is set to the radian mode.) f (x + h) = 2(x + h) 2 (in− hours) 3(x + that h) the + 1 student = 2(xstudies. 2 + 2xh He+ claims h 2 ) −that 3x the − 3h closer + 1to<br />
PAGE<br />
85 57. (a) Use a table to investigate lim<br />
2<br />
.<br />
(b) How close must x be to 2, so that f (x) is within 0.1 of the<br />
limit?<br />
Now<br />
(c) How close must x be to 2, so that f (x) is within 0.01 of the<br />
limit?<br />
58. (a) Use a table to investigate lim(5 − 2x).<br />
x→2<br />
(b) How close must x be to 2, so that f (x) is within f (x + 0.1h) of− thef (x)<br />
limit?<br />
h<br />
(c) How close must x be to 2, so that f (x) is within 0.01 of the<br />
x→2<br />
x − 8<br />
limit?<br />
59. First-Class Mail As of April<br />
2016, the U.S. Postal Service<br />
charged $0.47 postage for<br />
first-class letters weighing up to<br />
and including 1 ounce, plus a flat<br />
fee of $0.21 for each additional<br />
Summary or partial ounce up to and<br />
including 3.5 ounces. First-class<br />
Twoletter Basic rates Limits do not apply to letters<br />
weighing more than 3.5 ounces.<br />
• lim A = A, where A is a constant.<br />
x→c Source: U.S. Postal Service Notice 123<br />
• lim x = c<br />
x→c<br />
(a) Find a function C that models the first-class postage charged,<br />
in dollars, for a letter weighing w ounces. Assume w>0.<br />
Properties (b) What ofisLimits<br />
the domain of C?<br />
If f and (c) gGraph are functions the function for which C. lim f (x) and lim g(x) both exist,<br />
x→c x→c<br />
(d) Use the graph to investigate lim C(w) and lim C(w). Do<br />
and k is a constant, then<br />
w→2− w→2 +<br />
• Limit<br />
these<br />
of a<br />
suggest<br />
Sum or<br />
that<br />
a Difference:<br />
lim C(w) exists?<br />
w→2<br />
lim (e) [ Use f (x) the ± g(x)] graph = to investigate lim f (x) ± lim<br />
g(x) C(w).<br />
x→c x→c w→0 x→c +<br />
• Limit (f) Use of the a Product: graph tolim<br />
investigate [ f (x) · g(x)] lim = C(w). lim f (x) · lim g(x)<br />
x→c w→3.5 − x→c x→c<br />
• Limit of a Constant Times a Function: lim[kg(x)] = k lim g(x)<br />
60. First-Class Mail As of April 2016, the x→c U.S. Postal Service x→c<br />
charged $0.94 postage for first-class large envelope weighing up to<br />
and including 1 ounce, plus a flat fee of $0.21 for each additional<br />
or partial ounce up to and including 13 ounces. First-class rates do<br />
not apply to large envelopes weighing more than 13 ounces.<br />
1.2Source: Assess<br />
U.S.<br />
Your<br />
Postal Service<br />
Understanding<br />
Notice 123<br />
(a) Find a function C that models the first-class postage charged,<br />
Concepts and Vocabulary<br />
in dollars, for a large envelope weighing w ounces. Assume<br />
1. (a) lim w>0. (−3) = ; (b) lim π =<br />
x→4 x→0<br />
(b) What is the domain of C?<br />
2. If lim f (x) = 3, then lim[ f (x)] 5 = .<br />
x→c x→c<br />
3. If lim x→c<br />
f (x) = 64, then lim x→c<br />
3 √ f (x) = .<br />
The limit of the difference quotient<br />
tends to appear as a topic for a<br />
multiple-choice question on the exam.<br />
In the next chapter, students learn<br />
to recognize this as the definition of<br />
the derivative. Once this connection<br />
is made, the limit of the difference<br />
quotient can be found by simply taking<br />
the derivative of the function. This will<br />
be a great time saver.<br />
Kathryn Sidenstricker /Dreamstime.com<br />
AP® Exam Tip<br />
6 Find the Limit of a Difference Quotient<br />
seven hours one studies, the closer to 100% the student scores<br />
= 2x 2 + 4xh on+ the2h final. 2 − He 3x claims − 3h + that 1 studying significantly less than seven<br />
hours may cause one to be underprepared for the test, while<br />
studying significantly more than seven hours may cause<br />
“burnout.”<br />
f (x +h)− f (x) = (2x 2 +4xh+2h 2 −3x −3h +1)−(2x 2 −3x +1) = 4xh+2h 2 −3h<br />
(a) Write Professor Smith’s claim symbolically as a limit.<br />
Then, the difference quotient is<br />
(b) Write Professor Smith’s claim using the ε-δ definition<br />
of limit.<br />
= 4xh + 2h2 − 3h h(4x + 2h − 3)<br />
= = 4x + 2h − 3, h = 0<br />
h<br />
h<br />
Source: Submitted by the students of Millikin University.<br />
62. The definition of the slope of the tangent line to the graph of<br />
f (x + h) − f (x)<br />
(b) lim<br />
= lim(4x + 2h − 3) = 4x + 0 − 3 = 4x −f (x) 3 − f (c) ■<br />
h→0 h y = h→0 f (x) at the point (c, f (c)) is m tan = lim<br />
.<br />
x→c x − c<br />
Another way to express this slope is toNOW defineWORK a new variable Problem 71.<br />
h = x − c. Rewrite the slope of the tangent line m tan using h and c.<br />
63. If f (2) = 6, can you conclude anything about lim f (x)? Explain<br />
x→2<br />
your reasoning.<br />
64. If lim f (x) = 6, can you conclude anything about f (2)? Explain<br />
x→2 [ ] n<br />
• Limit your reasoning. of a Power: lim<br />
65. The graph of f (x) = x [ − f (x)] n 3 = lim f (x) x→c x→c<br />
where n ≥ 2 is an integer is a straight line with a point punched<br />
3 − x<br />
•<br />
out.<br />
√ √<br />
n<br />
Limit of a Root: lim f (x) = n lim f (x)<br />
x→c x→c<br />
provided (a) Whatf straight (x) >0line if nand ≥ 2what is even point?<br />
(b) Use the graph of f to investigate<br />
[<br />
the one-sided<br />
] m/n limits of f as<br />
• Limit of [f(x)]<br />
x approaches m/n : lim[ f (x)]<br />
3.<br />
m/n = lim f (x) x→c x→c<br />
provided (c) Does[ the f (x)] graph m/n is suggest defined that forlim<br />
positive f (x) integers exists? If mso, andwhat n is it?<br />
[ x→3 ] f (x)<br />
lim<br />
66. • f (x)<br />
x→c<br />
Limit (a) Use of a table Quotient: to investigate lim lim(1 = + x) 1/x .<br />
x→c g(x) x→0 lim g(x) x→c<br />
provided<br />
(b) Use graphing<br />
lim g(x)<br />
technology<br />
= 0<br />
to graph g(x) = (1 + x) 1/x .<br />
(c) What x→c do (a) and (b) suggest about lim(1 + x) 1/x ?<br />
• Limit of a Polynomial Function: lim P(x) x→0 = P(c)<br />
CAS (d) Find lim(1 + x) 1/x .<br />
x→c<br />
• Limit of a x→0 Rational Function: lim R(x) = R(c)<br />
x→c<br />
if c is in the domain of R<br />
Challenge Problems<br />
For Problems 67–70, investigate each of the following limits.<br />
{ 1 if x is an integer<br />
f (x) =<br />
4. (a) lim x = 0 ; (b) if xlim<br />
is not an integer<br />
x→−1 x→e<br />
67. 5. (a) lim f (x) 68. lim f (x) 69. lim f (x) x→2 lim (x − 2) = ; (b) lim (3 + x) =<br />
x→0<br />
x→1/2<br />
x→1/2<br />
x→3<br />
70. lim x→0<br />
6. (a) lim (−3x) = x→2<br />
; (b) lim (3x) =<br />
x→0<br />
7. True or False If p is a polynomial function,<br />
then lim p(x) = p(5).<br />
x→5<br />
Alternate Example<br />
Finding the Limit of a Difference<br />
Quotient<br />
For fx ( ) = x + 1 find the limit of the<br />
difference quotient as h approaches 0 + .<br />
Solution<br />
fx+ h −f x<br />
lim ( ) ( )<br />
+<br />
→0<br />
h<br />
h<br />
=<br />
=<br />
=<br />
=<br />
=<br />
lim<br />
+<br />
h→0<br />
lim<br />
+<br />
h→0<br />
lim<br />
+<br />
h→0<br />
lim<br />
+<br />
h→0<br />
lim<br />
+<br />
h→0<br />
( x+ h+ 1) − ( x + 1)<br />
h<br />
x+ h−<br />
x<br />
h<br />
( x+ h−<br />
x ) ( x+ h+<br />
x )<br />
⋅<br />
h<br />
( x+ h+<br />
x )<br />
x+ h−x<br />
h( x+ h+<br />
x)<br />
h<br />
h( x+ h+<br />
x)<br />
1<br />
= lim<br />
+<br />
h→0<br />
x+ h+<br />
x<br />
1<br />
=<br />
x+ 0 + x<br />
1<br />
=<br />
2 x<br />
Must-Do Problems for<br />
Exam Readiness<br />
AB: 35, 37, 39, 41, 43, 47, 48, 51–65<br />
odd, 67, 71, 73–79 odd, AP ® Practice<br />
Problems<br />
BC: 10, 19, 37, 39, 43, 59, 75, and all AP ®<br />
Practice Problems (especially 5)<br />
TRM Full Solutions to Section<br />
1.2 Problems and AP® Practice<br />
Problems<br />
Answers to Section 1.2<br />
Problems<br />
1. (a) −3<br />
(b) π<br />
2. 243<br />
3. 4<br />
4. (a) −1<br />
(b) e<br />
5. (a) −2<br />
(b) 7 2<br />
6. (a) −6<br />
(b) 0<br />
7. True<br />
Section 1.2 • Assess Your Understanding<br />
99<br />
TE_<strong>Sullivan</strong>_Chapter01_PART 0.indd 28<br />
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<strong>Sullivan</strong> AP˙<strong>Sullivan</strong>˙Chapter01 October 8, 2016 17:4<br />
<strong>Sullivan</strong><br />
100 Chapter 1 • Limits and Continuity<br />
8. 2<br />
9. False<br />
10. True<br />
11. 14<br />
12. –3<br />
13. 0<br />
14. 18<br />
15. 1<br />
16. 1<br />
17. 6<br />
18. 1 2<br />
19. 11<br />
20. 10<br />
21. 2 78<br />
22. 0<br />
23. 7+<br />
3<br />
24. 2<br />
25. 0<br />
26. 0<br />
27. 5<br />
28. 1<br />
29. − 31<br />
8<br />
30. –3<br />
31. 10<br />
32. 14 3<br />
33. 13 4<br />
34. 1 5<br />
35. 4<br />
8. If the domain of a rational function R is {x | x = 0},<br />
then lim R(x) = R( ).<br />
x→2<br />
9. True or False Properties of limits cannot be used for one-sided<br />
limits.<br />
(x + 1)(x + 2)<br />
10. True or False If f (x) = and g(x) = x + 2,<br />
x + 1<br />
then lim f (x) = lim g(x).<br />
x→−1 x→−1<br />
Skill Building<br />
In Problems 11–44, find each limit using properties of limits.<br />
11. lim [2(x + 4)] 12. lim [3(x + 1)]<br />
x→3 x→−2<br />
PAGE<br />
93 13. lim [x(3x−1)(x + 2)]<br />
x→−2<br />
14. lim [x(x − 1)(x + 10)]<br />
x→−1<br />
PAGE<br />
94 15. lim (3t − 2) 3<br />
t→1<br />
16. lim (−3x + 1) 2<br />
x→0<br />
17. lim (3 √ ( ) 1<br />
x) 18. lim<br />
3√ x x→4 x→8 4<br />
√<br />
PAGE<br />
95 19. lim 5x − 4<br />
x→3<br />
20. lim<br />
t→2<br />
√<br />
3t + 4<br />
21. lim t→2<br />
[t √ (5t + 3)(t + 4)] 22. lim<br />
t→−1 [t 3√ (t + 1)(2t − 1)]<br />
PAGE<br />
95 23. lim ( √ x + x + 4) 1/2<br />
x→3<br />
24. lim t→2<br />
(t √ 2t + 4) 1/3<br />
25. lim<br />
t→−1 [4t(t + 1)]2/3 26. lim x→0<br />
(x 2 − 2x) 3/5<br />
27. lim t→1<br />
(3t 2 − 2t + 4) 28. lim x→0<br />
(−3x 4 + 2x + 1)<br />
PAGE<br />
96 29. lim (2x 4 − 8x 3 + 4x − 5) 30. lim (27x 3 + 9x + 1)<br />
x→ 1 2<br />
x→− 1 3<br />
PAGE<br />
97 31.<br />
x 2 + 4<br />
lim √ x→4 x<br />
32.<br />
x 2 + 5<br />
lim √ x→3 3x<br />
PAGE<br />
97 33.<br />
2x 3 + 5x<br />
lim<br />
x→−2 3x − 2<br />
PAGE<br />
97 35.<br />
x 2 − 4<br />
lim x→2 x − 2<br />
x 3 − x<br />
37. lim<br />
x→−1 x + 1<br />
39. lim<br />
x→−8<br />
( 2x<br />
x + 8 + 16<br />
x + 8<br />
√ √<br />
PAGE<br />
x − 2<br />
98 41. lim x→2 x − 2<br />
43. lim x→4<br />
√<br />
x + 5 − 3<br />
(x − 4)(x + 1)<br />
)<br />
34. lim x→1<br />
2x 4 − 1<br />
3x 3 + 2<br />
x + 2<br />
36. lim<br />
x→−2 x 2 − 4<br />
x 3 + x 2<br />
38. lim<br />
x→−1 x 2 − 1<br />
40. lim x→2<br />
( 3x<br />
x − 2 − 6<br />
x − 2<br />
42. lim x→3<br />
√ x −<br />
√<br />
3<br />
x − 3<br />
44. lim x→3<br />
√<br />
x + 1 − 2<br />
x(x − 3)<br />
In Problems 45–50, find each one-sided limit using properties of limits.<br />
45. lim<br />
x→3 −(x2 − 4) 46. lim<br />
x→2 +(3x2 + x)<br />
x 2 − 9<br />
47. lim<br />
x→3 − x − 3<br />
x 2 − 9<br />
48. lim<br />
x→3 + x − 3<br />
√<br />
√<br />
49. lim<br />
x→3 −( 9 − x 2 + x) 2 50. lim<br />
x→2 +(2 x 2 − 4 + 3x)<br />
)<br />
In Problems 51–58, use the information below to find each limit.<br />
lim f (x) = 5<br />
x→c<br />
lim g(x) = 2<br />
x→c<br />
lim h(x) = 0<br />
x→c<br />
51. lim x→c<br />
[ f (x) − 3g(x)] 52. lim x→c<br />
[5 f (x)]<br />
53. lim [g(x)] 3 f (x)<br />
54. lim<br />
x→c x→c g(x) − h(x)<br />
55. lim x→c<br />
h(x)<br />
g(x)<br />
[ 1<br />
57. lim x→c g(x)<br />
56. lim x→c<br />
[4 f (x) · g(x)]<br />
] 2<br />
58. lim x→c<br />
3 √ 5g(x) − 3<br />
In Problems 59 and 60, use the graphs of the functions and properties<br />
of limits to find each limit, if it exists. If the limit does not exist, write,<br />
“the limit does not exist,” and explain why.<br />
59. (a) lim [ f (x) + g(x)]<br />
x→4<br />
y<br />
y h(x)<br />
10 y f (x)<br />
(b) lim { f (x) [g(x) − h(x)]}<br />
x→4 8<br />
(c) lim[ f (x) · g(x)]<br />
x→4 6<br />
(4, 6)<br />
(d) lim[2h(x)]<br />
x→4<br />
g(x)<br />
2<br />
(e) lim x→4 f (x)<br />
(4, 0)<br />
(f)<br />
8 x<br />
h(x)<br />
2<br />
lim<br />
y g(x)<br />
x→4<br />
(3, 2)<br />
f (x)<br />
60. (a) lim x→3<br />
{2 [ f (x) + h(x)]}<br />
(b)<br />
lim + h(x)]<br />
x→3−[g(x) (c) lim x→3<br />
3 √ h(x)<br />
(d) lim x→3<br />
f (x)<br />
h(x)<br />
(e) lim x→3<br />
[h(x)] 3<br />
(f)<br />
lim[ f (x) − 2h(x)] 3/2<br />
x→3<br />
y<br />
6<br />
4<br />
2<br />
2<br />
y f (x)<br />
(3, 2)<br />
y g(x)<br />
3<br />
(3, 6)<br />
y h(x)<br />
In Problems 61–66, for each function f, find the limit as x approaches c<br />
of the average rate of change of f from c to x. That is, find<br />
f (x) − f (c)<br />
lim<br />
x→c x − c<br />
61. f (x) = 3x 2 , c = 1 62. f (x) = 8x 3 , c = 2<br />
PAGE<br />
98 63. f (x) =−2x 2 + 4, c = 1 64. f (x) = 20 − 0.8x 2 , c = 3<br />
65. f (x) = √ x, c = 1 66. f (x) = √ 2x, c = 5<br />
In Problems 67–72, find the limit of the difference quotient for each<br />
f (x + h) − f (x)<br />
function f . That is, find lim<br />
.<br />
h→0 h<br />
67. f (x) = 4x − 3 68. f (x) = 3x + 5<br />
69. f (x) = 3x 2 + 4x + 1 70. f (x) = 2x 2 + x<br />
PAGE<br />
99 71. f (x) = 2 x<br />
72. f (x) = 3 x 2<br />
x<br />
36. − 1 4<br />
37. 2<br />
38. − 1 2<br />
39. 2<br />
40. 3<br />
2<br />
41.<br />
4<br />
3<br />
42.<br />
6<br />
1<br />
43.<br />
30<br />
1<br />
44.<br />
12<br />
45. 5<br />
46. 14<br />
47. 6<br />
48. 6<br />
49. 9<br />
50. 6<br />
51. –1<br />
52. 25<br />
53. 8<br />
54. 5 2<br />
55. 0<br />
56. 40<br />
57. 1 4<br />
3<br />
58. 7<br />
59. (a) 6<br />
(b) –16<br />
(c) –16<br />
(d) 0<br />
(e) − 1 4<br />
(f) 0<br />
60. (a) –4<br />
(b) 4<br />
3<br />
(c) − 2<br />
(d) 0<br />
(e) –8<br />
(f) 8<br />
61. 6<br />
62. 96<br />
63. –4<br />
64. –4.8<br />
65. 1 2<br />
1<br />
66.<br />
10<br />
67. 4<br />
68. 3<br />
69. 6x + 4<br />
70. 4x + 1<br />
2<br />
71. −<br />
x<br />
2<br />
6<br />
72. −<br />
x<br />
3<br />
100<br />
Chapter 1 • Limits and Continuity<br />
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Section 1.2 • Assess Your Understanding 101<br />
In Problems 73–80, find lim f (x) and lim f (x) for the<br />
x→c− x→c +<br />
given number c. Based on the results, determine whether lim f (x)<br />
x→c<br />
exists.<br />
2x − 3 if x ≤ 1<br />
PAGE<br />
93 73. f (x) =<br />
3 − x if x > 1<br />
at c = 1<br />
5x + 2 if x < 2<br />
74. f (x) =<br />
1 + 3x if x ≥ 2<br />
at c = 2<br />
⎧<br />
⎨ 3x − 1 if x < 1<br />
75. f (x) = 4 if x = 1 at c = 1<br />
⎩<br />
2x if x > 1<br />
⎧<br />
⎨ 3x − 1 if x < 1<br />
76. f (x) = 2 if x = 1 at c = 1<br />
⎩<br />
2x if x > 1<br />
x − 1 if x < 1<br />
77. f (x) = √<br />
x − 1 if x > 1<br />
at c = 1<br />
<br />
78. f (x) =<br />
9 − x<br />
<br />
2 if 0 < x < 3<br />
at c = 3<br />
x 2 − 9 if x > 3<br />
79.<br />
⎧<br />
⎨ x 2 − 9<br />
if x = 3<br />
f (x) = x − 3<br />
⎩<br />
6 if x = 3<br />
at c = 3<br />
⎧<br />
⎨ x − 2<br />
if x = 2<br />
80. f (x) = x 2 − 4<br />
at c = 2<br />
⎩<br />
1 if x = 2<br />
Applications and Extensions<br />
Heaviside Functions In Problems 81 and 82, find the limit, if it<br />
exists, of the given Heaviside function at c.<br />
0 if t < 1<br />
PAGE<br />
94 81. u 1(t) =<br />
1 if t ≥ 1<br />
at c = 1<br />
0 if t < 3<br />
82. u 3(t) =<br />
1 if t ≥ 3<br />
at c = 3<br />
In Problems 83–92, find each limit.<br />
(x + h) 2 − x 2<br />
83. lim h→0 h<br />
1<br />
85. lim<br />
x + h − 1 x<br />
h→0 h<br />
1 1<br />
87. lim x→0 x 4 + x − 1 <br />
4<br />
√ √ x + h − x<br />
84. lim h→0 h<br />
1<br />
(x + h)<br />
86. lim<br />
3 − 1 x 3<br />
h→0<br />
h<br />
2<br />
88. lim<br />
x→−1 x + 1<br />
x − 7<br />
x − 2<br />
89. lim √ 90. lim √ x→7 x + 2 − 3 x→2 x + 2 − 2<br />
91. lim x→1<br />
x 3 − 3x 2 + 3x − 1<br />
x 2 − 2x + 1<br />
1<br />
3 − 1 <br />
x + 4<br />
x 3 + 7x 2 + 15x + 9<br />
92. lim<br />
x→−3 x 2 + 6x + 9<br />
93. Cost of Water The Jericho Water District determines quarterly<br />
water costs, in dollars, using the following rate schedule:<br />
Water used<br />
(in thousands of gallons) Cost<br />
0 ≤ x ≤ 10 $9.00<br />
10 < x ≤ 30 $9.00 + 0.95 for each thousand<br />
gallons in excess of 10,000 gallons<br />
30 < x ≤ 100 $28.00 + 1.65 for each thousand<br />
gallons in excess of 30,000 gallons<br />
x > 100 $143.50 + 2.20 for each thousand<br />
gallons in excess of 100,000 gallons<br />
Source: Jericho Water District, Syosset, NY.<br />
(a) Find a function C that models the quarterly cost, in dollars, of<br />
using x thousand gallons of water.<br />
(b) What is the domain of the function C?<br />
(c) Find each of the following limits. If the limit does not exist,<br />
explain why.<br />
lim C(x) lim C(x)<br />
x→5 x→10<br />
(d) What is lim C(x)?<br />
x→0 +<br />
(e) Graph the function C.<br />
lim C(x)<br />
x→30<br />
lim C(x)<br />
x→100<br />
94. Cost of Electricity In June 2016, Florida Power and Light had<br />
the following monthly rate schedule for electric usage in<br />
single-family residences:<br />
Monthly customer charge $7.87<br />
Fuel charge<br />
≤ 1000 kWH $0.02173 per kWH<br />
> 1000 kWH $21.73 + 0.03173 for each kWH<br />
in excess of 1000<br />
Source: Florida Power and Light, Miami, FL.<br />
(a) Find a function C that models the monthly cost, in dollars, of<br />
using x kWH of electricity.<br />
(b) What is the domain of the function C?<br />
(c) Find lim C(x), if it exists. If the limit does not exist,<br />
x→1000<br />
explain why.<br />
(d) What is lim C(x)?<br />
x→0 +<br />
(e) Graph the function C.<br />
95. Low-Temperature Physics In thermodynamics, the average<br />
molecular kinetic energy (energy of motion) of a gas having<br />
molecules of mass m is directly proportional to its temperature T<br />
on the absolute (or Kelvin) scale. This can be expressed as<br />
1<br />
2 mv2 = 3 kT, where v = v(T ) is the speed of a typical molecule<br />
2<br />
at time t and k is a constant, known as the Boltzmann constant.<br />
(a) What limit does the molecular speed v approach as the gas<br />
temperature T approaches absolute zero (0 K or −273 ◦ C<br />
or −469 ◦ F)?<br />
(b) What does this limit suggest about the behavior of a gas as<br />
its temperature approaches absolute zero?<br />
88. 2 9<br />
89. 6<br />
90. 4<br />
91. 0<br />
92. –2<br />
93. (a)<br />
⎧<br />
⎪<br />
⎪<br />
Cx ( ) = ⎨<br />
⎪<br />
⎪<br />
⎩<br />
9.00 0≤x<br />
≤10<br />
9.00 + 0.95( x− 10) 10 < x ≤30<br />
28.00 + 1.65( x− 30) 30 < x ≤100<br />
143.50 + 2.20( x− 100) x > 100<br />
(b) { xx≥<br />
0}<br />
(c)<br />
lim Cx ( ) = 9.00, lim Cx ( ) = 9.00,<br />
x→5<br />
lim Cx ( ) = 28.00,<br />
x→30<br />
(d) 9.00<br />
(e)<br />
94. (a)<br />
Quarterly water cost<br />
(dollars)<br />
y<br />
250<br />
200<br />
150<br />
100<br />
50<br />
x→10<br />
lim Cx ( ) = 143.50<br />
x→100<br />
50 100 150 x<br />
Water usage (thousands of gallons)<br />
⎪<br />
x<br />
( ) = ⎨<br />
( x )<br />
C x<br />
⎧ 7.87 + 0.02173 if 0 ≤x<br />
≤ 1000<br />
+ − ><br />
⎩<br />
⎪<br />
29.60 0.3173 1000 if x 1000<br />
(b) { xx≥<br />
0}<br />
(c) lim Cx ( ) = 8.9565;<br />
−<br />
x→50<br />
lim Cx ( ) = 8.9565;<br />
+<br />
x→<br />
50<br />
lim Cx ( ) = 8.9565<br />
x→50<br />
(d) lim Cx ( ) = 7.87<br />
x→ 0<br />
+<br />
73. lim fx ( ) =−1,<br />
lim fx ( ) = 2,<br />
x→1<br />
−<br />
not exist.<br />
74. lim fx ( ) = 12,<br />
x→2<br />
−<br />
not exist.<br />
75. lim fx ( ) = 2,<br />
x→1<br />
−<br />
76. lim fx ( ) = 2,<br />
x→1<br />
−<br />
77. lim fx ( ) = 0,<br />
x→1<br />
−<br />
78. lim fx ( ) = 0,<br />
x→3<br />
−<br />
x→ 1<br />
+<br />
lim fx ( ) = 7,<br />
x→ 2<br />
+<br />
lim fx ( ) does<br />
x→1<br />
lim fx ( ) does<br />
x→ 2<br />
lim fx ( ) = 2, lim fx ( ) = 2<br />
x→ 1<br />
+<br />
x→1<br />
lim fx ( ) = 2, lim fx ( ) = 2<br />
x→ 1<br />
+<br />
x→1<br />
lim fx ( ) = 0, lim fx ( ) = 0<br />
x→ 1<br />
+<br />
x→1<br />
lim fx ( ) = 0, lim fx ( ) = 0<br />
x→ 3<br />
+<br />
x→3<br />
79. lim fx ( ) = 6,<br />
x→3<br />
−<br />
lim fx ( ) = 6,<br />
x→ 3<br />
+<br />
1<br />
80. lim fx ( ) =<br />
4 , lim fx ( )<br />
x→2<br />
−<br />
x→ 2<br />
+<br />
81. Limit does not exist.<br />
82. Limit does not exist.<br />
83. 2x<br />
1<br />
84.<br />
2 x<br />
1<br />
85. −<br />
x<br />
2<br />
86. −<br />
x<br />
3<br />
4<br />
87. − 1<br />
16<br />
=<br />
1<br />
4 ,<br />
lim fx ( ) = 6<br />
x→3<br />
lim fx ( ) =<br />
x→2<br />
1<br />
4<br />
(e)<br />
C<br />
50<br />
40<br />
30<br />
20<br />
10<br />
95. (a) 0<br />
1 2 3 4 5 6 7 8<br />
(b) As the temperature of a gas<br />
approaches zero, the molecules in<br />
the gas stop moving.<br />
x<br />
Section 1.2 • Assess Your Understanding<br />
101<br />
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<strong>Sullivan</strong><br />
88 102 Chapter 1 • Limits and Continuity<br />
96. (a) 3<br />
(b) –1<br />
(c) Limit does not exist.<br />
97. lim x = 0 since lim x = 0 and<br />
x→0<br />
x→0<br />
−<br />
lim = 0.<br />
x<br />
x→ 0<br />
+<br />
98. lim x = lim<br />
2<br />
x =<br />
2<br />
lim x = 0<br />
x→0<br />
x→0<br />
x→0<br />
= 0<br />
99. Answers will vary. Sample answer:<br />
⎧⎪<br />
1, if x < 2<br />
f( x)=<br />
⎨<br />
⎩⎪ 0, if x ≥ 2 ,<br />
⎧⎪<br />
1, if x < 2<br />
gx ( )= ⎨<br />
⎩⎪ 0, if x ≥ 2<br />
100. Answers will vary. Sample answer:<br />
⎧⎪<br />
fx ( ) = ⎨<br />
⎩⎪<br />
⎧⎪<br />
gx ( ) = ⎨<br />
⎩⎪<br />
0, if x < 2<br />
1, if x ≥ 2 ,<br />
1, if x < 2<br />
0, if x ≥ 2<br />
101. Answers will vary. Sample answer<br />
⎧⎪<br />
fx ( ) = ⎨<br />
⎩⎪<br />
⎧⎪<br />
gx ( ) = ⎨<br />
⎩⎪<br />
2, if x < 1<br />
−1, if x ≥1 ,<br />
1, if x < 1<br />
−1, if x ≥1<br />
102. Answers will vary. Sample answer:<br />
⎧⎪<br />
1, if x < 0<br />
f( x)=<br />
⎨<br />
⎩⎪ −1, if x ≥0<br />
103. See TSM.<br />
104. See TSM.<br />
−<br />
105. na n 1<br />
106. If n is an even positive integer, then<br />
limit does not exist. If n is an odd<br />
positive integer, then the limit equals<br />
−<br />
na n 1 .<br />
107. m n<br />
108. 1 3<br />
a+<br />
b<br />
109.<br />
2<br />
110.<br />
a1+ a2+ +<br />
a<br />
2<br />
111. 0<br />
n<br />
2x 2 if x < 3x 1 + 5 if x ≤ 2<br />
96. 33. For f (x) the = function<br />
3x 2 f (x) = at c = 1<br />
− 1 if x > 13 1 − x if x > 2 , find<br />
103.<br />
x 3 if x < −1<br />
34. f (x) = f (2<br />
x 2 + h) − f (2) at c =−1<br />
(a) lim − 1 if x > −1<br />
h→0 − h<br />
x 2 if x ≤ 0<br />
35. f (x) =<br />
f (2 + h) − f (2)<br />
(b) lim<br />
at c = 0<br />
h→0 + 2x + 1 h if x > 0<br />
⎧<br />
⎨ xf 2 (2 + h) if−x < f (2) 1<br />
(c) Does lim<br />
exist?<br />
36. f (x) = h→0 2 hif x = 1 at c = 1<br />
⎩ <br />
−3x + 2 if x > 1 if x ≥ 0<br />
97. Use the fact that |x| =<br />
to show that lim |x| =0.<br />
−x if x < 0 x→0<br />
Applications 98. Use the fact and that Extensions |x| = √ x 2 to show that lim |x| =0.<br />
x→0<br />
In Problems 37–40, sketch a graph of a function with the given<br />
99. Find functions f and g for which lim [ f (x) + g(x)] may exist<br />
properties. Answers will vary.<br />
x→c<br />
37. even lim f though (x) = 3; lim f lim (x) and f (x) lim= g(x) 3; do lim notf exist. (x) = 1;<br />
x→2 x→c x→3− x→c x→3 +<br />
100. f Find (2) = functions 3; f (3) f and = 1g for which lim [ f (x)g(x)] may exist even<br />
x→c<br />
38. though lim f (x) lim= f 0; (x) and lim limf g(x) =−2; do not exist. lim f (x) =−2;<br />
x→−1 x→c x→2 x→c − x→2 +<br />
<br />
f (−1) is not defined; f (2) =−2 f (x)<br />
101. Find functions f and g for which lim<br />
39. lim f (x) = 4; lim f (x) =−1; lim<br />
may exist even<br />
x→c g(x) f (x) = 0;<br />
x→1 x→0− x→0 +<br />
though lim f (x) and lim g(x) do not exist.<br />
f (0) =−1; x→c f (1) = x→c 2<br />
102.<br />
40. lim<br />
Find<br />
f<br />
a<br />
(x)<br />
function<br />
= 2;<br />
f<br />
lim<br />
for which<br />
f (x) =<br />
lim<br />
0;<br />
| f (x)| lim may<br />
f (x)<br />
exist<br />
= 1;<br />
even though<br />
x→c x→2 lim f (x) does not x→−1 exist.<br />
x→1<br />
f x→c (−1) = 1; f (2) = 3<br />
In Problems 41–50, use either a graph or a table to investigate<br />
each limit.<br />
|x − 5|<br />
|x − 5|<br />
41. AP® lim Practice Problems 42. lim<br />
43. lim<br />
x→5 + x − 5<br />
x→5 − x − 5<br />
<br />
x→ 12<br />
2x<br />
−<br />
PAGE<br />
PAGE<br />
93 1. Consider the piecewise-defined function f given by<br />
PAGE<br />
44. lim <br />
x→ 12<br />
2x ⎧45. lim<br />
+ <br />
x→ 23<br />
2x 46. lim<br />
− <br />
x→ 23<br />
2x<br />
⎨ −x − 2 if x < −1<br />
+<br />
f (x) = x<br />
2 if −1 ≤ x < 2<br />
⎩ <br />
47. lim |x|−x 48. −4x + lim 12 |x|−x if x ≥ 2<br />
x→2 + x→2 −<br />
Investigate the limits below and decide which limit does NOT<br />
3 3<br />
49. lim x−x 50. lim x−x<br />
x→2 + x→2 − PAGE<br />
exist.<br />
(A) lim f (x) (B) lim<br />
51. Slope of a Tangent Line For f (x) f (x)<br />
x→−1 + x→2 − = 3x 2 :<br />
(a) (C) Find limthe f (x) slope of the (D) secant lim line f (x)<br />
x→2 x→−1 containing the points (2, 12)<br />
PAGE<br />
and (3, 27).<br />
PAGE<br />
(5 − t) 2<br />
97 2. (b) limFind the slope of the secant line containing the points (2, 12)<br />
t→5<br />
and<br />
t −<br />
(x,<br />
5 =<br />
f (x)), x = 2.<br />
(c) (A) Create −5 a table (B) 0 to investigate (C) 1 the(D) slope 5 of the tangent line to the PAGE<br />
graph of f at 2 using the result from (b).<br />
PAGE<br />
f (x) − f (2)<br />
98 3. Find lim<br />
for the function f (x) = 3x<br />
(d) On the same set of axes, graph f , the tangent line 3 − 4.<br />
x→2 x − 2<br />
to the graph<br />
of f at the point (2, 12), and the secant line from (a).<br />
(A) 0 (B) 12 (C) 24 (D) 36<br />
52. Slope of a Tangent Line For f (x) = x 3 :<br />
PAGE<br />
x − s<br />
98 4. lim √ √ =<br />
(a) x→s<br />
Find x −<br />
the slope s<br />
of the secant line containing the points (2, 8)<br />
(A) and 2s (3, 27). (B) 2 √ √<br />
s (C) 2s (D) s<br />
(b) Find the slope of the secant line containing the points (2, 8)<br />
and (x, f (x)), x = 2.<br />
(c) Create a table to investigate the slope of the tangent line to the<br />
graph of f at 2 using the result from (b).<br />
(d) On the same set of axes, graph f , the tangent line to the graph<br />
of f at the point (2, 8), and the secant line from (a).<br />
Answers to AP® Practice Problems<br />
1. D<br />
2. B<br />
3. D<br />
4. B<br />
5. C<br />
6. C<br />
7. C<br />
8. B<br />
Prove that if g is a function for which lim<br />
53. Slope of a Tangent Line For f (x) = 1 g(x) exists and<br />
x→c<br />
2 x2 − 1:<br />
if k is any real number, then lim[kg(x)] exists and<br />
(a) Find the slope m sec of the x→c<br />
secant line containing the<br />
lim[kg(x)] = k lim<br />
points P = (2, f g(x).<br />
x→c x→c (2)) and Q = (2 + h, f (2 + h)).<br />
104. Prove that if the number c is in the domain of a rational<br />
(b) Use the result from (a) to complete the following table:<br />
function R(x) = p(x) , then lim R(x) = R(c).<br />
q(x) x→c<br />
h −0.5 −0.1 −0.001 0.001 0.1 0.5<br />
m sec<br />
Challenge Problems<br />
(c) Investigate x n − the a n<br />
105. Find lim , limit n a positive of the slope integer. of the secant line found in (a)<br />
as x→a h →x 0. − a<br />
(d) What isx the n + slope a n of the tangent line to the graph of f at the<br />
106. Find lim , n a positive integer.<br />
point x→−aP = x + (2, a f (2))?<br />
(e) On thex m same − 1set of axes, graph f and the tangent line to f at<br />
107. Find<br />
P<br />
lim x→1 = (2, x n f (2)).<br />
, m, n positive integers.<br />
√ − 1<br />
3<br />
1 + x − 1<br />
108. 54. Slope Find lim of a Tangent Line . For f (x) = x 2 − 1:<br />
x→0 x<br />
<br />
(a) Find the (1<br />
slope + ax)(1<br />
m sec of +<br />
the bx)<br />
secant − 1<br />
line containing the<br />
109. Findpoints lim P = (−1, f (−1)) and Q . = (−1 + h, f (−1 + h)).<br />
x→0 x<br />
(b) Use the result from (a) to complete the following table:<br />
(1 + a1x)(1 + a 2x) ···(1 + a nx) − 1<br />
110. Find lim<br />
.<br />
x→0 x<br />
h −0.1<br />
f (h) −<br />
−0.01<br />
f (0)<br />
−0.001 −0.0001 0.0001 0.001 0.01 0.1<br />
111. Findm lim h→0 sec<br />
if f (x) = x|x|.<br />
h<br />
(c) Investigate the limit of the slope of the secant line found<br />
in (a) as h → 0.<br />
(d) What is the slope of the tangent line to the graph of f at the<br />
point P = (−1, f (−1))?<br />
(e) On the same set of axes, graph f and the tangent line to f<br />
at P = (−1, f (−1)).<br />
ax<br />
93 5. For g(x) =<br />
2 − 5 if x < 2<br />
ax + b if x<br />
85 55. (a) Investigate lim cos π > 2 ,<br />
by using a table and evaluating the<br />
find values for a and x→0 b so that x<br />
function f (x) = cos π lim g(x) = 7.<br />
x→2<br />
(A) a = 1, b x at<br />
x =− 1 = 5<br />
2 , − 1 (B)<br />
4 , − 1 a<br />
8 , − 1 = 2, b<br />
10 , − 1 = 3<br />
(C) a = 3, b = 1 (D) a = 6, b =−5<br />
12 ,..., 1<br />
12 , 1 10 , 1 8 , 1 4 , 1 2 .<br />
<br />
95 6. lim<br />
x→4<br />
(b) +(5 x 2 − 16 + 3x) =<br />
Investigate lim cos π by using a table and evaluating the<br />
(A) −12 (B) x→0<br />
0 (C) x<br />
function f (x) = cos π 12 (D) The limit does not exist.<br />
<br />
[ f (x)] x at<br />
2<br />
x =−1, − 1 − 8x<br />
3 , − 1 + 3<br />
95 7. If lim<br />
5 , − 1 =<br />
7 , − 9, 1 then<br />
9 ,..., lim 1 9 , f 1 (x)<br />
7 , 1 =<br />
5 , 1 x→2 x + 1<br />
x→2<br />
√ √ 3 , 1.<br />
(A) 22 (B) 2 10 (C) 16 (D) 256<br />
(c) Compare the results from (a) and (b). What do you conclude<br />
95 8. lim[x −1/2 about (5xthe −limit? 7) 1/3 ] Why =<br />
x→3<br />
do you think this happens? What is<br />
your view about using a table to draw a conclusion about<br />
(A) 3 −1/2 2<br />
8<br />
limits? (B)<br />
3 1/2 (C)<br />
3 1/2 (D) 6 −1/2<br />
(d) Use technology to graph f . Begin with the x-window<br />
[−2π, 2π] and the y-window [−1, 1]. If you were finding<br />
lim f (x) using a graph, what would you conclude? Zoom in<br />
x→0<br />
on the graph. Describe what you see. (Hint: Be sure your<br />
calculator is set to the radian mode.)<br />
56. (a) Investigate lim cos π by using a table and evaluating the<br />
x→0 x2 function f (x) = cos π at x =−0.1, −0.01, −0.001,<br />
x2 −0.0001, 0.0001, 0.001, 0.01, 0.1.<br />
102<br />
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Section 1.3 • Continuity 103<br />
y<br />
c<br />
y f (x)<br />
(c, f (c))<br />
x<br />
y<br />
1.3 Continuity<br />
c<br />
(c, f (c))<br />
OBJECTIVES When you finish this section, you should be able to:<br />
1 Determine whether a function is continuous at a number (p. 103)<br />
2 Determine intervals on which a function is continuous (p. 106)<br />
3 Use properties of continuity (p. 108)<br />
4 Use the Intermediate Value Theorem (p. 110)<br />
Sometimes lim f (x) equals f (c) and sometimes it does not. In fact, f (c) may not even<br />
x→c<br />
be defined and yet lim f (x) may exist. In this section, we investigate the relationship<br />
x→c<br />
between lim f (x) and f (c). Figure 21 shows some possibilities.<br />
x→c<br />
y f (x)<br />
x<br />
y<br />
c<br />
y f (x)<br />
(b) lim f(x) lim f(x) f (c) (c) lim f (x) lim f(x)<br />
(a) lim f(x) lim f(x) f (c)<br />
x→c x→c x→c x→c x→c x→c<br />
f(c) is not defined.<br />
Figure 21<br />
x<br />
y<br />
c<br />
(c, f (c))<br />
y f (x)<br />
(d) lim f (x) lim f (x)<br />
x→c x→c<br />
f (c) is defined.<br />
x<br />
y<br />
c<br />
y f (x)<br />
(e) lim f (x) lim f (x)<br />
x→c x→c<br />
f (c) is not defined.<br />
Of these five graphs, the “nicest” one is Figure 21(a). There, lim<br />
x→c<br />
f (x) exists and is<br />
equal to f (c). Functions that have this property are said to be continuous at the number c.<br />
This agrees with the intuitive notion that a function is continuous if its graph can be drawn<br />
without lifting the pencil. The functions in Figures 21(b)–(e) are not continuous at c, since<br />
each has a break in the graph at c. This leads to the definition of continuity at a number.<br />
DEFINITION Continuity at a Number<br />
A function f is continuous at a number c if the following three conditions are met:<br />
• f (c) is defined (that is, c is in the domain of f )<br />
• lim f (x) exists<br />
x→c<br />
• lim f (x) = f (c)<br />
x→c<br />
If any one of these three conditions is not satisfied, then the function is discontinuous<br />
at c.<br />
NOW WORK AP® Practice Problems 1 and 2.<br />
1 Determine Whether a Function Is Continuous at a Number<br />
EXAMPLE 1<br />
Teaching Tip<br />
Another advantage to the left–right–center<br />
definition of continuity is that it will help students to<br />
identify the type of discontinuity a function exhibits.<br />
If lim fx ( ) ≠ lim fx ( ), then the function jumps<br />
− +<br />
x→c x→c<br />
from one value to another and therefore is<br />
classified as a jump discontinuity.<br />
If lim fx ( ) = lim fx ( ), then the function does not<br />
− +<br />
x→c x→c<br />
jump. That doesn’t mean it is continuous, though.<br />
We still have to determine if the left- and righthand<br />
limits equal fc (). If they do not, there is a tiny<br />
hole. This is called a removable discontinuity.<br />
Determining Whether a Function Is Continuous<br />
at a Number<br />
(a) Determine whether f (x) = 3x 2 − 5x + 4 is continuous at 1.<br />
(b) Determine whether g(x) = x 2 + 9<br />
is continuous at 2.<br />
x 2 − 4<br />
Solution (a) We begin by checking the conditions for continuity. First, 1 is in the<br />
domain of f and f (1) = 2. Second, lim f (x) = lim(3x 2 − 5x + 4) = 2, so lim f (x)<br />
x→1 x→1 x→1<br />
exists. Third, lim f (x) = f (1). Since the three conditions are met, f is continuous at 1.<br />
x→1<br />
(b) Since 2 is not in the domain of g, the function g is discontinuous at 2. ■<br />
x<br />
TRM Alternate Examples<br />
Section 1.3<br />
You can find the Alternate Examples for<br />
this section in PDF format in the Teacher’s<br />
Resource Materials.<br />
TRM AP® Calc Skill Builders<br />
Section 1.3<br />
You can find the AP ® Calc Skill Builders for<br />
this section in PDF format in the Teacher’s<br />
Resource Materials.<br />
Teaching Tip<br />
The definition of continuity can also be<br />
stated this way:<br />
A function is continuous at a number c if<br />
lim fx ( ) = lim fx ( ) = fc ().<br />
− +<br />
x→c x→c<br />
Students may remember this set of 3<br />
checks because they examine the limit<br />
from the left, the limit from the right, and<br />
then the center, the value at c.<br />
The three checks can become three<br />
2<br />
questions. For instance, for lim x<br />
x→3<br />
1. What is the limit of f (x) as x<br />
approaches 3 from the right?<br />
2. What is the limit of f (x) as x<br />
approaches 3 from the left?<br />
3. What is the value of f(x) at x = 3?<br />
Since these three values are 9, then f(x) =<br />
x 2 is continuous at x = 3.<br />
In contrast, consider the following function<br />
at c = 3:<br />
⎛<br />
⎜<br />
fx ( ) = ⎜<br />
⎜<br />
⎝<br />
2<br />
x , x < 3<br />
4, x = 3<br />
x+ 6, x > 3<br />
1. lim x 2 as x approaches 3 from the<br />
left = 9<br />
2. lim x + 6 as x approaches 3 from the<br />
right = 9<br />
3. f (3) = 4<br />
Since these three values are not the same,<br />
then f is not continuous.<br />
Teaching Tip<br />
Remind the students that if<br />
lim fx ( ) = lim fx ( ) then lim fx ( ) exists.<br />
− +<br />
x→c x→c<br />
x→c<br />
If it helps their understanding, consider<br />
using this phrase: If the limit of a function<br />
from the left equals the limit of the<br />
function from the right, then the limit in<br />
general exists.<br />
Section 1.3 • Continuity<br />
103<br />
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104 Chapter 1 • Limits and Continuity<br />
AP® CaLC skill builder<br />
for example 2<br />
Determining Whether a Function Is<br />
Continuous at a Number<br />
3<br />
x −25x<br />
Determine whether f( x)<br />
= is<br />
2<br />
continuous at x = 5. x −5x<br />
Solution<br />
We first check the conditions for continuity.<br />
To determine if 5 is in the domain of f,<br />
compute<br />
3<br />
5 −25(5)<br />
f (5) =<br />
2<br />
5 −5(5)<br />
This shows that f (5) is not defined,<br />
because both numerator and denominator<br />
are 0. Since 5 is not in the domain of f,<br />
then f is not continuous at 5.<br />
AP® CaLC skill builder<br />
for example 2<br />
Determining Whether a Function Is<br />
Continuous at a Number<br />
4<br />
( x −1)<br />
Use the graph of f( x)<br />
= to<br />
1−<br />
x<br />
determine whether f is continuous at<br />
x = 1. If f is discontinuous, what type of<br />
discontinuity does the function have?<br />
y<br />
y<br />
4<br />
2<br />
x<br />
f(x) 2 2<br />
x 2 4<br />
4 2 2 4<br />
2<br />
(0, <br />
4 )<br />
2<br />
4<br />
Figure 23 f is continuous at 0;<br />
f is discontinuous at −2 and 2.<br />
x<br />
Figure 22 shows the graphs of f and g from Example 1. Notice that f is continuous<br />
at 1, and its graph is drawn without lifting the pencil. But the function g is discontinuous<br />
at 2, and to draw its graph, you must lift your pencil at x = 2.<br />
10<br />
f (x) 3x 2 5x 4<br />
4<br />
Figure 22<br />
2<br />
y<br />
20<br />
15<br />
5<br />
(1, 2)<br />
2<br />
4<br />
x<br />
(a) f is continuous at 1. (b) g is discontinuous at 2.<br />
4<br />
2<br />
y<br />
10<br />
5<br />
5<br />
2<br />
g(x) x2 9<br />
x 2 4<br />
NOW WORK Problem 19 and AP® Practice Problem 3.<br />
Determining Whether a Function Is Continuous<br />
EXAMPLE 2 at a Number<br />
<br />
x<br />
2<br />
+ 2<br />
Determine if f (x) = is continuous at the numbers −2, 0, and 2.<br />
x 2 − 4<br />
Solution The domain of f is {x|x = −2, x = 2}. Since f is not defined at −2 and 2,<br />
the function f is not continuous at −2 and at 2. The number 0 is in the domain of f.<br />
√<br />
2<br />
That is, f is defined at 0, and f (0) =−<br />
4 . Also,<br />
<br />
x<br />
2<br />
+ 2<br />
lim f (x) = lim<br />
x→0 x→0 x 2 − 4<br />
<br />
lim x 2<br />
+ 2 lim<br />
= x→0<br />
lim (x 2 − 4) = lim<br />
x→0<br />
x→0<br />
(x 2 + 2)<br />
x→0 x 2 − lim<br />
x→0<br />
4<br />
√ √<br />
0 + 2 2<br />
=<br />
0 − 4 =− 4 = f (0)<br />
The three conditions of continuity at a number are met. So, the function f is continuous<br />
at 0. ■<br />
Figure 23 shows the graph of f.<br />
NOW WORK Problem 21.<br />
4<br />
x<br />
4<br />
24<br />
2<br />
22<br />
22<br />
24<br />
2<br />
4<br />
x<br />
CALC<br />
CLIP<br />
EXAMPLE 3<br />
Determining Whether a Piecewise-Defined Function<br />
Is Continuous<br />
Determine whether the function<br />
⎧<br />
x 2 − 9<br />
if x < 3<br />
⎪⎨ x − 3<br />
f (x) =<br />
9 if x = 3<br />
⎪⎩<br />
x 2 − 3 if x > 3<br />
is continuous at 3.<br />
Solution<br />
The graph of f reveals a hole at<br />
x = 1 because f (1) is undefined. We can<br />
conclude that f is discontinuous at x = 1.<br />
Since lim − fx ( ) = lim + fx ( ) =−4,<br />
x→1 x→1<br />
the function has a removable discontinuity<br />
at x = 1. If we define f (1) = − 4, then f is a<br />
continuous function.<br />
TRM Section 1.3: Worksheet 1<br />
This worksheet contains 7 functions along<br />
with their graphs. Students are asked to<br />
identify the point(s) of discontinuity as<br />
well as to state why the listed points are<br />
discontinuous.<br />
Calculator Tip<br />
Some students may be tempted to graph functions<br />
on their calculator to make decisions about<br />
continuity. Most students, though, use a default<br />
calculator setting in which holes cannot be seen<br />
on the graph that is made on the calculator screen.<br />
This might lead the students to (falsely) believe<br />
a function is continuous when it is not. Students<br />
mainly should determine points of discontinuity<br />
analytically and then use a graph to support their<br />
conclusion.<br />
104<br />
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Section 1.3 • Continuity 105<br />
2<br />
y<br />
20<br />
10<br />
(3, 9)<br />
2 4<br />
x 2 9<br />
if x 3<br />
x 3<br />
f (x) <br />
9 if x 3<br />
x 2 3 if x 3<br />
Figure 24 f is discontinuous at 3.<br />
IN WORDS Visually, the graph of a<br />
function f with a removable<br />
discontinuity has a hole at (c, lim x→c<br />
f (x)).<br />
The discontinuity is removable because<br />
we can fill in the hole by appropriately<br />
defining f . The function will then be<br />
continuous at x = c.<br />
NEED TO REVIEW? The floor function<br />
is discussed in Section P.2, p. 19.<br />
y<br />
2<br />
2<br />
2<br />
Figure 25 f (x) =x<br />
2<br />
4<br />
x<br />
x<br />
Solution Since f (3) = 9, the function f is defined at 3. To check the second condition,<br />
we investigate the one-sided limits.<br />
Teaching Tip<br />
x 2 − 9<br />
lim f (x) = lim<br />
x→3 −<br />
x→3 − x − 3 = lim (x − 3)(x + 3)<br />
= lim<br />
x→3 − x − 3 x→3−(x + 3) = 6<br />
↑<br />
Divide out x − 3<br />
lim f (x) = lim<br />
x→3 +<br />
x→3 +(x 2 − 3) = 9 − 3 = 6<br />
3<br />
Since lim f (x) = lim f (x), then lim f (x) exists. But, lim f (x) = 6 and f (3) = 9,<br />
x −25x<br />
x→3 −<br />
x→3 +<br />
x→3 x→3 fx ( ) =<br />
2<br />
so the third condition of continuity is not satisfied. The function f is discontinuous<br />
x −5x<br />
at 3. ■<br />
Figure 24 shows the graph of f .<br />
NOW WORK Problem 25 and AP® Practice Problems 5, 8, and 11.<br />
fx<br />
The discontinuity at c = 3 in Example 3 is called a removable discontinuity because<br />
we can redefine f at the number c to equal lim f (x) and make f continuous at c. So, in<br />
x→c fx<br />
Example 3, if f (3) is redefined to be 6, then f would be continuous at 3.<br />
DEFINITION Removable Discontinuity<br />
Let f be a function that is defined everywhere in an open interval containing c, except<br />
fx<br />
possibly at c. The number c is called a removable discontinuity of f if the function<br />
is discontinuous at c but lim f (x) exists. The discontinuity is removed by defining (or<br />
x→c<br />
redefining) the value of f at c to be lim f (x).<br />
x→c fx<br />
NOW WORK Problems 13 and 35 and AP® Practice Problem 4.<br />
Determining Whether a Function Is Continuous<br />
EXAMPLE 4 at a Number<br />
Determine whether the floor function f (x) =x is continuous at 1.<br />
Solution The floor function f (x) =x =the greatest integer ≤ x. The floor function f<br />
is defined at 1 and f (1) = 1. But<br />
lim f (x) = lim<br />
x→1 −<br />
x→1−x =0 and lim f (x) = lim<br />
x→1 +<br />
x→1 +x<br />
=1<br />
So, limx does not exist. Since limx does not exist, f is discontinuous at 1. ■<br />
x→1 x→1<br />
Figure 25 illustrates that the floor function is discontinuous at each integer. Also,<br />
none of the discontinuities of the floor function is removable. Since at each integer the<br />
value of the floor function “jumps” to the next integer, without taking on any intermediate<br />
values, the discontinuity at integer values is called a jump discontinuity.<br />
evaluate fc<br />
NOW WORK Problem 53.<br />
We have defined what it means for a function f to be continuous at a number. Now<br />
we define one-sided continuity at a number.<br />
Teaching Tip<br />
DEFINITION One-Sided Continuity at a Number<br />
Let f be a function defined on the interval (a, c]. Then f is continuous from the left<br />
at the number c if<br />
lim f (x) = f (c)<br />
x→c −<br />
Let f be a function defined on the interval [c, b). Then f is continuous from the right<br />
at the number c if<br />
lim f (x) = f (c)<br />
x→c +<br />
TRM Section 1.3: Worksheet 2<br />
This worksheet contains 5 piecewise functions.<br />
The students are asked to determine if the<br />
function is continuous at various values of each<br />
function.<br />
Continuity can also be taught using<br />
graphical and analytic methods. Students<br />
may be surprised to see that the function<br />
looks like a line.<br />
Notice the algebraic simplification:<br />
3<br />
x −25x<br />
( ) =<br />
2<br />
x −5x<br />
2<br />
xx ( −25)<br />
( ) =<br />
xx ( −5)<br />
( x− 5)( x+<br />
5)<br />
( ) =<br />
( x −5)<br />
( ) = x + 5<br />
The x’s and the x – 5’s were removed from<br />
the equation. The values, 0 and 5, were<br />
not in the domain of the original function.<br />
Since both were removed through algebraic<br />
simplification, they are both removable<br />
discontinuities. This function has no other<br />
points of discontinuity.<br />
Common Error<br />
Remind the students that when they<br />
(), they must substitute the<br />
value c into the original function, not the<br />
simplified version of the original function.<br />
If you present the left–right–center check<br />
for continuity at a point, consider having<br />
the students evaluate f() c first. If f()<br />
c<br />
is undefined, the function cannot be<br />
continuous at that point.<br />
Teaching Tip<br />
If a student is given a function and asked to<br />
identify all points of discontinuity, have the<br />
student begin by identifying the domain of<br />
the function. Points not in the domain of the<br />
function are points of discontinuity.<br />
AP® Exam Tip<br />
Although questions about continuity<br />
seldom appear directly on the multiplechoice<br />
portion of the exam, it is<br />
common to see continuity as part of a<br />
multiple-choice question. For example,<br />
a principle of continuity may appear as<br />
one of the answer choices. Also, in a<br />
problem about continuity, it is common<br />
to be given a piecewise function.<br />
Section 1.3 • Continuity 105<br />
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106 Chapter 1 • Limits and Continuity<br />
AP® CaLC skill builder<br />
for example 5<br />
Determining Whether a Function Is<br />
Continuous on a Closed Interval<br />
⎧<br />
⎪<br />
fx ( ) = ⎨<br />
⎪<br />
⎩<br />
2<br />
x −3x−4<br />
x − 4<br />
k<br />
x ≠ 4<br />
x = 4<br />
Determine the value of k that makes the<br />
function f defined above continuous at<br />
x = 4.<br />
Solution<br />
The function will be continuous at x = 4 if<br />
lim fx ( ) = lim fx ( ) = f (4).<br />
− +<br />
x→4 x→4<br />
In Example 4, we showed that the floor function f (x) =x is discontinuous at<br />
x = 1. But since<br />
f (1) =1 =1 and lim f (x) =x =1<br />
x→1 +<br />
the floor function is continuous from the right at 1. In fact, the floor function is<br />
discontinuous at each integer n, but it is continuous from the right at every integer n.<br />
(Do you see why?)<br />
2 Determine Intervals on Which a Function Is Continuous<br />
So far, we have considered only continuity at a number c. Now, we use one-sided<br />
continuity to define continuity on an interval.<br />
DEFINITION Continuity on an Interval<br />
• A function f is continuous on an open interval (a, b) if f is continuous at every<br />
number in (a, b).<br />
• A function f is continuous on an interval [a, b) if f is continuous on the open<br />
interval (a, b) and continuous from the right at the number a.<br />
• A function f is continuous on an interval (a, b] if f is continuous on the open<br />
interval (a, b) and continuous from the left at the number b.<br />
• A function f is continuous on a closed interval [a, b] if f is continuous on the<br />
open interval (a, b), continuous from the right at a, and continuous from the left<br />
at b.<br />
Figure 26 gives examples of graphs over different types of intervals.<br />
x<br />
lim fx ( ) = lim<br />
−<br />
−<br />
x→4 x→4<br />
−3x−4<br />
x −4<br />
( x+ 1)( x−4)<br />
= lim<br />
−<br />
x→4<br />
( x −4)<br />
= lim ( x + 1)<br />
−<br />
x→4<br />
2<br />
y<br />
f(a)<br />
y f (x)<br />
a<br />
b x<br />
lim f(x) f (a)<br />
x→a <br />
(a) f is continuous on [a, b).<br />
y<br />
f(a)<br />
y f (x)<br />
a<br />
b x<br />
lim f(x) f (a)<br />
x→a <br />
(b) f is continuous on (a, b).<br />
y<br />
f(b)<br />
y f (x)<br />
a<br />
b x<br />
lim f(x) f (b)<br />
x→b <br />
(c) f is continuous on (a, b].<br />
y<br />
y f (x)<br />
f(b)<br />
a<br />
b x<br />
lim f(x) f (b)<br />
x→b <br />
(d) f is continuous on (a, b).<br />
y<br />
f(b)<br />
y f (x)<br />
f(a)<br />
a<br />
b x<br />
lim f (x) f (a)<br />
x→a <br />
lim f (x) f (b)<br />
x→b <br />
(e) f is continuous on [a, b].<br />
= 5<br />
Figure 26<br />
x<br />
lim fx ( ) = lim<br />
+ +<br />
x→4 x→4<br />
−3x−4<br />
x −4<br />
( x+ 1)( x−4)<br />
= lim<br />
+<br />
x→4<br />
( x −4)<br />
= lim ( x + 1)<br />
+<br />
x→4<br />
= 5<br />
If we define f (4) = 5, the function f will be<br />
continuous at x = 4.<br />
2<br />
For example, the graph of the floor function f (x) =x in Figure 25 illustrates<br />
that f is continuous on every interval [n, n + 1), n an integer. In each interval, f is<br />
continuous from the right at the left endpoint n and is continuous at every number in the<br />
open interval (n, n + 1).<br />
EXAMPLE 5<br />
Determining Whether a Function Is Continuous<br />
on a Closed Interval<br />
Is the function f (x) = √ 4 − x 2 continuous on the closed interval [−2, 2]?<br />
Solution The domain of f is {x|−2 ≤ x ≤ 2}. So, f is defined for every number in<br />
the closed interval [−2, 2].<br />
For any number c in the open interval (−2, 2),<br />
lim<br />
x→c<br />
f (x) = lim<br />
x→c<br />
So, f is continuous on the open interval (−2, 2).<br />
√<br />
4 − x<br />
2<br />
= √ lim<br />
x→c<br />
(4 − x 2 ) = √ 4 − c 2 = f (c)<br />
AP® Exam Tip<br />
The topic of continuity on an interval<br />
is generally not directly tested on the<br />
exam. The concept, however, is still<br />
important because many important<br />
theorems that do appear on the exam<br />
are applicable only for functions that<br />
are continuous on a closed interval.<br />
106<br />
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Section 1.3 • Continuity 107<br />
22 21<br />
y<br />
2<br />
1<br />
Figure 27 f (x) = √ 4 − x 2 , −2 ≤ x ≤ 2<br />
y<br />
4<br />
2<br />
(0, 0)<br />
(1, 0)<br />
1<br />
2 4<br />
2<br />
x<br />
x<br />
f (x) x 2 (x 1)<br />
Figure 28 f is discontinuous at 0;<br />
f is continuous on [1, ∞).<br />
To determine whether f is continuous on [−2, 2], we investigate the limit from the<br />
right at −2 and the limit from the left at 2. Then,<br />
√<br />
lim f (x) = lim<br />
x→−2 +<br />
4 − x<br />
2<br />
= 0 = f (−2)<br />
x→−2 +<br />
So, f is continuous from the right at −2. Similarly,<br />
lim<br />
x→2 −<br />
√<br />
f (x) = lim 4 − x 2<br />
= 0 = f (2)<br />
x→2 −<br />
So, f is continuous from the left at 2. We conclude that f is continuous on the closed<br />
interval [−2, 2]. ■<br />
Figure 27 shows the graph of f .<br />
DEFINITION Continuity on a Domain<br />
NOW WORK Problem 37.<br />
A function f is continuous on its domain if it is continuous at every number c in its<br />
domain.<br />
EXAMPLE 6<br />
√<br />
Determining Whether f(x) = x 2 (x − 1) Is Continuous<br />
on Its Domain<br />
Determine if the function f (x) = √ x 2 (x − 1) is continuous on its domain.<br />
Solution The domain of f (x) = √ x 2 (x − 1) is {x|x = 0} ∪{x|x ≥ 1}. We need to<br />
determine whether f is continuous at the number 0 and whether f is continuous on the<br />
interval [1, ∞).<br />
At the number 0, there is an open interval containing 0 that contains no other number<br />
( )<br />
in the domain of f . [For example, use the interval − 1 2 , 1 2<br />
.] This means lim f (x)<br />
x→0<br />
does not exist. So, f is discontinuous at 0.<br />
and<br />
For all numbers c in the open interval (1, ∞) we have<br />
f (c) = √ c 2 (c − 1)<br />
√<br />
lim x 2<br />
(x − 1) = √ lim[x 2 (x − 1)] = √ c 2 (c − 1) = f (c)<br />
x→c<br />
x→c<br />
So, f is continuous on the open interval (1, ∞).<br />
Now, at the number 1,<br />
So, f is continuous from the right at 1.<br />
f (1) = 0 and lim<br />
x→1 + √<br />
x<br />
2<br />
(x − 1) = 0<br />
The function f (x) = √ x 2 (x − 1) is continuous on the interval [1, ∞), but it is<br />
discontinuous at 0. So, f is not continuous on its domain. ■<br />
Figure 28 shows the graph of f. The discontinuity at 0 is subtle. It is neither a<br />
removable discontinuity nor a jump discontinuity.<br />
When listing the properties of a function in Chapter P, we included the function’s<br />
domain, its symmetry, and its zeros. Now we add continuity to the list by asking, “Where<br />
is the function continuous?” We answer this question here for two important classes of<br />
functions: polynomial functions and rational functions.<br />
THEOREM<br />
• A polynomial function is continuous on its domain, all real numbers.<br />
• A rational function is continuous on its domain.<br />
WEB SITE<br />
MathIsFun: The Math Is Fun Web site<br />
has a very clear interactive explanation<br />
that can help the struggling student to<br />
grasp continuity. A link to this resource is<br />
available on the Chapter 1 Additional<br />
Resources document, available for<br />
download.<br />
AP® Calc Skill Builder<br />
for Example 7<br />
Identifying Where Functions Are<br />
Continuous<br />
⎧<br />
⎪<br />
⎪<br />
⎪<br />
fx ( ) = ⎨<br />
⎪<br />
⎪<br />
⎪<br />
⎩<br />
| x| x ≤ 1<br />
1<br />
1< x ≤3<br />
x + 1<br />
2 3< x < 4<br />
x<br />
x ≥ 4<br />
Let f be the function defined here. For what<br />
values of x is f not continuous?<br />
Solution<br />
We have to check each point where one<br />
piece of the function stops and the next<br />
one starts to determine if they start and<br />
stop at the same place. Find the one-sided<br />
limits at x = 1, 3, and 4.<br />
lim fx ( ) = |1| = 1<br />
x→1<br />
− 1 1<br />
lim fx ( ) = =<br />
x→ 1<br />
+ x + 1 2<br />
Since the one-sided limits are not equal, f is<br />
not continuous at x = 1.<br />
1 1<br />
lim fx ( ) = =<br />
x→3<br />
− x + 1 4<br />
lim fx ( ) = 2<br />
x→ 3<br />
+<br />
Since the one-sided limits are not equal, f is<br />
not continuous at x = 3.<br />
lim fx ( ) = 2<br />
x→4<br />
−<br />
lim fx ( ) = 4 = 2<br />
→ + f (4) = 4 = 2<br />
x 4<br />
f is continuous at x = 4.<br />
Section 1.3 • Continuity 107<br />
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108 Chapter 1 • Limits and Continuity<br />
Teaching Tip<br />
Consider teaching the overall principles<br />
without relying on rote learning of<br />
the theorems presented, such as the<br />
continuity of a sum, difference, product,<br />
and quotient presented on this page.<br />
Students may balk at the number of<br />
theorems presented, when in reality, they<br />
only have to understand the overarching<br />
principle of how to determine if a function<br />
is continuous at a point and how to identify<br />
points of discontinuity algebraically.<br />
Proof If P is a polynomial function, its domain is the set of real numbers. For a<br />
polynomial function,<br />
lim P(x) = P(c)<br />
x→c<br />
for any number c. That is, a polynomial function is continuous at every real number.<br />
If R(x) = p(x) is a rational function, then p(x) and q(x) are polynomials and the<br />
q(x)<br />
domain of R is {x|q(x) = 0}. The Limit of a Rational Function (p. 96) states that for<br />
all c in the domain of a rational function,<br />
lim R(x) = R(c)<br />
x→c<br />
So a rational function is continuous at every number in its domain. ■<br />
To summarize:<br />
• If a function is continuous on an interval, its graph has no holes or gaps on that<br />
interval.<br />
• If a function is continuous on its domain, it will be continuous at every number<br />
in its domain; its graph may have holes or gaps at numbers that are not in the<br />
domain.<br />
For example, the function R(x) = x 2 − 2x + 1<br />
is continuous on its domain<br />
x − 1<br />
{x|x = 1} even though the graph has a hole at (1, 0), as shown in Figure 29. The<br />
function f (x) = 1 is continuous on its domain {x|x = 0}, as shown in Figure 30.<br />
x<br />
Notice the behavior of the graph as x goes from negative numbers to positive numbers.<br />
y<br />
y<br />
2<br />
4<br />
2<br />
f (x) 1 x<br />
2<br />
2<br />
4<br />
x<br />
4<br />
2<br />
2<br />
4<br />
x<br />
2<br />
R(x) x2 2x 1<br />
x 1<br />
2<br />
4<br />
Figure 29 The graph of R has a hole at (1, 0).<br />
Figure 30 f is not defined at 0.<br />
3 Use Properties of Continuity<br />
So far we have shown that polynomial and rational functions are continuous on their<br />
domains. From these functions, we can build other continuous functions.<br />
THEOREM Continuity of a Sum, Difference, Product, and Quotient<br />
If the functions f and g are continuous at a number c, and if k is a real number, then<br />
the functions f + g, f − g, f · g, and kf are also continuous at c. If g(c) = 0, the<br />
function f is continuous at c.<br />
g<br />
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Section 1.3 • Continuity 109<br />
NEED TO REVIEW? Composite<br />
functions are discussed in Section P.3,<br />
pp. 27--30.<br />
The proofs of these properties are based on properties of limits. For example, the<br />
proof of the continuity of f + g is based on the Limit of a Sum property. That is, if<br />
lim f (x) and lim g(x) exist, then lim[ f (x) + g(x)] = lim f (x) + lim g(x).<br />
x→c x→c x→c x→c x→c<br />
EXAMPLE 7<br />
Identifying Where Functions Are Continuous<br />
Determine where each function is continuous:<br />
(a) F(x) = x 2 + 5 −<br />
x<br />
x 2 + 4<br />
(b) G(x) = x 3 + 2x + x 2<br />
x 2 − 1<br />
Solution First we determine the domain of each function.<br />
(a) F is the difference of the two functions f (x) = x 2 + 5 and g(x) =<br />
x<br />
x 2 + 4 , each<br />
of whose domain is the set of all real numbers. So, the domain of F is the set of all real<br />
numbers. Since f and g are continuous on their domains, the difference function F is<br />
continuous on its domain.<br />
(b) G is the sum of the two functions f (x) = x 3 + 2x, whose domain is the set of all<br />
real numbers, and g(x) = x 2<br />
, whose domain is {x|x = −1, x = 1} . Since f and g<br />
x 2 − 1<br />
are continuous on their domains, G is continuous on its domain, {x|x = −1, x = 1} . ■<br />
NOW WORK Problem 45.<br />
The continuity of a composite function depends on the continuity of its components.<br />
THEOREM Continuity of a Composite Function<br />
If a function g is continuous at c and a function f is continuous at g(c), then the<br />
composite function ( f ◦ g)(x) = f (g(x)) is continuous at c. That is,<br />
lim( f ◦ g)(x) = lim<br />
x→c x→c<br />
f (g(x)) = f [lim g(x)] = f (g(c))<br />
x→c<br />
EXAMPLE 8 Identifying Where Functions Are Continuous<br />
Determine where each function is continuous:<br />
(a) F(x) = √ x 2 + 4 (b) G(x) = √ x 2 − 1 (c) H(x) = x 2 − 1<br />
x 2 − 4 + √ x − 1<br />
Solution (a) F = f ◦ g is the composite of f (x) = √ x and g(x) = x 2 + 4. f is<br />
continuous for x ≥ 0 and g is continuous for all real numbers. The domain of F is all<br />
real numbers and F = ( f ◦ g)(x) = √ x 2 + 4 is continuous for all real numbers. That<br />
is, F is continuous on its domain.<br />
(b) G is the composite of f (x) = √ x and g(x) = x 2 − 1. f is continuous for x ≥ 0<br />
and g is continuous for all real numbers. The domain of G is {x|x ≥ 1}∪{x|x ≤−1}<br />
and G = ( f ◦ g)(x) = √ x 2 − 1 is continuous on its domain.<br />
(c) H is the sum of f (x) = x 2 − 1<br />
x 2 − 4 and the function g(x) = √ x − 1. The domain<br />
of f is {x|x = −2, x = 2}; f is continuous on its domain. The domain of g is<br />
x ≥ 1. The domain of H is {x|1 ≤ x < 2} ∪{x|x > 2}; H is continuous on its<br />
domain. ■<br />
NOW WORK Problem 47.<br />
Teaching Tip<br />
Students should have a strong knowledge<br />
of what all of the basic functions (as shown<br />
in Section P.2) look like. The following is<br />
a suggested list of functions the students<br />
should be familiar with. Students should<br />
also be comfortable with transformations of<br />
these functions.<br />
y = x y = sinx y = e<br />
2<br />
y = x y = cos x y = e<br />
−x<br />
3<br />
y = x y = tanx y = lnx<br />
1<br />
y = | x|<br />
y = x y =<br />
x<br />
Alternate Example<br />
Identifying Where Functions Are<br />
Continuous<br />
Which of the following functions are<br />
continuous at x = 0?<br />
1<br />
(a) fx ( ) =<br />
x<br />
(b) g(x) = tan x<br />
Solution<br />
(a) Since x = 0 is not in the domain of f,<br />
1<br />
fx ( ) = is discontinuous at x = 0.<br />
x<br />
(b) The function gx ( ) = tan x is defined<br />
and continuous on − π < x < π , so<br />
2 2<br />
gx ( ) = tanx<br />
is continuous at x = 0.<br />
x<br />
NEED TO REVIEW? Inverse functions<br />
are discussed in Section P.4, pp. 37--40.<br />
Recall that for any function f that is one-to-one over its domain, its inverse f −1<br />
is also a function, and the graphs of f and f −1 are symmetric with respect to the line<br />
y = x. It is intuitive that if f is continuous, then so is f −1 . See Figure 31 on page 110.<br />
The following theorem, whose proof is given in Appendix B, confirms this.<br />
Section 1.3 • Continuity 109<br />
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110 Chapter 1 • Limits and Continuity<br />
AP® Exam Tip<br />
y<br />
4<br />
y f 1 (x)<br />
y x<br />
The Intermediate Value Theorem<br />
sometimes appears on the multiplechoice<br />
portion of the exam. The<br />
Intermediate Value Theorem also<br />
appears sometimes as a portion of a<br />
free-response question. Even if it only<br />
appears sporadically on the exams, it<br />
is important that students learn it. For<br />
example, see the 2007 AP ® Calculus<br />
AB Exam, Question 3.<br />
2<br />
y f (x)<br />
4 2 2 4<br />
Figure 31<br />
THEOREM Continuity of an Inverse Function<br />
If f is a one-to-one function that is continuous on its domain, then its inverse<br />
function f −1 is also continuous on its domain.<br />
2<br />
4<br />
x<br />
AP® Exam Tip<br />
The three theorems that students<br />
must cite and use on the exam as a<br />
part of an answer or justification for<br />
an answer are the IVT (Intermediate<br />
Value Theorem), the EVT (Extreme<br />
Value Theorem, which will be learned<br />
in Section 4.2), and the MVT (Mean<br />
Value Theorem, which will be learned<br />
in Section 4.3). It is a good idea to have<br />
students practice them throughout the<br />
year.<br />
5000 m<br />
3765.6 m<br />
2000 m<br />
4 Use the Intermediate Value Theorem<br />
Functions that are continuous on a closed interval have many important properties. One<br />
of them is stated in the Intermediate Value Theorem. The proof of the Intermediate Value<br />
Theorem may be found in most books on advanced calculus.<br />
THEOREM The Intermediate Value Theorem<br />
Let f be a function that is continuous on a closed interval [a, b] and f (a) = f (b).<br />
If N is any number between f (a) and f (b), then there is at least one number c in the<br />
open interval (a, b) for which f (c) = N.<br />
To get a better idea of this result, suppose you climb a mountain, starting at an<br />
elevation of 2000 meters and ending at an elevation of 5000 meters. No matter how<br />
many ups and downs you take as you climb, at some time your altitude must be 3765.6<br />
meters, or any other number between 2000 and 5000.<br />
In other words, a function f that is continuous on a closed interval [a, b] must take<br />
on all values between f (a) and f (b). Figure 32 illustrates this. Figure 33 shows why<br />
the continuity of the function is crucial. Notice in Figure 33 that there is a hole in the<br />
graph of f at the point (c, N). Because of the discontinuity at c, there is no number c in<br />
the open interval (a, b) for which f (c) = N.<br />
y<br />
y<br />
Teaching Tip<br />
f(b)<br />
f(b)<br />
A picture is worth 1,000 words! The<br />
Intermediate Value Theorem is very much<br />
a commonsense theorem that can easily<br />
be conveyed using Figures 32 and 33.<br />
N<br />
f(a)<br />
a<br />
f (c) N<br />
y f (x)<br />
c<br />
b<br />
x<br />
N<br />
f(a)<br />
a<br />
y f (x)<br />
c<br />
b<br />
x<br />
∑ Mathematical Practices Tip<br />
MPAC 1: Reasoning with Definitions and<br />
Theorems<br />
Justifying answers will now include<br />
confirming that the hypotheses of theorems<br />
are met before stating the conclusions. The<br />
Web site AP ® Central, under the heading<br />
Teaching and Assessing AP ® Calculus,<br />
offers resources and short videos by AP ®<br />
Calculus teachers demonstrating how they<br />
prepare students. The module on continuity<br />
and differentiability is particularly good<br />
and can be accessed through a calculus<br />
teacher’s AP ® Audit account.<br />
Figure 32 f takes on every value<br />
between f (a) and f (b).<br />
Figure 33 A discontinuity at c results in no<br />
number c in (a, b) for which f (c) = N.<br />
The Intermediate Value Theorem is an existence theorem. It states that there is at<br />
least one number c for which f (c) = N, but it does not tell us how to find c. However, we<br />
can use the Intermediate Value Theorem to locate an open interval (a, b) that contains c.<br />
An immediate application of the Intermediate Value Theorem involves locating the<br />
zeros of a function. Suppose a function f is continuous on the closed interval [a, b] and<br />
f (a) and f (b) have opposite signs. Then by the Intermediate Value Theorem, there is<br />
TRM Section 1.3: Worksheet 3<br />
This worksheet contains an explanation of<br />
the Intermediate Value Theorem and one<br />
problem based upon a portion of a former<br />
free-response exam question.<br />
110<br />
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Section 1.3 • Continuity 111<br />
f 0, 3g 3 f22, 10g<br />
Figure 34 f (x) = x 3 + x 2 − x − 2<br />
at least one number c between a and b for which f (c) = 0. That is, f has at least one<br />
zero between a and b.<br />
EXAMPLE 9<br />
Using the Intermediate Value Theorem<br />
Use the Intermediate Value Theorem to show that<br />
f (x) = x 3 + x 2 − x − 2<br />
has a zero between 1 and 2.<br />
Solution Since f is a polynomial, it is continuous on the closed interval [1, 2]. Because<br />
f (1) =−1 and f (2) = 8 have opposite signs, the Intermediate Value Theorem states<br />
that f (c) = 0 for at least one number c in the interval (1, 2). That is, f has at least one<br />
zero between 1 and 2. Figure 34 shows the graph of f on a graphing utility. ■<br />
NOW WORK Problem 59 and AP® Practice Problems 6, 7, 9, and 10.<br />
The Intermediate Value Theorem can be used to approximate a zero in the interval<br />
(a, b) by dividing the interval [a, b] into smaller subintervals. There are two popular<br />
methods of subdividing the interval [a, b].<br />
The bisection method bisects [a, b], that is, divides [a, b] into two equal subintervals<br />
( ) b − a<br />
and compares the sign of f to the signs of the previously computed values<br />
2<br />
f (a) and f (b). The subinterval whose endpoints have opposite signs is then bisected,<br />
and the process is repeated.<br />
The second method divides [a, b] into 10 subintervals of equal length and compares<br />
the signs of f evaluated at each of the 11 endpoints. The subinterval whose endpoints<br />
have opposite signs is then divided into 10 subintervals of equal length and the process<br />
is repeated.<br />
We choose to use the second method because it lends itself well to the table feature<br />
of a graphing utility. You are asked to use the bisection method in Problems 107–114.<br />
EXAMPLE 10<br />
Using the Intermediate Value Theorem to Approximate<br />
a Real Zero of a Function<br />
The function f (x) = x 3 +x 2 −x −2 has a zero in the interval (1, 2). Use the Intermediate<br />
Value Theorem to approximate the zero correct to three decimal places.<br />
Solution Using the TABLE feature on a graphing utility, we subdivide the interval [1, 2]<br />
into 10 subintervals, each of length 0.1. Then we find the subinterval whose endpoints<br />
have opposite signs, or the endpoint whose value equals 0 (in which case, the exact<br />
zero is found). From Figure 35, since f (1.2) =−0.032 and f (1.3) = 0.587, by the<br />
Intermediate Value Theorem, a zero lies in the interval (1.2, 1.3). Correct to one decimal<br />
place, the zero is 1.2.<br />
Repeat the process by subdividing the interval [1.2, 1.3] into 10 subintervals, each<br />
of length 0.01. See Figure 36. We conclude that the zero is in the interval (1.20, 1.21),<br />
so correct to two decimal places, the zero is 1.20.<br />
Figure 35 Figure 36<br />
AP® Exam Tip<br />
Although problems like Example 10 have<br />
not appeared on previous AP ® Exams,<br />
this example can help the student better<br />
understand how to find the zeros of a function<br />
using technology. A much faster method using<br />
a TI-84 Plus calculator is pressing (2nd Trace<br />
then Zero), which will find the zero graphically.<br />
This is one of the four graphing calculator skills<br />
that students have to know to be successful on<br />
the AP ® Exam.<br />
AP® Calc Skill Builder<br />
for Example 9<br />
Using the Intermediate Value Theorem<br />
Let f be a continuous function on a<br />
closed interval [−5, 5]. Also suppose that<br />
f ( − 5) =− 3and f (5) = 3. Which of the<br />
following statements must be true?<br />
I. f() c= 0 for at least 1 c between −5 and 5.<br />
II. f (0) = 0<br />
III. f() c= 0 for at least 1 c between −3 and 3.<br />
Solution<br />
I. Since f ( − 5) < 0 and f (5) > 0according<br />
to the IVT, there is at least 1 number c in<br />
the open interval (−5, 5) for which fc () = 0.<br />
This answer is true.<br />
II. The IVT only tells us that f (x) = 0 for<br />
some value of x in (−5, 5), but we do not<br />
know where in the interval the function<br />
crosses the x-axis. So we do not know if<br />
f (0) = 0. This answer is not true.<br />
III. Also, we do not know if the function<br />
crosses the x-axis between −3 and 3. It<br />
might cross between −5 and −4. We do<br />
not know! This answer is not true.<br />
∑ Mathematical Practices Tip<br />
MPAC 1: Reasoning with Definitions and<br />
Theorems<br />
To use the IVT, the function must be<br />
continuous on a closed interval. Ask<br />
students to explain why the function must<br />
be continuous. Tell students that if you<br />
draw a curve with a pencil and do not<br />
lift the pencil from the page, the result is<br />
a continuous function. Ask students to<br />
explain why this pencil drawing shows the<br />
conclusion of the IVT.<br />
Section 1.3 • Continuity<br />
111<br />
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112 Chapter 1 • Limits and Continuity<br />
Must-Do Problems for<br />
Exam Readiness<br />
AB: 13–18, 19–35 odd, 59–63 odd, AP ®<br />
Practice Problems<br />
BC: 13–18, 24, 25, 29, 51–56, 79, and all<br />
AP ® Practice Problems<br />
TRM Full Solutions to Section<br />
1.3 Problems and AP® Practice<br />
Problems<br />
Answers to Section 1.3<br />
Problems<br />
1. True.<br />
2. False.<br />
3. fc () is defined, lim fx ( ) exists,<br />
x→c<br />
lim fx ( ) = fc ()<br />
x→c<br />
4. True.<br />
5. False.<br />
6. False.<br />
7. False.<br />
8. True.<br />
9. Discontinuous.<br />
10. Continuous.<br />
11. True.<br />
12. False.<br />
13. (a) Discontinuous at c =− 3.<br />
(b) lim ( ) ≠ f ( −3)<br />
fx<br />
x→−3<br />
(c) Removable.<br />
(d) f ( − 3) =−2<br />
14. (a) Continuous at c = 0.<br />
15. (a) Discontinuous at c = 2.<br />
(b) lim fx ( ) does not exist.<br />
x→2<br />
(c) Not removable.<br />
16. (a) Discontinuous at c = 3.<br />
(b) lim fx ( ) does not exist.<br />
x→3<br />
(c) Not removable.<br />
17. (a) Continuous at c = 4.<br />
18. (a) Discontinuous at c = 5.<br />
(b) f not defined at c = 5, and<br />
lim fx ( ) ≠ lim fx ( ) so lim fx ( ) does<br />
− +<br />
x→5 x→5<br />
x→5<br />
not exist.<br />
(c) Not removable.<br />
19. Continuous at c =− 1.<br />
20. Continuous at c = 5.<br />
21. Continuous at c =− 2.<br />
22. Discontinuous at c = 2.<br />
23. Continuous at c = 2.<br />
Figure 37<br />
1.3 Assess Your Understanding<br />
Now subdivide the interval [1.20, 1.21] into 10 subintervals, each of length 0.001.<br />
See Figure 37.<br />
We conclude that the zero of the function f is 1.205, correct to three decimal<br />
places. ■<br />
Notice that a benefit of the method used in Example 10 is that each additional<br />
iteration results in one additional decimal place of accuracy for the approximation.<br />
Concepts and Vocabulary<br />
PAGE<br />
105 13. c =−3 14. c = 0<br />
1. True or False A polynomial function is continuous at every 15. c = 2 16. c = 3<br />
real number.<br />
2. True or False Piecewise-defined functions are never continuous<br />
at numbers where the function changes equations.<br />
3. The three conditions necessary for a function f to be continuous<br />
17. c = 4 18. c = 5<br />
at a number c are , , and .<br />
4. True or False If f is continuous at 0, then g(x) = 1 f (x) is<br />
4<br />
continuous at 0.<br />
5. True or False If f is a function defined everywhere in an open<br />
interval containing c, except possibly at c, then the number c is<br />
called a removable discontinuity of f if the function f is not<br />
continuous at c.<br />
6. True or False If a function f is discontinuous at a number c,<br />
then lim f (x) does not exist.<br />
x→c<br />
7. True or False If a function f is continuous on an open interval<br />
(a, b), then it is continuous on the closed interval [a, b].<br />
8. True or False If a function f is continuous on the closed interval<br />
[a, b], then f is continuous on the open interval (a, b).<br />
In Problems 9 and 10, explain whether each function is continuous or<br />
discontinuous on its domain.<br />
9. The velocity of a ball thrown up into the air as a function of<br />
time, if the ball lands 5 seconds after it is thrown and stops.<br />
10. The temperature of an oven used to bake a potato as a function<br />
of time.<br />
11. True or False If a function f is continuous on a closed interval<br />
[a, b], then the Intermediate Value Theorem guarantees that the<br />
function takes on every value between f (a) and f (b).<br />
12. True or False If a function f is continuous on a closed interval<br />
[a, b] and f (a) = f (b), but both f (a) >0 and f (b) >0, then<br />
according to the Intermediate Value Theorem, f does not have a<br />
zero on the open interval (a, b).<br />
Skill Building<br />
In Problems 13–18, use the graph of y = f (x) (top right).<br />
(a) Determine if f is continuous at c.<br />
(b) If f is discontinuous at c, state which condition(s) of the definition<br />
of continuity is (are) not satisfied.<br />
(c) If f is discontinuous at c, determine if the discontinuity is<br />
removable.<br />
(d) If the discontinuity is removable, define (or redefine) f at c to<br />
make f continuous at c.<br />
24. Discontinuous at c = 0.<br />
25. Discontinuous at c = 1.<br />
26. Continuous at c = 1.<br />
27. Discontinuous at c = 1.<br />
28. Discontinuous at c = 1.<br />
29. Continuous at c = 0.<br />
30. Discontinuous at c =− 1.<br />
31. Discontinuous at c = 0.<br />
32. Discontinuous at c = 4.<br />
33. f (2) = 4<br />
34. f (3) = 7<br />
35. f (1) = 2<br />
36. f ( − 1) =−4<br />
(3, 1)<br />
NOW WORK Problem 65.<br />
y<br />
4<br />
2<br />
4 2 2 4<br />
2 (3, 1)<br />
4<br />
(2, 3)<br />
In Problems 19–32, determine whether the function f is continuous<br />
at c.<br />
PAGE<br />
104 19. f (x) = x 2 + 1 at c =−1 20. f (x) = x 3 − 5 at c = 5<br />
PAGE<br />
104 21. f (x) = x<br />
at c =−2 22. f (x) = x<br />
x − 2<br />
x 2 + 4<br />
<br />
2x + 5 if x ≤ 2<br />
23. f (x) =<br />
at c = 2<br />
4x + 1 if x > 2<br />
<br />
2x + 1 if x ≤ 0<br />
24. f (x) =<br />
at c = 0<br />
2x if x > 0<br />
⎧<br />
3x − 1 if x < 1<br />
⎪⎨<br />
PAGE<br />
105 25. f (x) = 4 if x = 1 at c = 1<br />
⎪⎩<br />
2x if x > 1<br />
⎧<br />
3x − 1 if x < 1<br />
⎪⎨<br />
26. f (x) = 2 if x = 1 at c = 1<br />
⎪⎩<br />
2x if x > 1<br />
<br />
3x − 1 if x < 1<br />
27. f (x) =<br />
at c = 1<br />
2x if x > 1<br />
⎧<br />
3x − 1 ⎪⎨<br />
if x < 1<br />
28. f (x) = 2 if x = 1 at c = 1<br />
⎪⎩<br />
3x if x > 1<br />
<br />
x 2 if x ≤ 0<br />
29. f (x) =<br />
at c = 0<br />
2x if x > 0<br />
⎧<br />
⎪⎨<br />
x 2 if x < −1<br />
30. f (x) = 2 if x =−1<br />
⎪⎩<br />
−3x + 2 if x > −1<br />
at c =−1<br />
37. Continuous on the given interval.<br />
38. Continuous on the given interval.<br />
39. Not continuous on the given interval.<br />
Continuous on { xx | 3}.<br />
40. Continuous on the given interval.<br />
41. Continuous on { xx | ≠ 0} .<br />
42. Continuous on the set of all real numbers.<br />
43. Continuous on the set of all real numbers.<br />
44. Continuous on { xx | ≥ 0} .<br />
45. Continuous on { xx | ≥0, x ≠ 9} .<br />
46. Continuous on { xx | ≥0, x ≠ 4} .<br />
47. Continuous on { xx | < 2} .<br />
48. Continuous on { x|| x| > 1} .<br />
x<br />
at c = 2<br />
112<br />
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⎧<br />
4 − 3x<br />
⎪⎨<br />
2 if x < 0<br />
4 if x = 0 at c = 0<br />
31. f (x) = <br />
⎪⎩ 16 − x 2<br />
if 0 < x < 4<br />
4 − x<br />
⎧√ ⎨ 4 + x if −4 ≤ x ≤ 4<br />
<br />
32. f (x) =<br />
⎩ x 2 − 3x − 4<br />
at c = 4<br />
if x > 4<br />
x − 4<br />
In Problems 33–36, each function f has a removable<br />
discontinuity at c. Define f (c) so that f is continuous at c.<br />
33. f (x) = x2 − 4<br />
x − 2 , c = 2<br />
34. f (x) = x2 + x − 12<br />
, c = 3<br />
x − 3<br />
⎧<br />
⎨ 1 + x if x < 1<br />
PAGE<br />
105 35. f (x) = 4 if x = 1 c = 1<br />
⎩<br />
2x if x > 1<br />
⎧<br />
⎨ x 2 + 5x if x < −1<br />
36. f (x) = 0 if x =−1 c =−1<br />
⎩<br />
x − 3 if x > −1<br />
In Problems 37–40, determine if each function f is<br />
continuous on the given interval. If the answer is no,<br />
state the interval, if any, on which f is continuous.<br />
PAGE<br />
107 37. f (x) = x2 − 9<br />
on the interval [−3, 3)<br />
x − 3<br />
38. f (x) = 1 + 1 on the interval [−1, 0)<br />
x<br />
39.<br />
1<br />
f (x) = on the interval [−3, 3]<br />
x 2 − 9<br />
<br />
40. f (x) = 9 − x 2 on the interval [−3, 3]<br />
In Problems 41–50, determine where each function f is continuous.<br />
First determine the domain of the function. Then support your decision<br />
using properties of continuity.<br />
41. f (x) = 2x 2 + 5x − 1 42. f (x) = x + 1 + 2x<br />
x<br />
x 2 + 5<br />
43. f (x) = (x − 1)(x 2 + x + 1) 44. f (x) = √ x(x 3 − 5)<br />
PAGE<br />
109 45. f (x) = √ x − 9<br />
46. f (x) = √ x − 4<br />
x − 3 x − 2<br />
<br />
PAGE<br />
109 47. f (x) =<br />
x 2 + 1<br />
2 − x<br />
48. f (x) =<br />
<br />
4<br />
x 2 − 1<br />
49. f (x) = (2x 2 + 5x − 3) 2/3 50. f (x) = (x + 2) 1/2<br />
In Problems 51–56, use the function<br />
⎧ √<br />
15 − 3x if x < 2<br />
⎪⎨ √<br />
5 if x = 2<br />
f (x) =<br />
9 − x ⎪⎩<br />
2 if 2 < x < 3<br />
x − 2 if 3 ≤ x<br />
51. Is f continuous at 0? Why or why not?<br />
52. Is f continuous at 4? Why or why not?<br />
PAGE<br />
105 53. Is f continuous at 3? Why or why not?<br />
49. Continuous on the set of all real numbers.<br />
50. Continuous on { xx | ≥− 2} .<br />
51. f is continuous at 0 because lim fx ( ) = f (0).<br />
x→0<br />
52. f is discontinuous at 4 because lim fx ( )<br />
x→4<br />
does not exist.<br />
53. f is discontinuous at 3 because lim fx ( ) does<br />
x → 3<br />
not exist.<br />
54. f is discontinuous at 2 because lim fx ( ) does<br />
x→2<br />
not exist.<br />
55. f is continuous at 1 because lim fx ( ) = f (1).<br />
x→1<br />
56. f is continuous at 2.5 because<br />
lim fx ( ) = f (2.5).<br />
x→2.5<br />
54. Is f continuous at 2? Why or why not?<br />
55. Is f continuous at 1? Why or why not?<br />
56. Is f continuous at 2.5? Why or why not?<br />
In Problems 57 and 58:<br />
(a) Use technology to graph f using a suitable scale on each axis.<br />
(b) Based on the graph from (a), determine where f is continuous.<br />
(c) Use the definition of continuity to determine where f is continuous.<br />
(d) What advice would you give a fellow student about using<br />
technology to determine where a function is continuous?<br />
57. f (x) = x3 − 8<br />
x − 2<br />
58. f (x) = x2 − 3x + 2<br />
3x − 6<br />
In Problems 59–64, use the Intermediate Value Theorem to determine<br />
which of the functions must have zeros in the given interval. Indicate<br />
those for which the theorem gives no information. Do not attempt to<br />
locate the zeros.<br />
PAGE<br />
111 59. f (x) = x 3 − 3x on [−2, 2]<br />
60. f (x) = x 4 − 1on[−2, 2]<br />
61.<br />
x<br />
f (x) = − 1 on [10, 20]<br />
(x + 1) 2<br />
62. f (x) = x 3 − 2x 2 − x + 2 on [3, 4]<br />
63. f (x) = x3 − 1<br />
on [0, 2]<br />
x − 1<br />
64. f (x) = x2 + 3x + 2<br />
x 2 on [−3, 0]<br />
− 1<br />
In Problems 65–72, verify that each function has a zero in the indicated<br />
interval. Then use the Intermediate Value Theorem to approximate the<br />
zero correct to three decimal places by repeatedly subdividing the<br />
interval containing the zero into 10 subintervals.<br />
PAGE<br />
112 65. f (x) = x 3 + 3x − 5; interval: [1, 2]<br />
66. f (x) = x 3 − 4x + 2; interval: [1, 2]<br />
67. f (x) = 2x 3 + 3x 2 + 4x − 1; interval: [0, 1]<br />
68. f (x) = x 3 − x 2 − 2x + 1; interval: [0, 1]<br />
69. f (x) = x 3 − 6x − 12; interval: [3, 4]<br />
70. f (x) = 3x 3 + 5x − 40; interval: [2, 3]<br />
71. f (x) = x 4 − 2x 3 + 21x − 23; interval: [1, 2]<br />
72. f (x) = x 4 − x 3 + x − 2; interval: [1, 2]<br />
In Problems 73 and 74,<br />
(a) Use the Intermediate Value Theorem to show that f has a zero in<br />
the given interval.<br />
(b) Use technology to find the zero rounded to three decimal places.<br />
<br />
73. f (x) = x 2 + 4x − 2 in [0, 1]<br />
74. f (x) = x 3 − x + 2in[−2, 0]<br />
Applications and Extensions<br />
Heaviside Functions In Problems 75 and 76, determine whether the<br />
given Heaviside function is continuous at c.<br />
0 if t < 1<br />
75. u 1(t) =<br />
1 if t ≥ 1<br />
c = 1<br />
0 if t < 3<br />
76. u 3(t) =<br />
1 if t ≥ 3<br />
c = 3<br />
57. (a) <br />
y<br />
18<br />
14<br />
10<br />
6<br />
2<br />
3 2 1 1 2 3 x<br />
(b) Based on the graph, it appears that the<br />
function is continuous for all real numbers.<br />
(c) f is actually continuous at all real numbers<br />
except x = 2.<br />
(d) Answers will vary. Sample answer:<br />
Conclusions drawn from graphing technology<br />
should always be confirmed using basic<br />
analysis.<br />
58. (a)<br />
23 22 21<br />
y<br />
1.5<br />
1<br />
0.5<br />
21<br />
21.5<br />
1 2 3<br />
(b) Based on the graph, it appears<br />
that the function is continuous on all<br />
real numbers.<br />
(c) f is continuous on { xx | ≠ 2}.<br />
(d) Answers will vary. Sample<br />
answer: Conclusions drawn from<br />
graphing technology should always be<br />
confirmed using analysis.<br />
59. Yes. A zero exists on the given<br />
interval.<br />
60. IVT gives no information.<br />
61. IVT gives no information.<br />
62. IVT gives no information.<br />
63. IVT gives no information.<br />
64. IVT gives no information.<br />
65. 1.154<br />
66. 1.675<br />
67. 0.211<br />
68. 0.445<br />
69. 3.134<br />
70. 2.137<br />
71. 1.157<br />
72. 1.308<br />
73. (a) Since f is continuous on [0,1],<br />
f (0) < 0, and f (1) > 0, the IVT<br />
guarantees that f has a zero on the<br />
interval (0,1).<br />
(b) 0.828<br />
74. (a) Since f is continuous on [ −2,0],<br />
f ( − 2) < 0, and f (2) > 0, the IVT<br />
guarantees that f has a zero on the<br />
interval (0,1).<br />
(b) −1.521<br />
75. Discontinuous at c = 1.<br />
76. Discontinuous at c = 3.<br />
x<br />
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114 Chapter 1 • Limits and Continuity<br />
77. Continuous on the set of all real<br />
numbers.<br />
y<br />
8<br />
6<br />
4<br />
2<br />
3 2 1 1 2 3<br />
78. Continuous on the set of all real<br />
numbers.<br />
y<br />
4<br />
3<br />
2<br />
1<br />
23 22 21 1 2 3<br />
⎧0.47 if 0< w ≤1<br />
⎪<br />
⎪0.68 if 1< w ≤2<br />
79. (a) C( w)=<br />
⎨<br />
⎪0.89 if 2< w ≤3<br />
⎪<br />
⎩⎪<br />
1.10 if 3< w ≤3.5<br />
{ }<br />
(b) w 0< w ≤3.5<br />
(c) Continuous on the intervals (0,1],<br />
(1,2], (2,3], and (3,3.5].<br />
(d) Jump discontinuities at w = 1,<br />
w = 2, and w = 3.<br />
(e) Answers will vary. Sample answer:<br />
It is in the consumer’s best interest<br />
to have letters weigh as close as<br />
possible to a whole number of ounces<br />
without going over.<br />
⎧<br />
⎪<br />
⎪<br />
⎪<br />
⎪<br />
⎪<br />
⎪<br />
⎪<br />
⎪<br />
80. (a) Cw ( ) = ⎨<br />
⎪<br />
⎪<br />
⎪<br />
⎪<br />
⎪<br />
⎪<br />
⎪<br />
⎪<br />
⎩<br />
0.94 if 0< w ≤1<br />
1.15 if 1< w ≤2<br />
1.36 if 2 < w ≤3<br />
1.57 if 3 < w ≤4<br />
1.78 if 4 < w ≤5<br />
1.99 if 5 < w ≤6<br />
2.20 if 6 < w ≤7<br />
2.41 if 7 < w ≤8<br />
2.62 if 8 < w ≤9<br />
2.83 if 9 < w ≤10<br />
3.04 if 10< w ≤11<br />
3.25 if 11< w ≤12<br />
3.46 if 12< w ≤13<br />
(b) { w|0< w ≤13}<br />
(c) C is continuous on the intervals<br />
(0, 1], (1, 2], (2, 3], (3, 4], (4, 5],<br />
(5, 6], (6, 7], (7, 8], (8, 9], (9, 10],<br />
(10, 11], (11, 12], and (12, 13].<br />
(d) C has jump discontinuities at<br />
w = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11,<br />
and 12.<br />
(e) Answers will vary. Sample answer:<br />
It is in the customer’s best interest<br />
x<br />
x<br />
In Problems 77 and 78, determine where each function is continuous.<br />
Graph each function.<br />
1 − x 2 if |x| ≤1<br />
77. f (x) =<br />
x 2 − 1 if |x| > 1<br />
<br />
78. f (x) =<br />
4 − x 2 if |x| ≤2<br />
| x | − 2 if |x| > 2<br />
79. First-Class Mail As of April 2016, the U.S. Postal Service<br />
charged $0.47 postage for first-class letters weighing up to and<br />
including 1 ounce, plus a flat fee of $0.21 for each additional or<br />
partial ounce up to 3.5 ounces. First-class letter rates do not apply<br />
to letters weighing more than 3.5 ounces.<br />
Source: U.S. Postal Service Notice 123.<br />
(a) Find a function C that models the first-class postage charged<br />
for a letter weighing w ounces. Assume w>0.<br />
(b) What is the domain of C?<br />
(c) Determine the intervals on which C is continuous.<br />
(d) At numbers where C is not continuous (if any), what type of<br />
discontinuity does C have?<br />
(e) What are the practical implications of the answer to (d)?<br />
80. First-Class Mail As of April 2016, the U.S. Postal Service<br />
charged $0.94 postage for first-class large envelopes weighing up<br />
to and including 1 ounce, plus a flat fee of $0.21 for each<br />
additional or partial ounce up to 13 ounces. First-class rates do not<br />
apply to large envelopes weighing more than 13 ounces.<br />
Source: U.S. Postal Service Notice 123.<br />
(a) Find a function C that models the first-class postage<br />
charged for a large envelope weighing w ounces. Assume<br />
w>0.<br />
(b) What is the domain of C?<br />
(c) Determine the intervals on which C is continuous.<br />
(d) At numbers where C is not continuous (if any), what type of<br />
discontinuity does C have?<br />
(e) What are the practical implications of the answer to (d)?<br />
81. Cost of Electricity In June 2016, Florida Power and Light<br />
had the following monthly rate schedule for electric usage in<br />
single-family residences:<br />
Monthly customer charge $7.87<br />
Fuel charge<br />
≤ 1000 kWH<br />
0.02173 per kWH<br />
> 1000 kWH $21.73 + 0.03173 for each<br />
kWH in excess of 1000<br />
Source: Florida Power and Light, Miami, FL.<br />
(a) Find a function C that models the monthly cost of<br />
using x kWH of electricity.<br />
(b) What is the domain of C?<br />
(c) Determine the intervals on which C is continuous.<br />
(d) At numbers where C is not continuous (if any), what type of<br />
discontinuity does C have?<br />
(e) What are the practical implications of the answer to (d)?<br />
82. Cost of Water The Jericho Water District determines quarterly<br />
water costs, in dollars, using the following rate schedule:<br />
to have packages that weigh as close as<br />
possible to a whole number of ounces<br />
without going over.<br />
81. (a)<br />
⎧⎪<br />
C x<br />
7.87 0.02173x<br />
if 0 x 1000<br />
( )= ⎨<br />
+ ≤ ≤<br />
⎩⎪ − 2.13+ 0.03173x<br />
if x><br />
1000<br />
(b) { x| x≥<br />
0}<br />
(c) C is continuous on its domain.<br />
(d) See (c).<br />
(e) Answers will vary. Sample answer: To<br />
minimize the monthly cost of electricity, it is in<br />
the consumer’s best interest to minimize the<br />
amount of electricity used.<br />
Water used<br />
(in thousands of gallons) Cost<br />
0 ≤ x ≤ 10 $9.00<br />
10 < x ≤ 30 $9.00 + 0.95 for each thousand<br />
gallons in excess of 10,000 gallons<br />
30 < x ≤ 100 $28.00 + 1.65 for each thousand<br />
gallons in excess of 30,000 gallons<br />
x > 100 $143.50 + 2.20 for each thousand<br />
gallons in excess of 100,000 gallons<br />
Source: Jericho Water District, Syosset, NY.<br />
(a) Find a function C that models the quarterly cost of<br />
using x thousand gallons of water.<br />
(b) What is the domain of C?<br />
(c) Determine the intervals on which C is continuous.<br />
(d) At numbers where C is not continuous (if any), what type of<br />
discontinuity does C have?<br />
(e) What are the practical implications of the answer to (d)?<br />
83. Gravity on Europa Europa, one of<br />
the larger satellites of Jupiter, has an icy<br />
surface and appears to have oceans<br />
beneath the ice. This makes it a<br />
candidate for possible extraterrestrial<br />
life. Because Europa is much smaller<br />
than most planets, its gravity is weaker.<br />
If we think of Europa as a sphere with<br />
uniform internal density, then inside the sphere, the gravitational<br />
field g is given by g(r) = Gm r, 0≤ r < R, where R is the<br />
R3 radius of the sphere, r is the distance from the center of the<br />
sphere, and G is the universal gravitation constant. Outside a<br />
uniform sphere of mass m, the gravitational field g is given by<br />
g(r) = Gm<br />
r 2 , R < r<br />
(a) For the gravitational field of Europa to be continuous<br />
at its surface, what must g(r) equal?<br />
[Hint: Investigate lim r→R<br />
g(r).]<br />
(b) Determine the gravitational field at Europa’s surface. This<br />
will indicate the type of gravity environment organisms will<br />
experience. Use the following measured values: Europa’s<br />
mass is 4.8 × 10 22 kilograms, its radius is 1.569 × 10 6<br />
meters, and G = 6.67 × 10 −11 .<br />
(c) Compare the result found in (b) to the gravitational field on<br />
Earth’s surface, which is 9.8 meter/second 2 . Is the gravity<br />
on Europa less than or greater than that on Earth?<br />
84. Find constants A and B so that the function below is continuous<br />
for all x. Graph the resulting function.<br />
⎧<br />
⎨ (x − 1) 2 if −∞ < x < 0<br />
f (x) = (A − x) 2 if 0 ≤ x < 1<br />
⎩<br />
x + B if 1 ≤ x < ∞<br />
85. Find constants A and B so that the function below is continuous<br />
for all x. Graph the resulting function.<br />
⎧<br />
⎨ x + A if −∞ < x < 4<br />
f (x) = (x − 1) 2 if 4 ≤ x ≤ 9<br />
⎩<br />
Bx + 1 if 9 < x < ∞<br />
82. (a) ⎧<br />
⎪<br />
⎪<br />
Cx ( ) = ⎨<br />
⎪<br />
⎪<br />
⎩<br />
9.00 if 0 ≤x<br />
≤10<br />
9.00 + 0.95( x− 10) if 10< x < 30<br />
28.00 + 1.65( x− 30) if 30 < x ≤100<br />
143.50 + 2.20( x− 100) if x > 100<br />
(b) { x| x≥<br />
0}<br />
NASA/JPL/DLR<br />
(c) C is continuous on its domain.<br />
(d) See (c).<br />
(e) Answers will vary. Sample answer: There<br />
is no penalty for going just a little over 10,000<br />
or 30,000 or 100,000 gallons.<br />
83. (a) Gm<br />
R 2<br />
(b) 1.3 m/s 2<br />
(c) The gravity on Europa is less than the<br />
gravity on Earth.<br />
Answers continue on p. 115<br />
114<br />
Chapter 1 • Limits and Continuity<br />
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Section 1.3 • Assess Your Understanding 115<br />
86. For the function f below, find k so that f is continuous at 2.<br />
⎧ √ √<br />
⎨ 2x + 5 − x + 7<br />
if x ≥− 5 f (x) = x − 2<br />
2 , x = 2<br />
⎩<br />
k if x = 2<br />
x 2 − 6x − 16<br />
87. Suppose f (x) =<br />
.<br />
(x 2 − 7x − 8) x 2 − 4<br />
(a) For what numbers x is f defined?<br />
(b) For what numbers x is f discontinuous?<br />
(c) Which discontinuities found in (b) are removable?<br />
88. Intermediate Value Theorem<br />
(a) Use the Intermediate Value Theorem to show that the<br />
function f (x) = sin x + x − 3 has a zero in the<br />
interval [0,π].<br />
(b) Approximate the zero rounded to three decimal places.<br />
89. Intermediate Value Theorem<br />
(a) Use the Intermediate Value Theorem to show that the<br />
function f (x) = e x + x − 2 has a zero in the interval [0, 2].<br />
(b) Approximate the zero rounded to three decimal places.<br />
In Problems 90–93, verify that each function intersects the given line in<br />
the indicated interval. Then use the Intermediate Value Theorem to<br />
approximate the point of intersection correct to three decimal places by<br />
repeatedly subdividing the interval into 10 subintervals.<br />
90. f (x) = x 3 − 2x 2 − 1; line: y =−1; interval: (1, 4)<br />
91. g(x) =−x 4 + 3x 2 + 3; line: y = 3; interval: (1, 2)<br />
92. h(x) = x3 − 5<br />
x 2 ; line: y = 1; interval: (1, 3)<br />
+ 1<br />
93. r(x) = x − 6<br />
x 2 ; line: y =−1; interval: (0, 3)<br />
+ 2<br />
94. Graph a function that is continuous on the closed interval [5, 12],<br />
that is negative at both endpoints and has exactly three distinct<br />
zeros in this interval. Does this contradict the Intermediate Value<br />
Theorem? Explain.<br />
95. Graph a function that is continuous on the closed interval [−1, 2],<br />
that is positive at both endpoints and has exactly two zeros in this<br />
interval. Does this contradict the Intermediate Value Theorem?<br />
Explain.<br />
96. Graph a function that is continuous on the closed interval [−2, 3],<br />
is positive at −2 and negative at 3 and has exactly two zeros in<br />
this interval. Is this possible? Does this contradict the<br />
Intermediate Value Theorem? Explain.<br />
97. Graph a function that is continuous on the closed interval [−5, 0],<br />
is negative at −5 and positive at 0 and has exactly three zeros in<br />
the interval. Is this possible? Does this contradict the Intermediate<br />
Value Theorem? Explain.<br />
98. (a) Explain why the Intermediate Value Theorem gives no<br />
information about the zeros of the function f (x) = x 4 − 1 on<br />
the interval [−2, 2].<br />
(b) Use technology to determine whether or not f has a zero on<br />
the interval [−2, 2].<br />
99. (a) Explain why the Intermediate Value Theorem gives no<br />
information about the zeros of the function<br />
f (x) = ln(x 2 + 2) on the interval [−2, 2].<br />
84. A = 1 and B =−1<br />
or A =− 1 and B = 3<br />
y<br />
6<br />
4<br />
2<br />
22 21 1 2 3<br />
y<br />
6<br />
4<br />
2<br />
22 21 1 2 3<br />
x<br />
x<br />
(b) Use technology to determine whether or not f has a zero on<br />
the interval [−2, 2].<br />
100. Intermediate Value Theorem<br />
(a) Use the Intermediate Value Theorem to show that the<br />
functions y = x 3 and y = 1 − x 2 intersect somewhere<br />
between x = 0 and x = 1.<br />
(b) Use technology to find the coordinates of the point of<br />
intersection rounded to three decimal places.<br />
(c) Use technology to graph both functions on the same set of<br />
axes. Be sure the graph shows the point of intersection.<br />
101. Intermediate Value Theorem An airplane is travelling at a<br />
speed of 620 miles per hour and then encounters a slight<br />
headwind that slows it to 608 miles per hour. After a few<br />
minutes, the headwind eases and the plane’s speed increases to<br />
614 miles per hour. Explain why the plane’s speed is 610 miles<br />
per hour on at least two different occasions during the flight.<br />
Source: Submitted by the students of Millikin University.<br />
102. Suppose a function f is defined and continuous on the closed<br />
interval [a, b]. Is the function h(x) = 1 also continuous on<br />
f (x)<br />
the closed interval [a, b]? Discuss the continuity of h<br />
on [a, b].<br />
103. Given the two functions f and h:<br />
f (x)<br />
f (x) = x 3 − 3x 2 if x = 3<br />
− 4x + 12 h(x) = x − 3<br />
p if x = 3<br />
(a) Find all the zeros of the function f .<br />
(b) Find the number p so that the function h is continuous at<br />
x = 3. Justify your answer.<br />
(c) Determine whether h, with the number found in (b), is even,<br />
odd, or neither. Justify your answer.<br />
104. The function f (x) = |x| is not defined at 0. Explain why it is<br />
x<br />
impossible to define f (0) so that f is continuous at 0.<br />
f<br />
105. Find two functions f and g that are each continuous at c, yet<br />
g<br />
is not continuous at c.<br />
106. Discuss the difference between a discontinuity that is removable<br />
and one that is nonremovable. Give an example of each.<br />
Bisection Method for Approximating Zeros of a Function Suppose<br />
the Intermediate Value Theorem indicates that a function f has a zero<br />
in the interval (a, b). The bisection method approximates the zero by<br />
evaluating f at the midpoint m 1 of the interval (a, b). If f (m 1) = 0,<br />
then m 1 is the zero we seek and the process ends. If f (m 1) = 0, then<br />
the sign of f (m 1) is opposite that of either f (a) or f (b) (but not both),<br />
and the zero lies in that subinterval. Evaluate f at the midpoint m 2 of<br />
this subinterval. Continue bisecting the subinterval containing the zero<br />
until the desired degree of accuracy is obtained.<br />
In Problems 107–114, use the bisection method three times to<br />
approximate the zero of each function in the given interval.<br />
107. f (x) = x 3 + 3x − 5; interval: [1, 2]<br />
108. f (x) = x 3 − 4x + 2; interval: [1, 2]<br />
109. f (x) = 2x 3 + 3x 2 + 4x − 1; interval: [0, 1]<br />
110. f (x) = x 3 − x 2 − 2x + 1; interval: [0, 1]<br />
111. f (x) = x 3 − 6x − 12; interval: [3, 4]<br />
85. A = 5 , B = 7<br />
80<br />
60<br />
40<br />
20<br />
y<br />
5 5 10 x<br />
86.<br />
1<br />
k =<br />
6<br />
87. (a) { x| x< 2} ∪ { x| x> 2, x ≠8}<br />
(b) f is discontinuous at x = 8 and on the<br />
interval [ − 2,2] .<br />
(c) The discontinuity at x = 8 is removable.<br />
88. (a) Since f is continuous on [0, π ], f (0) < 0,<br />
and f ( π ) > 0, the IVT guarantees that f has a<br />
zero on the interval (0, π ).<br />
(b) x ≈ 2.180<br />
89. (a) Since f is continuous<br />
on [0,2], f (0) < 0, and f (2) > 0, the<br />
IVT guarantees that f must have a zero<br />
on the interval (0,2).<br />
(b) x ≈ 0.443<br />
90. 2<br />
91. 1.732<br />
92. 2.219<br />
93. 1.561<br />
94. Graphs will vary. Sample graph:<br />
y<br />
4<br />
2<br />
22<br />
24<br />
2 4 6 8 10 12<br />
Does not contradict IVT because IVT<br />
provides no information when endpoint<br />
values have the same sign.<br />
95. Graphs will vary. Sample graph:<br />
y<br />
3<br />
2<br />
1<br />
22 21<br />
21<br />
1 2<br />
Does not contradict IVT. f ( −1) and f (2) are<br />
both positive, so IVT gives no information<br />
about zeros in the interval ( −1,2).<br />
96. Graphs will vary. Sample graph:<br />
y<br />
10<br />
5<br />
23 22 21<br />
25<br />
210<br />
215<br />
220<br />
1 2 3<br />
Does not contradict IVT, which guarantees<br />
that f must have at least 1 zero in the<br />
interval ( −2,3).<br />
97. Graphs will vary. Sample graph:<br />
25 24 23 22<br />
y<br />
20<br />
15<br />
10<br />
5<br />
21 1<br />
25<br />
210<br />
Does not contradict IVT, which guarantees<br />
that f must have at least 1 zero on the<br />
interval ( −5,0).<br />
98. (a) Since f ( −2) and f (2) are both<br />
positive, IVT gives no information<br />
about whether the function has a zero<br />
in the interval ( −2,2).<br />
Answers continue on p. 116<br />
x<br />
x<br />
x<br />
x<br />
Section 1.3 • Assess Your Understanding<br />
115<br />
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<strong>Sullivan</strong><br />
116 Chapter 1 • Limits and Continuity<br />
y<br />
3<br />
2<br />
1<br />
22 21<br />
21<br />
1 2<br />
The graph indicates that f has zeros<br />
at x =− 1 and x = 1 in the interval<br />
( −2,2).<br />
99. (a) Since f ( −2) and f (2) are both<br />
positive, there is no guarantee that<br />
the function has a zero in the interval<br />
( −2,2).<br />
(b)<br />
y<br />
2<br />
1<br />
x<br />
112. f (x) = 3x 3 + 5x − 40; interval: [2, 3]<br />
113. f (x) = x 4 − 2x 3 + 21x − 23; interval [1, 2]<br />
114. f (x) = x 4 − x 3 + x − 2; interval: [1, 2]<br />
115. Intermediate Value Theorem Use the Intermediate Value<br />
<br />
Theorem to show that the function f (x) = x 2 + 4x − 2 has a<br />
zero in the interval [0, 1]. Then approximate the zero correct to<br />
one decimal place.<br />
116. Intermediate Value Theorem Use the Intermediate Value<br />
Theorem to show that the function f (x) = x 3 − x + 2 has a zero<br />
in the interval [−2, 0]. Then approximate the zero correct to two<br />
decimal places.<br />
117. Continuity of a Sum If f and g are each continuous at c,<br />
prove that f + g is continuous at c.(Hint: Use the Limit of a<br />
Sum Property.)<br />
118. Intermediate Value Theorem Suppose that the functions f<br />
and g are continuous on the interval [a, b]. If f (a) g(b), prove that the graphs of y = f (x) and<br />
y = g(x) intersect somewhere between x = a and x = b.<br />
[Hint: Define h(x) = f (x) − g(x) and show h(x) = 0 for some<br />
x between a and b.]<br />
Challenge Problems<br />
119. Intermediate Value Theorem Let f (x) = 1<br />
x − 1 + 1<br />
x − 2 .<br />
Use the Intermediate Value Theorem to prove that there is a real<br />
number c between 1 and 2 for which f (c) = 0.<br />
120. Intermediate Value Theorem Prove that there is a real<br />
number c between 2.64 and 2.65 for which c 2 = 7.<br />
f (a + h) − f (a)<br />
121. Show that the existence of lim<br />
implies f is<br />
h→0 h<br />
continuous at x = a.<br />
122. Find constants A, B, C, and D so that the function below is<br />
continuous for all x. Sketch the graph of the resulting function.<br />
⎧<br />
x 2 + x − 2<br />
if −∞ < x < 1<br />
x − 1<br />
⎪⎨ A if x = 1<br />
f (x) =<br />
B (x − C) 2 if 1 < x < 4<br />
D<br />
⎪⎩<br />
if x = 4<br />
2x − 8 if 4 < x < ∞<br />
123. Let f be a function for which 0 ≤ f (x) ≤ 1 for all x in [0, 1].<br />
If f is continuous on [0, 1], show that there exists at least one<br />
number c in [0, 1] such that f (c) = c.<br />
[Hint: Let g(x) = x − f (x).]<br />
2 1 1 2<br />
x<br />
The graph indicates that f does not<br />
have a zero in the interval ( −2,2).<br />
100. If the functions intersect, then any<br />
point of intersection must be a solution<br />
3 2<br />
of fx ( ) = x + x − 1=<br />
0. This f is<br />
continuous on [0,1] , with f (0) < 0 and<br />
f (1) > 0, so IVT guarantees a solution,<br />
and hence an intersection of the two<br />
original functions, on (0,1).<br />
(b) (0.755,0.430)<br />
(c)<br />
y<br />
1.0<br />
0.8<br />
0.6<br />
0.4<br />
0.2<br />
101. See TSM.<br />
(0.755, 0.430)<br />
0.2 0.4 0.6 0.8 1.0<br />
102. h is continuous anywhere on<br />
[ ab , ] except where fx ( ) = 0 . If<br />
fx ( ) ≠ 0 anywhere on [ ab , ] , then<br />
h is continuous everywhere on that<br />
interval.<br />
103. (a) x =− 2, x = 2, x = 3<br />
(b) p = 5<br />
(c) h is an even function.<br />
104. This is because lim fx ( ) does not<br />
x → 0<br />
exist (because lim fx ( ) ≠ lim fx ( )).<br />
−<br />
x→0 x→+<br />
0<br />
105. Answers will vary. Sample answer:<br />
fx ( ) = x<br />
2 −1,<br />
gx ( ) = x − 3, c = 3.<br />
106. Answers will vary. Sample answer:<br />
2<br />
x −9<br />
fx ( ) = has a removable<br />
x −3<br />
discontinuity at x = 3, and<br />
⎧<br />
⎪ x+ 3, x≤3<br />
gx ( ) = ⎨<br />
has a<br />
2<br />
⎩⎪ 9 − x , x><br />
3<br />
nonremovable discontinuity at x = 3.<br />
x<br />
AP® Practice Problems<br />
PAGE<br />
103 1. The graph of a function f is shown below. Where on the open<br />
interval (−3, 5) is f discontinuous?<br />
23<br />
21<br />
(A) 3 only (B) −1 and 3 only<br />
(C) 1 only (D) −1, 2, and 3<br />
y<br />
3<br />
1<br />
y 5 f (x)<br />
2 3 5 x<br />
PAGE<br />
103 2. The graph of a function f is shown below.<br />
22 21<br />
y<br />
3<br />
1<br />
y 5 f (x)<br />
1 2 3 4<br />
If lim x→c<br />
f (x) exists and if f is not continuous at c, then c =<br />
(A) −1 (B) 1 (C) 2 (D) 3<br />
107. (1.125,1.25) 108. (1.625,1.75)<br />
109. (0.125,0.25) 110. (0.375,0.5)<br />
111. (3.125,3.25) 112. (2.125,2.25)<br />
113. (1.125,1.25) 114. (1.25,1.375)<br />
115. Since f is continuous on [0, 1],<br />
f (0) < 0, and f (1) > 0, IVT guarantees that f<br />
has a zero on (0, 1). Correct to one decimal<br />
place, the zero is x = 0.8.<br />
116. Since f is continuous on [−2, 0], f ( − 2) < 0,<br />
and f (0) > 0, IVT guarantees that f has a zero<br />
on ( −2,0). Correct to two decimal places, the<br />
zero is x =− 1.52.<br />
117. See TSM.<br />
118. See TSM.<br />
119. See TSM.<br />
120. See TSM.<br />
x<br />
PAGE<br />
104 3. If the function f (x) = x2 − 25<br />
then f (−5) =<br />
x + 5<br />
is continuous at −5,<br />
(A) −10 (B) −5 (C) 0 (D) 10<br />
⎧ √ √<br />
⎨ 2x + 5 − x + 15<br />
if x = 10<br />
PAGE<br />
105 4. If f (x) = x − 10<br />
⎩<br />
k if x = 10<br />
and if f is continuous at x = 10, then k =<br />
1<br />
(A) 0 (B) (C) 1 (D) 10<br />
10<br />
PAGE<br />
105 5. If lim f (x) = L, where L is a real number, which of the<br />
x→c<br />
following must be true?<br />
(A) f is defined at x = c. (B) f is continuous at x = c.<br />
(C) f (c) = L. (D) None of the above.<br />
PAGE<br />
111 6. If f (x) = x 3 − 2x + 5 and if f (c) = 0 for only one real<br />
number c, then c is between<br />
(A) −4 and −2 (B) −2 and −1 (C) −1 and 1 (D) 1 and 3<br />
PAGE<br />
111 7. The function f is continuous at all real numbers, and f (−8) = 3<br />
and f (−1) =−4. If f has only one real zero (root), then which<br />
number x could satisfy f (x) = 0?<br />
(A) −10 (B) −5 (C) 0 (D) 2<br />
121. See TSM.<br />
122. A = 3, B = 1 , C = 4, D = 0.<br />
3<br />
123. See TSM.<br />
y<br />
6<br />
4<br />
2<br />
22 2 4 6<br />
Answers to AP® Practice<br />
Problems<br />
1. D 2. C 3. A 4. B 5. D<br />
6. A 7. B<br />
x<br />
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Section 1.4 • Limits and Continuity of Trigonometric, Exponential, and Logarithmic Functions 117<br />
PAGE<br />
105 8. Let f be the function defined by<br />
⎧<br />
x<br />
⎪⎨<br />
2 if x < 0<br />
√ x if 0 ≤ x < 1<br />
f (x) =<br />
2 − x if 1 ≤ x < 2<br />
⎪⎩<br />
x − 3 if x ≥ 2<br />
For what numbers x is f NOT continuous?<br />
(A) 1 only<br />
(C) 0 and 2 only<br />
(B) 2 only<br />
(D) 1 and 2 only<br />
PAGE<br />
111 9. The function f is continuous on the closed interval [−2, 6].<br />
If f (−2) = 7 and f (6) =−1, then the Intermediate Value<br />
Theorem guarantees that<br />
(A) f (0) = 0.<br />
(B) f (c) = 2 for at least one number c between −2 and 6.<br />
(C) f (c) = 0 for at least one number c between −1 and 7.<br />
(D) −1 ≤ f (x) ≤ 7 for all numbers in the closed<br />
interval [−2, 6].<br />
PAGE<br />
111 10. The function f is continuous on the closed interval [−2, 2].<br />
Several values of the function f are given in the table below.<br />
x −2 0 2<br />
f (x) 3 c 2<br />
The equation f (x) = 1 must have at least two solutions in the<br />
interval [−2, 2] if c =<br />
(A)<br />
1<br />
2<br />
(B) 1 (C) 3 (D) 4<br />
PAGE<br />
105 11. The function f is defined by f (x) =<br />
<br />
x 2 − 2x + 3 if x ≤ 1<br />
−2x + 5 if x > 1<br />
(a) Is f continuous at x = 1?<br />
(b) Use the definition of continuity to explain your answer.<br />
8. B<br />
9. B<br />
10. A<br />
11. (a) No (b) Answers will vary.<br />
1.4 Limits and Continuity of Trigonometric,<br />
Exponential, and Logarithmic Functions<br />
OBJECTIVES When you finish this section, you should be able to:<br />
1 Use the Squeeze Theorem to find a limit (p. 117)<br />
2 Find limits involving trigonometric functions (p. 119)<br />
3 Determine where the trigonometric functions are continuous (p. 122)<br />
4 Determine where an exponential or a logarithmic function is continuous (p. 124)<br />
Until now we have found limits using the basic limits<br />
lim<br />
x→c A = A<br />
lim x = c<br />
x→c<br />
and properties of limits. But there are many limit problems that cannot be found by<br />
directly applying these techniques. To find such limits requires different results, such as<br />
the Squeeze Theorem ∗ , or basic limits involving trigonometric and exponential functions.<br />
TRM Alternate Examples Section 1.4<br />
You can find the Alternate Examples for<br />
this section in PDF format in the Teacher’s<br />
Resource Materials.<br />
Teaching Tip<br />
This section can be taught in conjunction<br />
with Section 1.2 when the students learn<br />
how to find limits analytically (that is,<br />
algebraically).<br />
y<br />
L<br />
c<br />
y h(x)<br />
y g(x)<br />
y f (x)<br />
lim f(x) L, lim h(x) L, lim g(x) L<br />
x→c x→c x→c<br />
x<br />
1 Use the Squeeze Theorem to Find a Limit<br />
To use the Squeeze Theorem to find lim<br />
x→c<br />
g(x), we need to know, or be able to find, two<br />
functions f and h that “sandwich” the function g between them for all x close to c. That<br />
is, in some interval containing c, the functions f, g, and h satisfy the inequality<br />
f (x) ≤ g(x) ≤ h(x)<br />
Then if f and h have the same limit L as x approaches c, the function g is “squeezed”<br />
to the same limit L as x approaches c. See Figure 38.<br />
We state the Squeeze Theorem here. The proof is given in Appendix B.<br />
Figure 38<br />
∗ The Squeeze Theorem is also known as the Sandwich Theorem and the Pinching Theorem.<br />
Section 1.4 • Limits and Continuity of Trigonometric, Exponential, and Logarithmic Functions<br />
117<br />
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<strong>Sullivan</strong><br />
118 Chapter 1 • Limits and Continuity<br />
AP® CaLC skill builder<br />
for example 1<br />
Using the Squeeze Theorem to Find a<br />
Limit<br />
⎛ ⎞<br />
Find lim<br />
⎝<br />
⎜ x cos 1<br />
x→0<br />
x ⎠<br />
⎟ .<br />
Solution<br />
Recall that −1≤cos x ≤ 1. We will use that<br />
fact as a springboard to write inequalities<br />
involving the original function. The original<br />
problem is not simply about cos x, but<br />
⎛ ⎞<br />
about cos 1 ⎝<br />
⎜ x ⎠<br />
⎟ . However, the cosine of any<br />
value is still a number between −1 and 1.<br />
⎛ ⎞<br />
−1≤cos 1 ⎝<br />
⎜ x ⎠<br />
⎟ ≤ 1<br />
We continue to build inequalities with<br />
⎛ ⎞<br />
x cos 1 ⎝<br />
⎜ x ⎠<br />
⎟ in the center. Next, we multiply<br />
all three parts by x.<br />
− x ⎛<br />
≤ x ⎞<br />
cos 1<br />
⎝<br />
⎜ x ⎠<br />
⎟ ≤ x<br />
We have now built inequalities with the<br />
original expression in the center. Now we<br />
take the limit of all three parts.<br />
⎛ ⎞<br />
lim ( −x) ≤ lim xcos 1 ⎝<br />
⎜<br />
⎠<br />
⎟ ≤ lim ( x)<br />
x→0 x→0 x x→0<br />
Finally, use the Squeeze Theorem.<br />
lim ( − x) = 0<br />
x→0<br />
lim ( x) = 0<br />
x→0<br />
⎛ ⎞<br />
Therefore lim x cos 1<br />
x→0<br />
⎝<br />
⎜ x ⎠<br />
⎟ = 0.<br />
Teaching Tip<br />
Often, students do not know how to begin<br />
a Squeeze Theorem problem. Most of the<br />
time, we begin with the fact that sin x or<br />
cos x falls between −1 and 1. Then<br />
consider guiding the students through<br />
building the original problem as shown in<br />
AP ® Calc Skill Builder for Example 1.<br />
y<br />
2<br />
1<br />
y x 2<br />
2 1<br />
1 2 3<br />
1<br />
2<br />
y x 2<br />
y f (x)<br />
x<br />
THEOREM Squeeze Theorem<br />
Suppose the functions f , g, and h have the property that for all x in an open interval<br />
containing c, except possibly at c,<br />
If<br />
then<br />
f (x) ≤ g(x) ≤ h(x)<br />
lim f (x) = lim h(x) = L<br />
x→c x→c<br />
lim<br />
x→c g(x) = L<br />
For example, suppose we wish to find lim f (x), and we know that −x 2 ≤ f (x) ≤ x 2<br />
x→0<br />
for all x = 0. Since lim(−x 2 ) = 0 and lim x 2 = 0, the Squeeze Theorem tells us that<br />
x→0 x→0<br />
lim f (x) = 0. Figure 39 illustrates how f is “squeezed” between y = x 2 and y =−x 2<br />
x→0<br />
near 0.<br />
Figure 39 EXAMPLE 1 Using the Squeeze Theorem to Find a Limit<br />
Use the Squeeze Theorem to find lim<br />
(x sin 1 )<br />
.<br />
x→0 x<br />
g<br />
g<br />
Solution If x = 0, then g(x) = x sin 1 is defined. We seek two functions that “squeeze”<br />
x<br />
g(x) = x sin 1 near 0. Since −1 ≤ sin x ≤ 1 for all x, we begin with the inequality<br />
x ∣ sin 1 x ∣ ≤ 1 x = 0<br />
Since x = 0 and we seek to squeeze g(x) = x sin 1 , we multiply both sides of the<br />
x<br />
inequality by |x|, x = 0. Since |x| > 0, the direction of the inequality is preserved. [Note<br />
that if we multiply<br />
∣ sin 1 x ∣ ≤ 1 by x, we would not know whether the inequality symbol<br />
would remain the same or be reversed since we do not know whether x > 0 or x < 0.]<br />
|x|<br />
∣ sin 1 x ∣ ≤|x| Multiply both sides by |x| > 0.<br />
∣ x sin 1 x ∣ ≤|x|<br />
|a|·|b| =|ab|.<br />
−|x| ≤x sin 1 ≤|x| |a|≤b is equivalent to −b ≤ a ≤ b.<br />
x<br />
Now use the Squeeze Theorem with f (x) = −|x|, g(x) = x sin 1 , and h(x) =|x|.<br />
x<br />
Since f (x) ≤ g(x) ≤ h(x) and<br />
lim<br />
x→0<br />
it follows that<br />
f (x) = lim<br />
x→0<br />
(−|x|) = 0 and lim<br />
x→0<br />
h(x) = lim |x| =0<br />
x→0<br />
(<br />
lim g(x) = lim x · sin 1 )<br />
= 0 ■<br />
x→0 x→0 x<br />
2<br />
p,<br />
p<br />
4 4 3 f 20.6, 0.6g<br />
Figure 40 g(x) = x sin 1 is squeezed<br />
x<br />
Figure 40 illustrates how g(x) = x sin 1 is squeezed between y = −|x| and y =|x|.<br />
x<br />
between f (x) = −|x| and h(x) =|x|. NOW WORK Problem 5 and AP® Practice Problem 11.<br />
TRM AP® Calc Skill Builders<br />
Section 1.4<br />
You can find the AP ® Calc Skill Builders for<br />
this section in PDF format in the Teacher’s<br />
Resource Materials.<br />
118<br />
Chapter 1 • Limits and Continuity<br />
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Section 1.4 • Limits and Continuity of Trigonometric, Exponential, and Logarithmic Functions 119<br />
2 Find Limits Involving Trigonometric Functions<br />
Knowing lim A = A and lim x = c helped us to find the limits of many algebraic<br />
x→c x→c<br />
functions. Knowing several basic trigonometric limits can help to find many limits<br />
involving trigonometric functions.<br />
THEOREM Two Basic Trigonometric Limits<br />
lim<br />
x→0<br />
lim sin x = 0<br />
x→0<br />
lim cos x = 1<br />
x→0<br />
The graphs of y = sin x and y = cos x in Figure 41 suggest that lim sin x = 0 and<br />
x→0<br />
cos x = 1. The proofs of these limits both use the Squeeze Theorem. Problem 63<br />
provides an outline of the proof that lim<br />
x→0<br />
sin x = 0.<br />
sin θ<br />
A third basic trigonometric limit, lim = 1, is important in calculus. In<br />
θ→0 θ<br />
sin θ<br />
Section 1.1, a table suggested that lim = 1. The function f (θ) = sin θ , whose<br />
θ→0 θ<br />
θ<br />
graph is shown in Figure 42, is defined for all real numbers θ = 0. The graph suggests<br />
lim<br />
θ→0<br />
sin θ<br />
θ<br />
= 1.<br />
AP® Exam Tip<br />
Students should memorize this limit to<br />
save time on the exam.<br />
ax<br />
lim sin( ) = a<br />
x→0<br />
x<br />
Consider working with variations of this<br />
problem as well, such as<br />
ax a<br />
lim sin( ) = .<br />
x→0<br />
bx b<br />
These limits can be found using the<br />
Squeeze Theorem, but memorization<br />
will be a big time saver.<br />
y<br />
y<br />
y<br />
1.0<br />
1<br />
y sin x<br />
1<br />
y cos x<br />
0.5<br />
2π 3π<br />
π<br />
2<br />
π<br />
<br />
2<br />
π<br />
2<br />
π<br />
3π<br />
2<br />
2π x<br />
2π<br />
π<br />
π<br />
2π<br />
x<br />
2π π π 2π θ<br />
1<br />
1<br />
0.5<br />
Figure 41<br />
Figure 42 f (θ) = sin θ<br />
θ<br />
THEOREM<br />
If θ is measured in radians, then<br />
sin θ<br />
lim = 1<br />
θ→0 θ<br />
Proof Although sin θ<br />
sin θ<br />
is a quotient, we have no way to divide out θ. To find lim ,<br />
θ<br />
θ→0 + θ<br />
we let θ be a positive acute central angle of a unit circle, as shown in Figure 43(a) on<br />
page 120. Notice that COP is a sector of the circle. We add the point B = (cos θ,0) to<br />
the graph and form triangle BOP. Next we extend the terminal side of angle θ until it<br />
intersects the line x = 1 at the point D, forming a second triangle COD. The x-coordinate<br />
of D is 1. Since the length of the line segment OC is OC = 1, then the y-coordinate of<br />
D is CD = CD = tan θ. So, D = (1, tan θ). See Figure 43(b).<br />
OC<br />
Section 1.4 • Limits and Continuity of Trigonometric, Exponential, and Logarithmic Functions<br />
119<br />
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<strong>Sullivan</strong><br />
120 Chapter 1 • Limits and Continuity<br />
x 2 y 2 1<br />
y<br />
1<br />
P (cos θ, sin θ)<br />
x 2 y 2 1<br />
y<br />
1<br />
D (1, tan θ)<br />
P<br />
1<br />
O<br />
θ<br />
C (1, 0)<br />
1 x<br />
1<br />
θ<br />
O<br />
B (cos θ, 0)<br />
1<br />
C<br />
x<br />
1<br />
1<br />
Figure 43<br />
π<br />
(a) 0 θ <br />
2<br />
(b)<br />
We see from Figure 43(b) that<br />
area of triangle BOP ≤ area of sector COP ≤ area of triangle COD (1)<br />
Each of these areas can be expressed in terms of θ as<br />
• area of triangle BOP = 1 2 cos θ · sin θ area of a triangle = 1 base × height<br />
2<br />
• area of sector COP = θ 2<br />
• area of triangle COD = 1 tan θ<br />
· 1 · tan θ =<br />
2 2<br />
area of a sector = 1 2 r 2 θ; r = 1<br />
area of a triangle = 1 base × height<br />
2<br />
So, for 0
<strong>Sullivan</strong> AP˙<strong>Sullivan</strong>˙Chapter01 October 8, 2016 17:4<br />
Section 1.4 • Limits and Continuity of Trigonometric, Exponential, and Logarithmic Functions 121<br />
Figure 44<br />
So,<br />
sin θ sin (−θ) sin θ<br />
lim = lim = lim = 1<br />
θ→0 − θ θ→0 + −θ θ→0 + θ<br />
sin θ<br />
It follows that lim = 1. ■<br />
θ→0 θ<br />
sin θ<br />
Since lim = 1, the ratio sin θ is close to 1 for values of θ close to 0. That is,<br />
θ→0 θ<br />
θ<br />
sin θ ≈ θ for values of θ close to 0. Figure 44 illustrates this property.<br />
sin θ<br />
The basic limit lim = 1 can be used to find the limits of similar expressions.<br />
θ→0 θ<br />
EXAMPLE 2<br />
Find:<br />
(a) lim<br />
θ→0<br />
sin(3θ)<br />
θ<br />
Finding the Limit of a Trigonometric Function<br />
(b) lim<br />
θ→0<br />
sin(5θ)<br />
sin(2θ)<br />
sin(3θ)<br />
sin θ<br />
Solution (a) Since lim is not in the same form as lim , we multiply the<br />
θ→0 θ<br />
θ→0 θ<br />
numerator and the denominator by 3, and make the substitution t = 3θ.<br />
(<br />
sin(3θ) 3 sin(3θ)<br />
lim = lim<br />
lim 3 sin t )<br />
sin t<br />
= 3 lim = (3)(1) = 3<br />
θ→0 θ θ→0 3θ<br />
t<br />
t→0 t ↑<br />
=<br />
↑ t→0<br />
t = 3θ<br />
t → 0 as θ → 0<br />
sin t<br />
lim<br />
t→0 t<br />
(b) We begin by dividing the numerator and the denominator by θ. Then<br />
sin(5θ)<br />
sin(5θ)<br />
sin(2θ) = θ<br />
sin(2θ)<br />
θ<br />
Now we follow the approach in (a) on the numerator and on the denominator.<br />
sin(5θ)<br />
lim = lim<br />
θ→0 θ<br />
sin(2θ)<br />
lim = lim<br />
θ→0 θ<br />
5 sin(5θ)<br />
θ→0 5 θ<br />
2 sin(2θ)<br />
θ→0 2θ<br />
lim<br />
=<br />
↑ t→0<br />
t = 5θ<br />
t → 0 as θ → 0<br />
=<br />
↑ t→0<br />
t = 2θ<br />
t → 0 as θ → 0<br />
5 sin t<br />
t<br />
( 2 sin t<br />
lim<br />
t<br />
= 1<br />
( ) sin t<br />
= 5 lim = 5<br />
t→0 t<br />
)<br />
sin t<br />
= 2 lim = 2<br />
t→0 t<br />
sin(5θ)<br />
sin(5θ)<br />
lim<br />
lim<br />
θ→0 sin(2θ) = θ→0 θ<br />
= 5 sin(2θ) 2 ■<br />
lim<br />
θ→0 θ<br />
NOW WORK Problems 23 and 25 and AP® Practice Problems 1, 2, 4, 6, 7, and 10.<br />
AP® CaLC skill builder<br />
for example 3<br />
Finding the Limit of a Trigonometric<br />
Function<br />
⎛ (cot x−<br />
csc x)sinx<br />
⎞<br />
Find lim<br />
x→0⎝<br />
⎜ x ⎠<br />
⎟<br />
Solution<br />
To find this limit, we first rewrite the<br />
trigonometric expressions in terms of<br />
sines and cosines. For x ≠ 0,<br />
(cot x−<br />
csc x)sinx<br />
x<br />
⎛ cos x 1 ⎞<br />
−<br />
⎝<br />
⎜ x x⎠<br />
⎟ sin x<br />
sin sin<br />
=<br />
x<br />
cosx<br />
−1<br />
=<br />
x<br />
Since we know<br />
x −<br />
lim cos 1 = 0, then<br />
x→0<br />
x<br />
x−<br />
x x<br />
lim (cot csc )sin = 0.<br />
x→0<br />
x<br />
Example 3 establishes an important limit used in Chapter 2.<br />
EXAMPLE 3<br />
AP® CaLC skill builder<br />
for example 2<br />
Establish the formula<br />
Finding the Limit of a Trigonometric<br />
Function<br />
x<br />
Find lim sin(3 )<br />
xcos(4 x) .<br />
x→0<br />
where θ is measured in radians.<br />
Solution<br />
x<br />
x<br />
lim sin(3 ) lim sin(3 ) 1<br />
= ⋅lim<br />
x→0 xcos(4 x)<br />
x→0 x x→0<br />
cos(4 x)<br />
x<br />
= lim 3sin(3 ) ⋅limsec(4 x)<br />
x→0 3x<br />
x→0<br />
x<br />
= 3 lim sin(3 ) ⋅sec(0)<br />
x→0<br />
3x<br />
= 311 ⋅ ⋅ = 3<br />
Finding a Basic Trigonometric Limit<br />
cos θ − 1<br />
lim = 0<br />
θ→0 θ<br />
AP® CaLC skill builder<br />
for example 3<br />
Finding the Limit of a Trigonometric<br />
Function<br />
x x+<br />
x<br />
Find lim 2 csc 2<br />
csc .<br />
x→0<br />
2<br />
xcsc<br />
x<br />
Solution<br />
x x+<br />
x<br />
lim 2 csc 2<br />
csc<br />
x→0<br />
2<br />
xcsc<br />
x<br />
⎛ 1 ⎞<br />
= lim +<br />
→ ⎝<br />
⎜ 2<br />
x 0 xcsc<br />
x⎠<br />
⎟<br />
⎛ sin x ⎞<br />
= lim +<br />
⎝<br />
⎜ 2<br />
⎠<br />
⎟ = 2 + 1 = 3<br />
x→0<br />
x<br />
Section 1.4 • Limits and Continuity of Trigonometric, Exponential, and Logarithmic Functions<br />
121<br />
TE_<strong>Sullivan</strong>_Chapter01_PART II.indd 4<br />
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Teaching Tip<br />
Alternate Example<br />
Finding the Limit of a Trigonometric<br />
Function<br />
x x+<br />
x<br />
Find lim 2 csc 2<br />
csc .<br />
x→0<br />
2<br />
xcsc<br />
x<br />
Solution<br />
This time, we rewrite in terms of sines:<br />
x x+<br />
x<br />
lim 2 csc 2<br />
csc<br />
x→0<br />
2<br />
xcsc<br />
x<br />
⎛ 2x<br />
1 ⎞<br />
⎜ 2 ⎟<br />
= lim sin x<br />
⎜ + sin x<br />
⎟<br />
x→0<br />
⎜<br />
x x<br />
⎟<br />
⎝<br />
2 2<br />
sin x sin x ⎠<br />
⎛ x x x ⎞<br />
= lim ⎜<br />
2 sin 2 2<br />
sin<br />
+<br />
x→0<br />
2 ⎟<br />
⎝ xsin<br />
x xsin<br />
x⎠<br />
⎛ sin x ⎞<br />
= lim +<br />
⎝<br />
⎜ 2<br />
⎠<br />
⎟ = 2 + 1 = 3<br />
x→0<br />
x<br />
Chapter objective 3: Determine where the<br />
trigonometric functions are continuous is a<br />
good review. This knowledge is a calculus<br />
prerequisite, so if the students are well<br />
prepared, you may be able to address this<br />
objective briefly to save time.<br />
AP® CaLC skill builder<br />
for example 4<br />
Finding the Limit of a Trigonometric<br />
Function<br />
<strong>Sullivan</strong> AP˙<strong>Sullivan</strong>˙Chapter01 October 8, 2016 17:4<br />
122 Chapter 1 • Limits and Continuity<br />
2π π π π<br />
<br />
π<br />
2 2<br />
y<br />
1<br />
1<br />
y<br />
1<br />
1<br />
y cos x<br />
y sin x<br />
3π<br />
2<br />
2π π π π π 3π<br />
<br />
2π x<br />
2 2<br />
2<br />
Figure 45<br />
x<br />
<strong>Sullivan</strong><br />
Solution First we rewrite the expression cos θ − 1 as the product of two terms whose<br />
θ<br />
limits are known. For θ = 0,<br />
( )( )<br />
cos θ − 1 cos θ − 1 cos θ + 1<br />
=<br />
θ<br />
θ cos θ + 1<br />
= cos2 θ − 1<br />
θ(cos θ + 1) = − sin 2 ( )<br />
θ sin θ (− sin θ)<br />
↑ θ(cos θ + 1) = θ cos θ + 1<br />
sin 2 θ+ cos 2 θ = 1<br />
Now we find the limit.<br />
[( )( )] [ ][<br />
]<br />
cos θ − 1 sin θ (− sin θ)<br />
sin θ − sin θ<br />
lim = lim<br />
= lim lim<br />
θ→0 θ θ→0 θ cos θ + 1 θ→0 θ θ→0 cos θ + 1<br />
lim(− sin θ)<br />
θ→0<br />
= 1 ·<br />
lim (cos θ + 1) = 0 2 = 0<br />
■<br />
θ→0<br />
NOW WORK Problem 31 and AP® Practice Problem 8.<br />
3 Determine Where the Trigonometric Functions<br />
Are Continuous<br />
The graphs of f (x) = sin x and g(x) = cos x in Figure 45 suggest that f and g are<br />
continuous on their domains, the set of all real numbers.<br />
EXAMPLE 4 Showing f(x) = sin x Is Continuous at 0<br />
• f (0) = sin 0 = 0, so f is defined at 0.<br />
• lim f (x) = lim sin x = 0, so the limit at 0 exists.<br />
x→0 x→0<br />
• lim sin x = sin 0 = 0.<br />
x→0<br />
Since all three conditions of continuity are satisfied, f (x) = sin x is continuous at 0. ■<br />
In a similar way, we can show that g(x) = cos x is continuous at 0. That is,<br />
lim cos x = cos 0 = 1.<br />
x→0<br />
You are asked to prove the following theorem in Problem 67.<br />
THEOREM<br />
• The sine function y = sin x is continuous on its domain, all real numbers.<br />
• The cosine function y = cos x is continuous on its domain, all real numbers.<br />
Based on this theorem the following two limits can be added to the list of basic<br />
limits.<br />
lim sin x = sin c for all real numbers c<br />
x→c<br />
lim cos x = cos c for all real numbers c<br />
x→c<br />
Following are some function values for two<br />
continuous functions, f and g.<br />
x fx ( ) gx ( )<br />
1 2 −1<br />
2 1 0<br />
sin( gx ( )) + fx ( )<br />
Find lim .<br />
x→2<br />
gf (( x))<br />
Solution<br />
Here we will use the continuity of f, g, and<br />
the sine function to find the limit.<br />
sin( gx ( )) + fx ( )<br />
lim<br />
x→2<br />
gf (( x))<br />
∑ Mathematical Practices Tip<br />
MPAC 1: Reasoning with Definitions and<br />
Theorems<br />
Use the continuity of a function to find a limit.<br />
Ask students to use the definition of continuity<br />
to explain why lim cosx<br />
=−1.<br />
Have students<br />
x→π<br />
explain by using a graph how the continuity<br />
of a function f at a point x = c shows that<br />
lim f( x) = f( c).<br />
x→c<br />
sin( g(2)) + f(2)<br />
= lim<br />
x→2<br />
gf ((2))<br />
+ +<br />
= lim sin(0) 1 0 1 =<br />
− =− 1<br />
x→2<br />
g(1)<br />
1<br />
122<br />
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Section 1.4 • Limits and Continuity of Trigonometric, Exponential, and Logarithmic Functions 123<br />
Using the facts that the sine and cosine functions are continuous for all real numbers,<br />
we can use basic trigonometric identities to determine where the remaining four<br />
trigonometric functions are continuous:<br />
• y = tan x: Since tan x = sin x , from the Continuity of a Quotient property,<br />
cos x<br />
y = tan x is continuous at all real numbers except those for which cos x = 0.<br />
That is, y = tan x is continuous on its domain, all real numbers except odd<br />
multiples of π 2 .<br />
• y = sec x: Since sec x = 1 , from the Continuity of a Quotient property,<br />
cos x<br />
y = sec x is continuous at all real numbers except those for which cos x = 0.<br />
That is, y = sec x is continuous on its domain, all real numbers except odd<br />
multiples of π 2 .<br />
• y = cot x: Since cot x = cos x , from the Continuity of a Quotient property,<br />
sin x<br />
y = cot x is continuous at all real numbers except those for which sin x = 0.<br />
That is, y = cot x is continuous on its domain, all real numbers except integer<br />
multiples of π.<br />
• y = csc x: Since csc x = 1 , from the Continuity of a Quotient property,<br />
sin x<br />
y = csc x is continuous at all real numbers except those for which sin x = 0.<br />
That is, y = csc x is continuous on its domain, all real numbers except integer<br />
multiples of π.<br />
Teaching Tip<br />
The abbreviation of the function as “cos”<br />
is derived from the word “cosine,” which is<br />
also an abbreviation—of the longer name<br />
“complement of sine.” In a right triangle,<br />
the cosine of an angle is the same as the<br />
sine of the complementary angle. The<br />
same explanation holds for cotangent and<br />
cosecant. The cotangent of an angle is the<br />
same as the tangent of the complementary<br />
angle, and the cosecant of an angle is the<br />
same as the secant of the complementary<br />
angle.<br />
NEED TO REVIEW? Inverse<br />
trigonometric functions are discussed in<br />
Section P.7, pp. 61--66.<br />
Recall that a one-to-one function that is continuous on its domain has an inverse<br />
function that is continuous on its domain.<br />
Since each of the six trigonometric functions is continuous on its domain, then each<br />
is continuous on the restricted domain used to define its inverse trigonometric function.<br />
This means the inverse trigonometric functions are continuous on their domains. These<br />
results are summarized in Table 10.<br />
TABLE 10 Continuity of the Trigonometric Functions and Their Inverses<br />
Function Domain Properties<br />
Sine all real numbers continuous on the interval (−∞, ∞)<br />
Cosine all real numbers continuous on the interval (−∞, ∞)<br />
{<br />
Tangent<br />
x|x = odd multiples of π }<br />
continuous at all real numbers except odd multiples of π 2<br />
2<br />
Cosecant {x|x = multiples of π} continuous at all real numbers except multiples of π<br />
{<br />
Secant<br />
x|x = odd multiples of π }<br />
continuous at all real numbers except odd multiples of π 2<br />
2<br />
Cotangent {x|x = multiples of π} continuous at all real numbers except multiples of π<br />
Inverse sine −1 ≤ x ≤ 1 continuous on the closed interval [−1, 1]<br />
Inverse cosine −1 ≤ x ≤ 1 continuous on the closed interval [−1, 1]<br />
Inverse tangent all real numbers continuous on the interval (−∞, ∞)<br />
Inverse cosecant |x| ≥1 continuous on the set (−∞, −1] ∪ [1, ∞)<br />
Inverse secant |x| ≥1 continuous on the set (−∞, −1] ∪ [1, ∞)<br />
Inverse cotangent all real numbers continuous on the interval (−∞, ∞)<br />
Section 1.4 • Limits and Continuity of Trigonometric, Exponential, and Logarithmic Functions<br />
123<br />
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<strong>Sullivan</strong><br />
124 Chapter 1 • Limits and Continuity<br />
Teaching Tip<br />
Section objective 4: Determine where<br />
an exponential or a logarithmic function<br />
is continuous is a good review. This<br />
knowledge is a calculus prerequisite, so if<br />
the students are well prepared, you may<br />
be able to address this objective briefly to<br />
save time.<br />
4 Determine Where an Exponential or a Logarithmic<br />
Function Is Continuous<br />
The graphs of an exponential function y = a x and its inverse function y = log a x are<br />
shown in Figure 46. The graphs suggest that an exponential function and a logarithmic<br />
function are continuous on their domains. We state the following theorem without<br />
proof.<br />
y a x<br />
1<br />
(1, )<br />
a<br />
y<br />
3<br />
(0, 1)<br />
(a, 1)<br />
(1, a)<br />
y x<br />
y y a x<br />
3<br />
(1, a)<br />
(0, 1) 1<br />
(1, )<br />
a<br />
(a, 1)<br />
(1, 0)<br />
y x<br />
y log a x<br />
Students should be familiar with the<br />
graphs of y = e x and y = ln x, so that<br />
they can easily evaluate the following<br />
limits:<br />
x<br />
lim e = 0<br />
x→−∞<br />
x<br />
lim e = 1<br />
x→0<br />
x<br />
lim e<br />
x→∞<br />
=∞<br />
lim lnx<br />
=−∞<br />
+<br />
x→0<br />
lim lnx<br />
= 0<br />
x→1<br />
lim lnx<br />
=∞<br />
x→∞<br />
AP® Exam Tip<br />
AP® CaLC skill builder<br />
for example 5<br />
Show a Composite Function Is<br />
Continuous<br />
Which of the following functions is (are)<br />
continuous at x = 0?<br />
2<br />
x /2<br />
I. fx ( ) = e<br />
II. gx ( ) =<br />
3 cos x<br />
III. hx ( ) = e −2x<br />
tanx<br />
Solution<br />
2<br />
x /2<br />
I. fx ( ) = e is continuous for all real<br />
numbers, so the function is<br />
continuous at x = 0.<br />
II. gx ( ) = 3 cos x is continuous on its<br />
domain, the set of all real numbers.<br />
Since the composition of two<br />
continuous functions is a continuous<br />
function, g is continuous at x = 0.<br />
III. The function hx ( ) = e −2x<br />
tanx<br />
is<br />
defined and continuous on<br />
π π<br />
− < x < , so h is continuous at<br />
2 2<br />
x = 0.<br />
CALC<br />
CLIP<br />
3 (1, 0)<br />
3 x<br />
Figure 46<br />
3<br />
(a) 0 a 1<br />
1<br />
( , 1)<br />
a<br />
y log a x<br />
3 3<br />
3<br />
(b) a 1<br />
1<br />
( , 1)<br />
a<br />
THEOREM Continuity of Exponential and Logarithmic Functions<br />
• An exponential function is continuous on its domain, all real numbers.<br />
• A logarithmic function is continuous on its domain, all positive real numbers.<br />
Based on this theorem, the following two limits can be added to the list of basic<br />
limits.<br />
lim<br />
x→c ax = a c for all real numbers c, a > 0, a = 1<br />
and<br />
lim log<br />
x→c<br />
a x = log a c for any real number c > 0, a > 0, a = 1<br />
EXAMPLE 5<br />
Show that:<br />
Showing a Composite Function Is Continuous<br />
(a) f (x) = e 2x is continuous for all real numbers.<br />
(b) F(x) = 3√ ln x is continuous for x > 0.<br />
Solution (a) The domain of the exponential function is the set of all real numbers, so<br />
f is defined for any number c. That is, f (c) = e 2c . Also for any number c,<br />
[<br />
lim f (x) = lim<br />
x→c x→c e2x = lim(e x ) 2 = lim ex] 2<br />
= (e c ) 2 = e 2c = f (c)<br />
x→c x→c<br />
Since lim f (x) = f (c) for any number c, then f is continuous at all numbers c.<br />
x→c<br />
(b) The logarithmic function f (x) = ln x is continuous on its domain, the set of<br />
all positive real numbers. The function g(x) = 3√ x is continuous on its domain, the<br />
set of all real numbers. Then for any real number c > 0, the composite function<br />
F(x) = (g ◦ f )(x) = 3√ ln x is continuous at c. That is, F is continuous at all real<br />
numbers x > 0. ■<br />
x<br />
124<br />
Chapter 1 • Limits and Continuity<br />
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Section 1.4 • Assess Your Understanding 125<br />
Figures 47(a) and 47(b) illustrate the graphs of f and F.<br />
y<br />
8<br />
4<br />
y<br />
1<br />
1<br />
2<br />
4<br />
x<br />
Must-Do Problems for<br />
Exam Readiness<br />
AB: 9–45 odd, AP ® Practice Problems<br />
BC: 11, 23, 25, 35, 43, 45, 55, all AP ®<br />
Practice Problems<br />
Summary Basic Limits<br />
• lim<br />
x→c<br />
A = A, where A is a constant<br />
• lim sin x = sin c • lim cos x = cos c<br />
x→c x→c<br />
• lim a x = a c , a > 0, a = 1<br />
x→c<br />
1.4 Assess Your Understanding<br />
Concepts and Vocabulary<br />
1. lim x→0<br />
sin x =<br />
Figure 47<br />
• lim<br />
x→c<br />
x = c<br />
2 2<br />
(a) f (x) e 2x<br />
sin θ<br />
cos θ − 1<br />
• lim = 1 • lim = 0<br />
θ→0 θ<br />
θ→0 θ<br />
• lim<br />
x→c<br />
log a x = log a c, c > 0, a > 0, a = 1<br />
cos x − 1<br />
2. True or False lim = 1<br />
x→0 x<br />
3. The Squeeze Theorem states that if the functions f, g, and h<br />
have the property f (x) ≤ g(x) ≤ h(x) for all x in<br />
an open interval containing c, except possibly at c, and<br />
if lim f (x) = lim h(x) = L, then lim g(x) = .<br />
x→c x→c x→c<br />
4. True or False f (x) = csc x is continuous for all real numbers<br />
except x = 0.<br />
Skill Building<br />
In Problems 5–8, use the Squeeze Theorem to find each limit.<br />
PAGE<br />
118 5. Suppose −x 2 + 1 ≤ g(x) ≤ x 2 + 1 for all x in an open<br />
interval containing 0. Find lim g(x).<br />
x→0<br />
6. Suppose −(x − 2) 2 − 3 ≤ g(x) ≤ (x − 2) 2 − 3 for all x<br />
in an open interval containing 2. Find lim x→2<br />
g(x).<br />
7. Suppose cos x ≤ g(x) ≤ 1 for all x in an open interval<br />
containing 0. Find lim x→0<br />
g(x).<br />
8. Suppose −x 2 + 1 ≤ g(x) ≤ sec x for all x in an open interval<br />
containing 0. Find lim x→0<br />
g(x).<br />
In Problems 9–22, find each limit.<br />
9. lim x→0<br />
(x 3 + sin x) 10. lim x→0<br />
(x 2 − cos x)<br />
x<br />
2<br />
3<br />
(b) F(x) ln x<br />
NOW WORK Problem 45 and AP® Practice Problems 3, 5, and 9.<br />
11. lim (cos x + sin x) 12. lim (sin x − cos x)<br />
x→π/3 x→π/3<br />
13.<br />
cos x<br />
sin x<br />
lim<br />
14. lim<br />
x→0 1 + sin x<br />
x→0 1 + cos x<br />
15.<br />
3<br />
e x − 1<br />
lim x→0 1 + e x 16. lim x→0 1 + e x<br />
17. lim(e x sin x) 18. lim(e −x tan x)<br />
x→0 x→0<br />
( ) e x<br />
( x<br />
)<br />
19. lim ln<br />
20. lim ln<br />
x→1 x<br />
x→1 e x<br />
e 2x<br />
1 − e x<br />
21. lim x→0 1 + e x 22. lim x→0 1 − e 2x<br />
In Problems 23–34, find each limit.<br />
PAGE<br />
121 23.<br />
sin(7x)<br />
lim x→0 x<br />
PAGE<br />
121 25.<br />
θ + 3 sin θ<br />
lim θ→0 2θ<br />
27.<br />
sin θ<br />
lim θ→0 θ + tan θ<br />
29.<br />
5<br />
lim θ→0 θ · csc θ<br />
PAGE<br />
122 31.<br />
1 − cos 2 θ<br />
lim θ→0 θ<br />
24. lim x→0<br />
sin x 3<br />
x<br />
26.<br />
2x − 5 sin(3x)<br />
lim x→0 x<br />
28.<br />
tan θ<br />
lim θ→0 θ<br />
30.<br />
sin(3θ)<br />
lim θ→0 sin(2θ)<br />
[<br />
33. lim(θ · cot θ) 34. lim sin θ<br />
θ→0 θ→0<br />
cos(4θ)− 1<br />
32. lim θ→0 2θ<br />
( )]<br />
cot θ − csc θ<br />
θ<br />
TRM Full Solutions to Section<br />
1.4 Problems and AP® Practice<br />
Problems<br />
Answers to Section 1.4<br />
Problems<br />
1. 0<br />
2. False.<br />
3. L<br />
4. False.<br />
5. 1<br />
6. −3<br />
7. 1<br />
8. 1<br />
9. 0<br />
10. −1<br />
3 1<br />
11. +<br />
2 2<br />
3 1<br />
12. −<br />
2 2<br />
13. 1<br />
14. 0<br />
15. 3 2<br />
16. 0<br />
17. 0<br />
18. 0<br />
19. 1<br />
20. −1<br />
21. 1 2<br />
TRM Section 1.4: Worksheet 2<br />
This worksheet includes 9 limit questions to be<br />
solved analytically.<br />
22. 1 2<br />
23. 7<br />
24. 1 3<br />
25. 2<br />
26. −13<br />
27. 1 2<br />
28. 1<br />
29. 5<br />
30. 3 2<br />
31. 0<br />
32. 0<br />
33. 1<br />
34. 0<br />
Section 1.4 • Assess Your Understanding<br />
125<br />
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<strong>Sullivan</strong><br />
35. f is continuous at c = 0.<br />
36. f is discontinuous at c = 0.<br />
37. f is continuous at c = π<br />
4 .<br />
38. f is discontinuous at c = 1.<br />
39. f is continuous on { x| x ≠ 4}.<br />
40. f is continuous on { x| x ≠ 0}<br />
41. f is continuous on<br />
⎧⎪<br />
3π<br />
⎫⎪<br />
⎨xx≠ + 2 kπ<br />
⎬ , k any integer.<br />
⎩⎪ 2<br />
⎭⎪<br />
42. f is continuous on the set of all real<br />
numbers.<br />
43. f is continuous on { x| x > 0, x ≠3}.<br />
44. f is continuous on the set of all real<br />
numbers.<br />
45. f is continuous on the set of all real<br />
numbers.<br />
46. f is continuous on the set of all real<br />
numbers.<br />
47. 0<br />
48. 0<br />
49. 0<br />
50. 0<br />
51. See TSM.<br />
52. See TSM.<br />
53. See TSM.<br />
54. See TSM.<br />
55. See TSM.<br />
56. See TSM.<br />
57. See TSM.<br />
58. f (0) =π<br />
59. f (0) = π, f (1) = π<br />
60. Yes.<br />
61. Yes.<br />
62. See TSM.<br />
63. See TSM.<br />
64. See TSM.<br />
65. Explanations will vary. For sample<br />
explanation, see TSM.<br />
66. Explanations will vary. For sample<br />
explanation, see TSM.<br />
67. See TSM.<br />
68. 0<br />
69. See TSM.<br />
70. See TSM.<br />
126 Chapter 1 • Limits and Continuity<br />
In Problems 35–38, determine whether f is continuous at the number c.<br />
⎧<br />
⎨ 3 cos x if x < 0<br />
35. f (x) = 3 if x = 0 at c = 0<br />
⎩<br />
x + 3 if x > 0<br />
⎧<br />
⎨ cos x if x < 0<br />
36. f (x) = 0 if x = 0 at c = 0<br />
⎩<br />
e x if x > 0<br />
⎧<br />
⎨ sin θ if θ ≤ π 4<br />
37. f (θ) =<br />
⎩ cos θ if θ> π at c = π 4<br />
4<br />
tan −1 x if x < 1<br />
38. f (x) =<br />
at c = 1<br />
ln x if x ≥ 1<br />
In Problems 39–46, determine where f is continuous.<br />
<br />
<br />
x 2 − 4x<br />
x 2 − 5x + 1<br />
39. f (x) = sin<br />
x − 4<br />
40. f (x) = cos<br />
2x<br />
41.<br />
1<br />
1<br />
f (θ) =<br />
42. f (θ) =<br />
1 + sin θ<br />
1 + cos 2 θ<br />
43. f (x) = ln x<br />
x − 3<br />
44. f (x) = ln(x 2 + 1)<br />
PAGE<br />
125 45. f (x) = e −x sin x 46.<br />
e x<br />
f (x) =<br />
1 + sin 2 x<br />
Applications and Extensions<br />
In Problems 47–50, use the Squeeze Theorem to find each limit.<br />
47. lim<br />
x 2 sin 1 <br />
<br />
48. lim x 1 − cos 1 <br />
x→0 x<br />
x→0 x<br />
49.<br />
<br />
lim x 2 1 − cos 1 <br />
<br />
1<br />
50. lim x x→0 x<br />
+ 3x 2 sin x→0 x<br />
In Problems 51–54, show that each statement is true.<br />
sin(ax)<br />
51. lim x→0 sin(bx) = a cos(ax)<br />
; b = 0 52. lim<br />
b x→0 cos(bx) = 1<br />
sin(ax)<br />
53. lim = a x→0 bx b ; b = 0<br />
1 − cos(ax)<br />
54. lim<br />
= 0; a = 0, b = 0<br />
x→0 bx<br />
1 − cos x<br />
55. Show that lim x→0 x 2 = 1 2 .<br />
56. Squeeze Theorem If 0 ≤ f (x) ≤ 1 for every number x, show<br />
that lim[x 2 f (x)] = 0.<br />
x→0<br />
57. Squeeze Theorem If 0 ≤ f (x) ≤ M for every x, show that<br />
lim<br />
x→0 [x2 f (x)] = 0.<br />
sin(π x)<br />
58. The function f (x) = is not defined at 0. Decide how to<br />
x<br />
define f (0) so that f is continuous at 0.<br />
sin(π x)<br />
59. Define f (0) and f (1) so that the function f (x) =<br />
x(1 − x) is<br />
continuous on the interval [0, 1].<br />
⎧<br />
⎨ sin x<br />
if x = 0<br />
60. Is f (x) = x<br />
continuous at 0?<br />
⎩<br />
1 if x = 0<br />
⎧<br />
⎨ 1 − cos x<br />
if x = 0<br />
61. Is f (x) = x<br />
continuous at 0?<br />
⎩<br />
0 if x = 0<br />
1<br />
62. Squeeze Theorem Show that lim x n sin = 0, where n<br />
x→0 x<br />
is a positive integer. (Hint: Look first at Problem 56.)<br />
63. Prove lim sin θ = 0.<br />
θ→0<br />
(Hint: Use a unit circle as<br />
shown in the figure, first<br />
assuming 0
<strong>Sullivan</strong> AP˙<strong>Sullivan</strong>˙Chapter01 October 8, 2016 17:4<br />
Section 1.5 • Infinite Limits; Limits at Infinity; Asymptotes 127<br />
AP® Practice Problems<br />
⎧<br />
⎨ sin(2x)<br />
if x = 0<br />
PAGE<br />
121 1. The function g(x) = 2x<br />
⎩<br />
k if x = 0<br />
is continuous at x = 0. What is the value of k?<br />
1<br />
(A) 0 (B) (C) 1 (D) 2<br />
2<br />
PAGE<br />
sin(4x)<br />
121 2. lim x→0 2x<br />
(A) 0<br />
=<br />
(B)<br />
1<br />
2<br />
PAGE<br />
125 3. The function f (x) =<br />
(C) 1 (D) 2<br />
x 3 + 2x 2 if x ≤−2<br />
e 2x+4 if x > −2 .<br />
Find lim f (x) if it exists.<br />
x→−2<br />
(A) 0 (B) 1 (C) 16 (D) The limit does not exist.<br />
PAGE<br />
1 − cos 2 (3x)<br />
121 4. lim x→0 x 2 =<br />
(A) 0 (B) 1 (C) 3 (D) 9<br />
PAGE<br />
125 5. Which of the following functions are continuous for all real<br />
numbers x?<br />
I. f (x) = x 1/3<br />
II. g(x) = sec x<br />
III. h(x) = e −x<br />
(A) I only<br />
(B) I and II only<br />
(C) I and III only (D) I, II, and III<br />
RECALL The symbols ∞ (infinity)<br />
and −∞ (negative infinity) are not<br />
numbers. The symbol ∞ expresses<br />
unboundedness in the positive direction,<br />
and −∞ expresses unboundedness in<br />
the negative direction.<br />
PAGE<br />
1<br />
121 6. Find lim if it exists.<br />
x→0 x csc x<br />
(A) −1 (B) 0 (C) 1 (D) The limit does not exist.<br />
<br />
sin x − π <br />
PAGE<br />
121 7. lim<br />
3<br />
x→π/3 x − π =<br />
3<br />
(A) − π π<br />
(B) 0 (C) 1 (D)<br />
3<br />
3<br />
PAGE<br />
1 − cos x<br />
122 8. lim x→0 3 sin 2 =<br />
x<br />
(A)<br />
1<br />
6<br />
PAGE<br />
125 9. If f (x) =<br />
(B)<br />
<br />
1<br />
3<br />
(C)<br />
1<br />
2<br />
(D) 1<br />
ln x if 0 < x < 3<br />
,<br />
(2x − 3) ln 3 if x ≥ 3<br />
then lim f (x) =<br />
x→3<br />
(A) ln 3 (B) 3 (C) ln 9 (D) The limit does not exist.<br />
PAGE<br />
tan(2x)<br />
121 10. lim =<br />
x→0 3x<br />
(A)<br />
1<br />
3<br />
PAGE<br />
118 11. lim<br />
x 3 sin 1 x→0 x<br />
1<br />
(B)<br />
2<br />
<br />
=<br />
(C)<br />
2<br />
3<br />
(D) 2<br />
(A) −1 (B) 0 (C) 1 (D) The limit does not exist.<br />
1.5 Infinite Limits; Limits at Infinity; Asymptotes<br />
OBJECTIVES When you finish this section, you should be able to:<br />
1 Investigate infinite limits (p. 128)<br />
2 Find the vertical asymptotes of a graph (p. 131)<br />
3 Investigate limits at infinity (p. 131)<br />
4 Find the horizontal asymptotes of a graph (p. 137)<br />
5 Find the asymptotes of the graph of a rational function (p. 138)<br />
We have described lim f (x) = L by saying if a function f is defined everywhere in<br />
x→c<br />
an open interval containing c, except possibly at c, then the value f (x) can be made as<br />
close as we please to L by choosing numbers x sufficiently close to c. Here c and L are<br />
real numbers. In this section, we extend the language of limits to allow c to be ∞ or<br />
−∞ (limits at infinity) and to allow L to be ∞ or −∞ (infinite limits). These limits are<br />
useful for locating asymptotes that aid in graphing some functions.<br />
We begin with infinite limits.<br />
Answers to AP® Practice<br />
Problems<br />
1. C<br />
2. D<br />
3. D<br />
4. D<br />
5. C<br />
6. C<br />
7. C<br />
8. A<br />
9. D<br />
10. C<br />
11. B<br />
AP® Exam Tip<br />
Limits at infinity, as well as vertical and<br />
horizontal asymptotes, are concepts<br />
that commonly appear on the multiplechoice<br />
portion of the exam.<br />
Teaching Tip<br />
Students may be familiar with finding<br />
horizontal and vertical asymptotes from<br />
prior math classes. In this section, they will<br />
learn the formal definition of a horizontal<br />
and vertical asymptote and how to use<br />
limits to find all horizontal and vertical<br />
asymptotes of a function.<br />
TRM Alternate Examples Section 1.5<br />
You can find the Alternate Examples for<br />
this section in PDF format in the Teacher’s<br />
Resource Materials.<br />
TRM AP® Calc Skill Builders<br />
Section 1.5<br />
You can find the AP ® Calc Skill Builders for<br />
this section in PDF format in the Teacher’s<br />
Resource Materials.<br />
Section 1.5 • Infinite Limits; Limits at Infinity; Asymptotes<br />
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128 Chapter 1 • Limits and Continuity<br />
Teaching Tip<br />
1<br />
Note that when finding the limit of<br />
x<br />
2<br />
as x approaches zero, the authors take<br />
a tabular and graphical approach. There<br />
is no algebraic manipulation that can be<br />
1<br />
performed to simplify the function , so a<br />
x<br />
2<br />
table or graph is the preferred method for<br />
this problem. Some students may prefer to<br />
create a table mentally when a written table<br />
is not required for the answer. For example,<br />
since zero cannot be substituted in this<br />
situation, the students may find it helpful<br />
to think, What does 1 divided by a really<br />
small positive number squared equal? Then<br />
they should think, What does 1 divided by<br />
a really small negative number squared<br />
equal?<br />
In both cases, the answer comes out to<br />
be an exceptionally large number, which<br />
grows increasingly large as x gets closer<br />
and closer to zero.<br />
Therefore. lim 1 =∞.<br />
x→0<br />
2<br />
x<br />
Teaching Tip<br />
It may be helpful for the students to keep<br />
these two limits in mind. They appear often<br />
as the students find one-sided limits and<br />
define vertical asymptotes.<br />
1<br />
lim<br />
=∞<br />
x→ 0<br />
+ very small positive value<br />
1<br />
lim<br />
=−∞<br />
x→0<br />
− very small negative value<br />
Numerical examples help here as well:<br />
1/0.1 = 10, 1/0.01 = 100, and so forth.<br />
For example,<br />
1<br />
=<br />
1 1<br />
1<br />
=<br />
0.1 10<br />
1<br />
=<br />
0.01 100<br />
1<br />
=<br />
0.001 1000<br />
1<br />
=<br />
0.0001 10,000<br />
y<br />
4<br />
1 Investigate Infinite Limits<br />
Consider the function f (x) = 1 . Table 11 lists values of f for selected numbers x that<br />
x<br />
2<br />
are close to 0 and Figure 48 shows the graph of f .<br />
Figure 48 f (x) = 1 x 2 As x approaches 0, the value of 1 x increases without bound. Since the value of 1 2 x is 2<br />
not approaching a single real number, the limit of f (x) as x approaches 0 does not exist.<br />
TABLE 11<br />
x approaches 0 x approaches 0<br />
2<br />
from the left<br />
−−−−−−−−−−−−−−−→<br />
from the right<br />
←−−−−−−−−−−−−−−−−<br />
x −0.01 −0.001 −0.0001 → 0 ← 0.0001 0.001 0.01<br />
f (x) = 1 10,000 10 6 10 8 f (x) becomes unbounded 10 8 10 6 10,000<br />
4 2<br />
2 4 x<br />
x 2<br />
x 0<br />
y<br />
4<br />
2<br />
2<br />
4 2 4<br />
4<br />
Figure 49 f (x) = 1 x<br />
2<br />
x<br />
However, since the numbers are increasing without bound, we describe the behavior of<br />
the function near zero by writing<br />
1<br />
lim<br />
x→0 x =∞ 2<br />
and say that f (x) = 1 has an infinite limit as x approaches 0.<br />
x<br />
2<br />
In other words, a function f has an infinite limit at c if f is defined everywhere in<br />
an open interval containing c, except possibly at c, and f (x) becomes unbounded when<br />
x is sufficiently close to c. ∗<br />
EXAMPLE 1<br />
1 1<br />
Investigating lim and lim<br />
x→0 − x x→0 + x<br />
Consider the function f (x) = 1 . Table 12 shows values of f for selected numbers x<br />
x<br />
that are close to 0 and Figure 49 shows the graph of f .<br />
TABLE 12<br />
x approaches 0 x approaches 0<br />
from the left<br />
−−−−−−−−−−−−−−−−→<br />
from the right<br />
←−−−−−−−−−−−−−−−−−<br />
x −0.01 −0.001 −0.0001 → 0 ← 0.0001 0.001 0.01 0.1<br />
f (x) = 1 x<br />
−100 −1000 −10,000 f (x) becomes unbounded 10,000 1000 100 10<br />
Here as x gets closer to 0 from the right, the value of f (x) = 1 can be made as<br />
x<br />
large as we please. That is, 1 becomes unbounded in the positive direction. So,<br />
x<br />
Similarly, the notation<br />
1<br />
lim<br />
x→0 + x =∞<br />
1<br />
lim<br />
x→0 − x = −∞<br />
is used to indicate that 1 becomes unbounded in the negative direction as x approaches 0<br />
x<br />
from the left. ■<br />
∗ A precise definition of an infinite limit is given in Section 1.6.<br />
128<br />
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Section 1.5 • Infinite Limits; Limits at Infinity; Asymptotes 129<br />
y<br />
2<br />
2<br />
4<br />
Figure 51 f (x) = ln x<br />
y f (x)<br />
lim<br />
x→c<br />
c x<br />
c x<br />
x→c x→c <br />
(a) lim f (x) ∞ (b) lim f(x) ∞<br />
Figure 50<br />
2 4<br />
x<br />
So, there are four possible one-sided infinite limits of a function f at c:<br />
x→c<br />
x→c<br />
x→c +<br />
f (x) =∞, lim f (x) = −∞, lim f (x) =∞, lim<br />
− − +<br />
See Figure 50 for illustrations of these possibilities.<br />
y f (x)<br />
c<br />
y f (x)<br />
(c) lim f (x) ∞<br />
x→c <br />
x<br />
c<br />
f (x) = −∞<br />
(d) lim f (x) ∞<br />
x→c <br />
y f (x)<br />
When we know the graph of a function, we can use it to investigate infinite limits.<br />
EXAMPLE 2<br />
Investigate lim ln x.<br />
x→0 +<br />
Investigating an Infinite Limit<br />
Solution The domain of f (x) = ln x is {x|x > 0}. Notice that the graph of f (x) = ln x<br />
in Figure 51 decreases without bound as x approaches 0 from the right. The graph<br />
suggests that<br />
lim ln x = −∞<br />
x→0 +<br />
■<br />
x<br />
NOW WORK Problem 11.<br />
Based on the graphs of the trigonometric functions in Figure 52, we have the<br />
following infinite limits:<br />
lim tan x =∞ lim<br />
x→π/2− lim csc x = −∞ lim<br />
x→0− tan x = −∞ lim<br />
x→π/2 +<br />
csc x =∞ lim<br />
x→0 +<br />
sec x =∞ lim<br />
x→π/2− cot x = −∞ lim<br />
x→0− sec x = −∞<br />
x→π/2 +<br />
cot x =∞<br />
x→0 +<br />
Teaching Tip<br />
Students should have sound knowledge<br />
of the graphical representations of the<br />
basic functions. In particular, they should<br />
be able to quickly sketch and identify the<br />
asymptotes of the following functions:<br />
y = tanx<br />
y = csc x<br />
y = sec x<br />
y = cot x<br />
y =<br />
y<br />
e x<br />
= e −x<br />
y = lnx<br />
1<br />
y =<br />
x<br />
Teaching Tip<br />
Consider presenting the students with the<br />
following graph:<br />
y<br />
y<br />
5<br />
5<br />
π<br />
x 2<br />
Figure 52<br />
y<br />
y<br />
y<br />
y tan x y sec x<br />
y csc x y cot x<br />
π<br />
x<br />
5<br />
5<br />
π<br />
x<br />
π<br />
<br />
2<br />
5<br />
5<br />
π<br />
x x 0<br />
x 0<br />
2<br />
π<br />
2<br />
Limits of quotients, in which the limit of the numerator is a nonzero number and<br />
the limit of the denominator is 0, often result in infinite limits. (If the limits of both<br />
the numerator and the denominator are 0, the quotient is called an indeterminate form.<br />
Limits of indeterminate forms are discussed in Section 4.5.)<br />
x<br />
π<br />
<br />
2<br />
5<br />
5<br />
π<br />
2<br />
x<br />
Ask them to draw a function such that<br />
1. lim fx ( ) =∞<br />
x→c<br />
−<br />
2. lim fx ( ) =∞<br />
x→ c<br />
+<br />
3. lim fx ( ) =−∞<br />
x→c<br />
−<br />
4. lim fx ( ) =−∞<br />
x→ c<br />
+<br />
A few possible answers are at the top of<br />
page 129 of the student edition. Students<br />
can show their graphs to students around<br />
them and troubleshoot misconceptions.<br />
c<br />
x<br />
Section 1.5 • Infinite Limits; Limits at Infinity; Asymptotes<br />
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130 Chapter 1 • Limits and Continuity<br />
Teaching Tip<br />
Ask your students to think through the<br />
limits of rational functions near vertical<br />
asymptotes mentally. For Example 3(a),<br />
the thought process would be: We would<br />
like to substitute a number in place of x<br />
that is slightly larger than 3. When we do<br />
so, the numerator will be essentially 7. The<br />
denominator will be a very small positive<br />
number because, for example, 3.00000001<br />
− 3 = 0.00000001. Therefore,<br />
x<br />
lim 2 + 1<br />
7<br />
.<br />
x − 3<br />
≈ A very small positive number<br />
=∞<br />
→ +<br />
x 3<br />
Teaching Tip<br />
Remind students of the graph of y = ln x<br />
when working on Example 3b.<br />
y<br />
The graph of y = ln x crosses the x-axis at<br />
x = 1. Therefore,<br />
and<br />
2<br />
1<br />
21<br />
22<br />
lim lnx<br />
= a small negative number<br />
x→1<br />
−<br />
lim lnx<br />
= a small positive number.<br />
x→ 1<br />
+<br />
2 4 6 8 x<br />
EXAMPLE 3<br />
Investigate: (a)<br />
Investigating an Infinite Limit<br />
2x + 1<br />
lim<br />
x→3 + x − 3<br />
(b)<br />
x 2<br />
lim<br />
x→1 − ln x<br />
2x + 1<br />
Solution (a) lim is the right-hand limit of the quotient of two functions,<br />
x→3 + x − 3<br />
f (x) = 2x + 1 and g(x) = x − 3. Since lim<br />
x→3 +(x<br />
− 3) = 0, we cannot use the Limit of a<br />
Quotient. So we need a different strategy. As x approaches 3 + , we have<br />
lim<br />
x→3<br />
+(2x<br />
+ 1) = 7 and lim<br />
+(x<br />
− 3) = 0<br />
From arithmetic, we know that a fraction p is arbitrarily large if p is a fixed nonzero<br />
q<br />
number and q is arbitrarily close to 0. Here, the limit of the numerator is 7. The<br />
denominator is positive and approaching 0. So the quotient is positive and becoming<br />
unbounded. We conclude that<br />
2x + 1<br />
lim<br />
x→3 + x − 3 =∞<br />
(b) As in (a), we cannot use the Limit of a Quotient because lim ln x = 0. Here we<br />
x→1− have<br />
lim x 2 = 1 lim ln x = 0<br />
x→1 − x→1 −<br />
We are seeking a left-hand limit, so for numbers close to 1 but less than 1, the denominator<br />
ln x < 0. So the quotient is negative and becoming unbounded. We conclude that<br />
x 2<br />
lim<br />
x→1 − ln x = −∞<br />
■<br />
x→3<br />
The discussion below may prove useful when finding lim<br />
x→c −<br />
is a nonzero number and lim g(x) = 0.<br />
x→c− • lim f (x) = L, L > 0, and lim g(x) = 0<br />
x→c −<br />
x→c −<br />
If g(x) 0, and lim g(x) = 0<br />
x→c −<br />
x→c −<br />
If g(x) >0 for numbers x close to c, but less than c, then<br />
f (x)<br />
lim<br />
x→c − g(x) =∞<br />
• lim f (x) = L, L < 0, and lim g(x) = 0<br />
x→c −<br />
x→c −<br />
If g(x) 0 for numbers x close to c, but less than c, then<br />
f (x)<br />
lim<br />
x→c − g(x) = −∞<br />
f (x)<br />
, where lim<br />
g(x) f (x)<br />
x→c −<br />
Similar arguments are valid for right-hand limits.<br />
NOW WORK Problem 27 and AP® Practice Problems 9 and 11.<br />
AP® CaLC skill builder<br />
for example 3<br />
Investigating an Infinite Limit<br />
⎛ −1<br />
⎞<br />
Find lim<br />
⎝<br />
⎜ − ⎠<br />
⎟ .<br />
x→5<br />
− x 5<br />
Solution<br />
Think x ≈ 4.999999999<br />
⎛ −1<br />
⎞<br />
⎝<br />
⎜ − ⎠<br />
⎟ ≈ −1<br />
lim<br />
→ − x 5 4.999999999 − 5<br />
x 5<br />
−1<br />
=<br />
a very small negative number<br />
As the values of x get closer to 5 from the<br />
left, the denominator decreases without<br />
⎛ −1<br />
⎞<br />
bound and lim<br />
⎝<br />
⎜ − ⎠<br />
⎟ =∞ .<br />
x→5<br />
− x 5<br />
Graphical Approach<br />
Solution<br />
Also, consider how this limit may be<br />
explained using the graph. Explain that since<br />
⎛ −1<br />
⎞<br />
fx ( ) =<br />
⎝<br />
⎜ x − 5⎠<br />
⎟ is a rational function, its graph<br />
has asymptotes. Show the graph and observe that<br />
there is a vertical asymptote at x = 5 because as<br />
x approaches 5 from the right or from the left, the<br />
function is infinite.<br />
y<br />
4<br />
2<br />
22<br />
24<br />
2 4 6 8<br />
Also note that as x approaches infinity, the limit<br />
approaches zero.<br />
x<br />
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Section 1.5 • Infinite Limits; Limits at Infinity; Asymptotes 131<br />
y<br />
x c<br />
y f (x)<br />
c x<br />
(a) lim f(x) ∞<br />
x→c<br />
Figure 53<br />
y<br />
y f (x)<br />
x c<br />
c<br />
x<br />
(b) lim f(x) ∞<br />
x→c <br />
2 Find the Vertical Asymptotes of a Graph<br />
Figure 53 illustrates the possibilities that can occur when a function has an infinite limit<br />
at c. In each case, notice that the graph of f has a vertical asymptote x = c.<br />
y<br />
x c<br />
y f (x)<br />
c<br />
x<br />
(c) lim f(x) ∞<br />
x→c <br />
lim<br />
x→c<br />
y<br />
x c y x c y x c<br />
c x<br />
c x<br />
c<br />
y f (x)<br />
y f (x)<br />
(d) lim f(x) ∞<br />
x→c<br />
x→c<br />
x→c<br />
(e) lim f(x) ∞<br />
x→c <br />
y f (x)<br />
(f) lim f(x) ∞<br />
x→c <br />
DEFINITION Vertical Asymptote<br />
The line x = c is a vertical asymptote of the graph of the function f if any of the<br />
following is true:<br />
f (x) =∞ lim f (x) =∞ lim f (x) = −∞ lim f (x) = −∞<br />
− + −<br />
x→c +<br />
For rational functions, a vertical asymptote may occur where the denominator<br />
equals 0.<br />
EXAMPLE 4 Finding a Vertical Asymptote<br />
x<br />
Find any vertical asymptote(s) of the graph of f (x) =<br />
(x − 3) . 2<br />
Solution The domain of f is {x|x = 3}. Since 3 is the only number for which the<br />
denominator of f equals zero, we construct Table 13 and investigate the one-sided limits<br />
of f as x approaches 3. Table 13 suggests that<br />
x<br />
lim<br />
x→3 (x − 3) =∞ 2<br />
So, x = 3 is a vertical asymptote of the graph of f .<br />
x<br />
AP® CaLC skill builder<br />
for example 4<br />
Finding a Vertical Asymptote<br />
Find any vertical asymptote(s) of the graph<br />
x<br />
of fx ( ) = .<br />
2<br />
x −1<br />
Solution<br />
The domain of f is {x | x ≠ ±1}. Since the<br />
denominator is zero for x = −1 and x = 1,<br />
we investigate the one-sided limits at x =<br />
−1 and x = 1.<br />
For lim , think x ≈ −1.0000001:<br />
x→−1<br />
x−<br />
− 2<br />
x 1<br />
x<br />
− ≈ −1.0000001<br />
lim<br />
2 2<br />
x 1 ( −1.0000001) −1<br />
−<br />
x→−1<br />
−1<br />
=<br />
very small positive<br />
x<br />
so lim<br />
− =−∞ .<br />
− 2<br />
x→−1<br />
x 1<br />
TABLE 13<br />
x approaches 3 from the left<br />
x approaches 3 from the right<br />
−−−−−−−−−−−−−−−−−−→<br />
←−−−−−−−−−−−−−−−−−−−<br />
x 2.9 2.99 2.999 → 3 ← 3.001 3.01 3.1<br />
x<br />
f (x) =<br />
290 29,900 2,999,000 f (x) becomes unbounded 3,001,000 30,100 310<br />
(x − 3) 2<br />
y<br />
12<br />
8<br />
4<br />
Figure 54 shows the graph of f (x) =<br />
x<br />
and its vertical asymptote.<br />
(x − 3)<br />
2<br />
NOW WORK Problems 15 and 63 (find any vertical asymptotes)<br />
and AP® Practice Problem 5.<br />
x<br />
3 Investigate Limits at Infinity<br />
2<br />
4<br />
x 3<br />
Now we investigate what happens as x becomes unbounded in either the positive direction<br />
or the negative direction. Suppose as x becomes unbounded, the value of a function f<br />
x<br />
Figure 54 f (x) =<br />
approaches some real number L. Then the number L is called the limit of f at infinity.<br />
(x − 3) 2<br />
common error<br />
Sometimes, students think that a vertical<br />
asymptote exists only if both the left-hand and<br />
right-hand limits are infinite. Note the precise<br />
definition of a vertical asymptote. If any of the<br />
one-sided limits are infinite at x = c, then x = c is a<br />
vertical asymptote of the graph of the function. It<br />
is not necessary for both sides to be infinite, and it<br />
is not necessary for the two sides to be equal.<br />
TRM Section 1.5: Worksheet 1<br />
This worksheet contains 4 rational functions and<br />
their corresponding graphs to help the students<br />
identify the vertical asymptotes of each of the<br />
functions.<br />
■<br />
For<br />
lim , think x ≈ −0.9999999:<br />
2<br />
x 1<br />
x<br />
− ≈ −0.9999999<br />
lim<br />
+ 2 2<br />
x→−1<br />
x 1 ( −0.9999999) −1<br />
x→− 1<br />
x−<br />
+<br />
−1<br />
=<br />
very small negative<br />
x<br />
so lim<br />
− =∞ .<br />
+ 2<br />
x→−1<br />
x 1<br />
The line x = −1 is a vertical asymptote of<br />
the graph of f because the one-sided limits<br />
are infinite.<br />
Note: Once one of these limits is found<br />
to be infinite, the value x = −1 has been<br />
shown to be a vertical asymptote. It is not<br />
necessary to show that both one-sided<br />
limits are infinite.<br />
Repeating the same procedure for x = 1,<br />
we get<br />
x<br />
lim<br />
x − 1<br />
=−∞ and lim<br />
2<br />
x→1<br />
−<br />
x→ 1<br />
+<br />
x<br />
x − 1<br />
=∞ .<br />
2<br />
The line x = 1 is a vertical asymptote of the<br />
graph of f because the one-sided limits are<br />
infinite.<br />
Section 1.5 • Infinite Limits; Limits at Infinity; Asymptotes<br />
131<br />
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132 Chapter 1 • Limits and Continuity<br />
Teaching Tip<br />
When studying the limit of functions as x<br />
approaches ∞ or −∞, it helps students to<br />
keep the following generalizations in mind:<br />
Infinity in the denominator:<br />
2<br />
y<br />
4<br />
2<br />
4 2 4<br />
2<br />
x<br />
For example, the graph of f (x) = 1 in Figure 55 suggests that, as x becomes<br />
x<br />
unbounded in either the positive direction or the negative direction, the values f (x) get<br />
closer to 0. Table 14 illustrates this for a few numbers x.<br />
TABLE 14<br />
x ±100 ±1000 ±10,000 ±100,000 x becomes unbounded<br />
f (x) = 1 x<br />
±0.01 ±0.001 ±0.0001 ±0.00001 f (x) approaches 0<br />
1. lim 1 = 0<br />
x→∞ x<br />
4<br />
2. lim 1 = 0<br />
x→−∞ x<br />
Infinity in the numerator:<br />
x<br />
3. =∞<br />
xlim<br />
→∞ 2<br />
x<br />
4. =−∞<br />
xlim<br />
→−∞ 2<br />
Square root and natural log of infinity:<br />
5. ∞≈∞ lim x =∞<br />
x→∞<br />
Figure 55 f (x) = 1 x<br />
Since f can be made as close as we please to 0 by choosing x sufficiently large, we<br />
write<br />
1<br />
lim<br />
x→∞ x = 0<br />
and say that the limit as x approaches infinity of f is equal to 0 . Similarly, as x becomes<br />
unbounded in the negative direction, the function f (x) = 1 can be made as close as we<br />
x<br />
please to 0, and we write<br />
lim<br />
x→−∞<br />
1<br />
x = 0<br />
1<br />
Although we do not prove lim = 0 here, it can be proved using the ε-δ<br />
x→−∞ x<br />
Definition of a Limit at Infinity, and is left as an exercise in Section 1.6.<br />
The limit properties discussed in Section 1.2 hold for infinite limits. Although these<br />
properties are stated below for limits as x → ∞, they are also valid for limits as<br />
x → −∞.<br />
6. ln( ∞) ≈∞ lim lnx =∞<br />
x→∞<br />
THEOREM Properties of Limits at Infinity<br />
If k is a real number, n ≥ 2 is an integer, and the functions f and g approach real<br />
numbers as x →∞, then the following properties are true:<br />
Alternate Example<br />
Finding Limits at Infinity<br />
Following are alternate ways to solve<br />
Example 5:<br />
a. lim 4 = 4 lim 1<br />
x→−∞<br />
2<br />
x x→−∞<br />
2<br />
x<br />
= 4 lim 1 ⋅ lim 1 = 400 ⋅ ⋅ = 0<br />
x→−∞<br />
x x→−∞<br />
x<br />
• lim A = A where A is a constant<br />
x→∞<br />
• lim [kf(x)] = k lim f (x)<br />
x→∞ x→∞<br />
• lim [ f (x) ± g(x)] = lim f (x) ± lim g(x)<br />
x→∞ x→∞ x→∞<br />
[ ][ ]<br />
• lim [ f (x)g(x)] = lim f (x) lim g(x)<br />
x→∞ x→∞ x→∞<br />
f (x) lim<br />
• f (x)<br />
lim<br />
x→∞ g(x) = x→∞<br />
, provided lim g(x) = 0<br />
lim g(x) x→∞<br />
x→∞<br />
[ ] n<br />
• lim [ f<br />
x→∞ (x)]n = lim f (x)<br />
x→∞<br />
√ √<br />
n<br />
• lim f (x) = n lim f (x), where f (x) >0 if n is even<br />
x→∞<br />
x→∞<br />
b. ⎛ 10 ⎞<br />
⎜ − ⎟<br />
⎝ ⎠<br />
=− ⎛<br />
lim<br />
10 lim<br />
x→∞<br />
x→∞⎜<br />
x ⎝<br />
=−10⋅ 0=<br />
0<br />
1 ⎞<br />
⎟<br />
x ⎠<br />
EXAMPLE 5<br />
Find:<br />
(a)<br />
lim<br />
x→−∞<br />
Finding Limits at Infinity<br />
(<br />
4<br />
(b) lim − 10 )<br />
√<br />
x 2 x→∞ x<br />
(c)<br />
(<br />
lim 2 + 3 )<br />
x→∞ x<br />
⎛ 3 ⎞<br />
c. +<br />
⎝<br />
⎜<br />
⎠<br />
⎟ = + ⎛ ⎞<br />
lim 2 lim 2 3 lim 1<br />
x→∞ x x→∞ x→∞⎝<br />
⎜ x⎠<br />
⎟<br />
⎛ ⎞<br />
→∞<br />
= 2+<br />
3 lim lim 1<br />
x<br />
⎝<br />
⎜<br />
⎠<br />
⎟ = 2 + 3 ⋅ 0 = 2<br />
x→∞<br />
lim x<br />
x→∞<br />
132<br />
Chapter 1 • Limits and Continuity<br />
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Section 1.5 • Infinite Limits; Limits at Infinity; Asymptotes 133<br />
Solution We use the properties of limits at infinity.<br />
<br />
4<br />
1 2 <br />
(a) lim<br />
x→−∞ x = 4 lim<br />
1 2<br />
= 4 lim = 4 · 0 = 0<br />
2 x→−∞ x<br />
x→−∞ x ↑<br />
1<br />
lim<br />
x→−∞ x = 0<br />
<br />
(b) lim − 10 <br />
1<br />
1<br />
√ =−10 lim √ =−10 · lim<br />
x→∞ x x→∞ x x→∞ x<br />
<br />
1<br />
=−10 · lim<br />
x→∞ x =−10 ↑<br />
· 0 = 0<br />
1<br />
lim<br />
x→∞ x = 0<br />
<br />
(c) lim 2 + 3 <br />
= lim<br />
x→∞ x<br />
2 + lim 3<br />
x→∞ x→∞ x = 2 + 3 · lim 1<br />
x→∞ x = 2 + 3 · 0 = 2<br />
EXAMPLE 6<br />
Find:<br />
(a)<br />
lim<br />
3x 2 − 2x + 8<br />
x→∞ x 2 + 1<br />
Finding Limits at Infinity<br />
(b)<br />
lim<br />
4x 2 − 5x<br />
x→−∞ x 3 + 1<br />
■<br />
NOW WORK Problem 45.<br />
Solution (a) We find this limit by dividing each term of the numerator and the<br />
denominator by the term with the highest power of x that appears in the denominator, in<br />
this case, x 2 . Then<br />
3x 2 − 2x + 8<br />
lim<br />
x→∞ x 2 + 1<br />
(b)<br />
= <br />
lim<br />
x→∞<br />
⏐<br />
Divide the numerator and<br />
denominator by x 2<br />
3x 2 − 2x + 8<br />
x 2<br />
x 2 + 1<br />
x 2<br />
3 − 2<br />
= lim<br />
x + 8 x 2<br />
x→∞<br />
1 + 1 = ⏐⏐⏐<br />
x 2<br />
Limit of a<br />
Quotient<br />
lim<br />
3 − 2<br />
x→∞ x + 8 <br />
x<br />
1 2 + 1 <br />
x 2<br />
lim<br />
x→∞<br />
<br />
lim 3 − lim 2<br />
x→∞ x→∞<br />
=<br />
x + lim 8<br />
1<br />
3 − 2 lim<br />
x→∞ x 2<br />
x→∞ x + 8 1<br />
lim<br />
x→∞ x<br />
=<br />
lim 1 + lim 1<br />
<br />
1 2<br />
x→∞ x→∞ x 2 1 + lim<br />
x→∞ x<br />
= 3 − 0 + 0 = 3<br />
↑ 1 + 0<br />
1<br />
lim<br />
x = 0<br />
x→∞<br />
lim<br />
4x 2 − 5x<br />
= lim<br />
x→−∞ x 3 + 1 ↑<br />
x→−∞<br />
Divide the numerator and<br />
denominator by x 3<br />
TRM Section 1.5: Worksheet 2<br />
This worksheet contains 6 questions in which the<br />
student finds the limit at infinity analytically.<br />
4x 2 − 5x 4<br />
x 3<br />
x 3 = lim<br />
x − 5 x 2<br />
+ 1 x→−∞<br />
1 + 1 =<br />
x 3 x 3<br />
lim<br />
x→−∞<br />
lim<br />
x→−∞<br />
4<br />
x − 5 x 2 <br />
2<br />
1 + 3 x 3 = 0 1 = 0<br />
■<br />
AP® CaLC skill builder<br />
for example 6<br />
Finding Limits at Infinity<br />
x − x+<br />
Find lim 2 3<br />
10 .<br />
x→∞<br />
4<br />
5−<br />
x<br />
Solution<br />
To find the limit at infinity, divide each term<br />
by the highest term in the denominator, that<br />
is, x 4 :<br />
3<br />
2x<br />
x 10<br />
− +<br />
x − x+<br />
lim 2 3<br />
10<br />
4 4 4<br />
= lim x x x<br />
x→∞<br />
4<br />
5 − x<br />
x→∞<br />
4<br />
5 x<br />
−<br />
4 4<br />
x x<br />
= lim<br />
x→∞<br />
2 1 10<br />
− +<br />
3 4<br />
x x x<br />
5<br />
−1<br />
4<br />
x<br />
lim 2 − lim 1 + lim 10<br />
x→∞ →∞ →∞<br />
=<br />
x x<br />
3<br />
x x<br />
4<br />
x<br />
lim 5 − lim 1<br />
x→∞<br />
4<br />
x x→∞<br />
0<br />
= − 0 + 0<br />
= 0<br />
0−1<br />
Alternate Example<br />
Finding Limits at Infinity<br />
Find<br />
x − x+<br />
a. lim 3 2<br />
2 8<br />
x→∞<br />
2<br />
x + 1<br />
x − x<br />
b.<br />
xlim 4 2<br />
5<br />
→−∞<br />
3<br />
x + 1<br />
Solution<br />
Another way to evaluate the limits in<br />
Example 6 is to consider only the leading<br />
terms in the numerator and denominator.<br />
For instance, in (a), consider only 3x 2 in the<br />
numerator and x 2 in the denominator:<br />
x − x+<br />
x<br />
lim 3 2<br />
2 8 = lim 3<br />
x→∞<br />
2<br />
x + 1 x→∞<br />
2<br />
x<br />
= lim 3=<br />
3<br />
x→∞<br />
x − x x<br />
lim 4 2<br />
5 = lim 4 2<br />
= lim 4 = 0<br />
→−∞<br />
3<br />
x + 1 →−∞<br />
3<br />
x →−∞ x<br />
x x x<br />
Your students may wonder why this method<br />
works. Go back to Example 6 and observe<br />
that every expression drops out except for<br />
the leading terms. This solution method is<br />
presented in Example 8 on page 135.<br />
2<br />
Section 1.5 • Infinite Limits; Limits at Infinity; Asymptotes<br />
133<br />
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134 Chapter 1 • Limits and Continuity<br />
The graphs of the functions g(x) = 3x 2 − 2x + 8<br />
and f (x) = 4x 2 − 5x<br />
are<br />
x 2 + 1<br />
x 3 + 1<br />
shown in Figures 56(a) and 56(b), respectively. In each graph, notice how the graph of<br />
the function behaves as x becomes unbounded.<br />
y<br />
y<br />
8<br />
6<br />
4<br />
g(x)<br />
3x 2 2x 8<br />
x 2 1<br />
y 3<br />
y 0<br />
4<br />
4<br />
f(x)<br />
4x 2 5x<br />
x 3 1<br />
4 8<br />
x<br />
2<br />
4<br />
Alternate Example<br />
Finding the Limit at Infinity<br />
Find<br />
x +<br />
lim 4 2<br />
10<br />
x→∞<br />
x − 5<br />
Solution<br />
Another way to find this limit at infinity is to<br />
consider the variable terms only, because<br />
the constant 10 in the numerator and −5 in<br />
the denominator contribute very little as<br />
x → ∞:<br />
x +<br />
x<br />
lim 4 2 10 = lim 4 2<br />
x→∞<br />
x − 5 x→∞<br />
x<br />
x<br />
= lim 2| | = lim 2=<br />
2<br />
x→∞<br />
x x→∞<br />
Alternate Example<br />
Finding the Limit at Infinity<br />
x<br />
Find lim sin .<br />
x→∞<br />
x<br />
Solution<br />
The numerator, sin x, fluctuates between<br />
−1 and 1 over its entire domain.<br />
Since the numerator is bounded, we can<br />
use the following argument to show the<br />
limit exists.<br />
sinx<br />
Some number between −1 and 1<br />
lim ≈<br />
x→∞<br />
x a very large number<br />
x<br />
and thus lim sin = 0<br />
x→∞<br />
x<br />
This problem can also be solved formally<br />
using the Squeeze Theorem.<br />
We can also take a graphical approach.<br />
y<br />
1<br />
20<br />
10<br />
y<br />
20<br />
10<br />
10<br />
Figure 57 f (x) =<br />
lim f (x) 2<br />
x →∞<br />
y 2<br />
10 20 x<br />
√<br />
4x 2 + 10<br />
x − 5<br />
NEED TO REVIEW? The end behavior<br />
of a polynomial function is discussed in<br />
Section P.2, p. 21.<br />
CALC<br />
CLIP<br />
Figure 56<br />
4<br />
8<br />
x<br />
(a) lim g(x) 3 (b) lim<br />
f(x) 0<br />
x→∞ x→ ∞<br />
Limits at infinity have the following property.<br />
If p > 0 is a rational number and k is any real number, then<br />
provided x p is defined when x < 0.<br />
k<br />
lim = 0 and lim<br />
x→∞ x<br />
p<br />
5<br />
−6<br />
For example, lim = 0, lim = 0, and<br />
x→∞ x<br />
3 x→∞ x lim<br />
2/3<br />
EXAMPLE 7 Finding the Limit at Infinity<br />
√<br />
4x<br />
2<br />
+ 10<br />
Find lim<br />
.<br />
x→∞ x − 5<br />
NOW WORK Problems 47 and 49.<br />
x→−∞<br />
x→−∞<br />
k<br />
x = 0 p<br />
4<br />
= 0.<br />
x<br />
8/3<br />
Solution Divide each term of the numerator and the denominator by x, the term with the<br />
highest power of x that appears in the denominator. But remember, since √ x 2 =|x| =x<br />
when x ≥ 0, in the numerator the divisor in the square root will be x 2 .<br />
lim<br />
x→∞<br />
√<br />
4x<br />
2<br />
+ 10<br />
x − 5<br />
= lim<br />
x→∞<br />
√<br />
4x 2 + 10<br />
x 2<br />
x − 5<br />
= lim<br />
x→∞<br />
x<br />
√<br />
√<br />
lim 4 + 10<br />
x→∞<br />
= [ x 2<br />
lim 1 − 5 ] =<br />
x→∞ x<br />
lim<br />
x→∞<br />
√<br />
4x 2<br />
x + 10<br />
2 x 2<br />
1 − 5 x<br />
[<br />
4 + 10 ]<br />
x 2<br />
1<br />
= √ 4 = 2<br />
√<br />
4x<br />
2<br />
+ 10<br />
The graph of f (x) =<br />
x − 5<br />
and its behavior as x →∞is shown in Figure 57.<br />
NOW WORK Problem 57 and AP® Practice Problem 6.<br />
An alternative method for finding limits at infinity for rational functions uses the<br />
end behavior of a polynomial function.<br />
■<br />
220<br />
20<br />
x<br />
21<br />
x→∞<br />
x<br />
x<br />
lim sin<br />
= 0.<br />
134<br />
Chapter 1 • Limits and Continuity<br />
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Section 1.5 • Infinite Limits; Limits at Infinity; Asymptotes 135<br />
Recall that for large values of x, either positive or negative, the graph of the<br />
polynomial function<br />
f (x) = a n x n + a n−1 x n−1 +···+a 1 x + a 0<br />
resembles the graph of the power function y = a n x n .<br />
From this fact, we conclude that<br />
As examples,<br />
lim f (x) = lim a nx n (1)<br />
x→±∞ x→±∞<br />
Teaching Tip<br />
When finding limits at infinity of rational<br />
functions, the terms with the largest<br />
degree in the numerator and denominator<br />
determine the limit.<br />
• lim (3x 3 − x 2 + 5x + 1) = lim 3x 3 = 3 lim x 3 =∞<br />
x→∞ ↑ x→∞ x→∞<br />
(1)<br />
• lim (−x 5 + 2x) = lim (−x 5 ) = −∞<br />
x→∞ ↑ x→∞<br />
(1)<br />
• lim (2x 4 − x 2 + 2x + 8) = lim (2x 4 ) =∞<br />
x→−∞ x→−∞<br />
• lim (x 3 + 10x + 7) = lim (x 3 ) = −∞<br />
x→−∞ x→−∞<br />
We conclude that for a polynomial function, as x becomes unbounded in either the<br />
positive direction or the negative direction, the polynomial function is also unbounded.<br />
In other words, polynomial functions have an infinite limit at infinity.<br />
Since a rational function R is the ratio of two polynomial functions p and q where<br />
q is not the zero polynomial, we can find a limit at infinity for a rational function using<br />
only the leading terms of p and q. That is, if the leading term of p(x) is a n x n and the<br />
leading term of q(x) is b m x m , then<br />
lim R(x) = lim p(x)<br />
x→±∞ x→±∞ q(x) = lim a n x n<br />
(2)<br />
x→±∞ b m x m<br />
EXAMPLE 8<br />
Find:<br />
(a) lim<br />
x→∞<br />
3x 2 − 2x + 8<br />
x 2 + 1<br />
Finding Limits at Infinity of a Rational Function<br />
(b)<br />
lim<br />
4x 2 − 5x<br />
x→−∞ x 3 + 1<br />
(c)<br />
5x 4 − 3x 2<br />
lim<br />
x→∞ 2x 2 + 1<br />
Solution We use the alternative method of finding a limit at infinity of a rational function.<br />
y<br />
3x 2 − 2x + 8 3x 2<br />
(a) lim<br />
= lim<br />
x→∞ x 2 + 1 ↑ x→∞ x 2<br />
(2)<br />
= lim<br />
x→∞ 3 = 3<br />
1<br />
R(x) 5x4 3x 2<br />
2x 2 1<br />
(b)<br />
lim<br />
x→−∞<br />
4x 2 − 5x<br />
x 3 + 1 = ↑<br />
lim<br />
(2)<br />
x→−∞<br />
4x 2<br />
x 3<br />
x 2<br />
x→−∞<br />
= 4 lim<br />
x 3 = 4 lim<br />
x→−∞<br />
1<br />
x = 4 · 0 = 0<br />
2<br />
2<br />
x<br />
5x 4 − 3x 2<br />
(c) lim<br />
x→∞ 2x 2 + 1<br />
= lim 5x 4<br />
x→∞ 2x = 5 2 2 lim x 4<br />
x→∞ x = 5 2 2 lim x 2 =∞<br />
↑<br />
x→∞<br />
(2)<br />
■<br />
1<br />
Figure 58 lim R(x) =∞<br />
x→∞<br />
Observe<br />
The function R(x) = 5x 4 − 3x 2<br />
has an infinite limit at infinity. The graph of R is<br />
2x 2 + 1<br />
shown in Figure 58.<br />
that the limits found in Example 8(a) and 8(b) are the same as the limits<br />
found in Example 6(a) and 6(b), as we should expect. Compare the two methods and<br />
decide which you prefer to use.<br />
NOW WORK Problem 59 and AP® Practice Problems 2 and 3.<br />
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136 Chapter 1 • Limits and Continuity<br />
Teaching Tip<br />
Students may not find it intuitive that the<br />
x<br />
lim e = 0.<br />
x→−∞<br />
Consider showing them the following steps:<br />
lim e<br />
x<br />
x→−∞<br />
= lim 1 Rewritethe expression.<br />
x→−∞<br />
e<br />
− x<br />
= 0 As x approaches negative<br />
infinity,the limit approaches<br />
zero.<br />
y<br />
4<br />
2<br />
(0, 1)<br />
2<br />
Figure 59 f (x) = e x<br />
2<br />
x<br />
Other Infinite Limits at Infinity<br />
We have seen that all polynomial functions have infinite limits at infinity, and some<br />
rational functions have infinite limits at infinity. Other functions also have an infinite<br />
limit at infinity.<br />
For example, consider the function f (x) = e x .<br />
The graph of the exponential function, shown in Figure 59, suggests that<br />
lim<br />
x→−∞ ex = 0<br />
lim<br />
x→∞ ex =∞<br />
These limits are supported by the information in Table 15.<br />
TABLE 15<br />
x −1 −5 −10 −20 x approaches −∞<br />
f (x) = e x 0.36788 0.00674 0.00005 −2 × 10 −9 f (x) approaches 0<br />
x 1 5 10 20 x approaches ∞<br />
f (x) = e x e ≈ 2.71828 148.41 22,026 4.85 × 10 8 f (x) becomes unbounded<br />
y<br />
2<br />
EXAMPLE 9<br />
Find lim ln x.<br />
x→∞<br />
Finding the Limit at Infinity of g(x) =ln x<br />
Solution Table 16 and the graph of g(x) = ln x in Figure 60 suggest that g(x) = ln x<br />
has an infinite limit at infinity. That is,<br />
2 4<br />
x<br />
lim ln x =∞<br />
x→∞<br />
2<br />
4<br />
Figure 60 g(x) = ln x<br />
y<br />
4<br />
TABLE 16<br />
x e 10 e 100 e 1000 e 10,000 e 100,000 → x becomes unbounded<br />
g(x) = ln x 10 100 1000 10,000 100,000 → g(x) becomes unbounded<br />
Now let’s compare the graph of f (x) = e x in Figure 59, the graph of g(x) = ln x<br />
in Figure 60, and the graph of h(x) = x 2 in Figure 61. As x becomes unbounded in<br />
the positive direction, all three of the functions increase without bound. But in Table 17,<br />
f (x) = e x approaches infinity more rapidly than h(x) = x 2 , which approaches infinity<br />
more rapidly than g(x) = ln x.<br />
■<br />
24<br />
22<br />
2<br />
21<br />
2<br />
4<br />
x<br />
TABLE 17<br />
x 10 50 100 1000 10,000<br />
f (x) = e x 22,026 5.185 × 10 21 2.688 × 10 43 e 1000 e 10,000<br />
h(x) = x 2 100 2500 = 2.5 × 10 3 1.0 × 10 4 1.0 × 10 6 1.0 × 10 8<br />
g(x) = ln x 2.303 3.912 4.605 6.908 9.210<br />
Figure 61 h(x) = x 2<br />
EXAMPLE 10<br />
Application: Decomposition of Salt in Water<br />
Salt (NaCl) dissolves in water into sodium (Na + ) ions and chloride (Cl − ) ions according<br />
to the law of uninhibited decay<br />
A(t) = A 0 e kt<br />
where A = A(t) is the amount (in kilograms) of undissolved salt present at time t (in<br />
hours), A 0 is the original amount of undissolved salt, and k is a negative number that<br />
represents the rate of dissolution.<br />
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NEED TO REVIEW? Solving exponential<br />
equations is discussed in Section P.5, pp.<br />
48--49.<br />
(a) If initially there are 25 kilograms (kg) of undissolved salt and after 10 hours (h)<br />
there are 15 kg of undissolved salt remaining, how much undissolved salt is left<br />
after one day?<br />
(b) How long will it take until 1 kg of undissolved salt remains?<br />
2<br />
(c) Find lim A(t).<br />
t→∞<br />
(d) Interpret the answer found in (c).<br />
Solution (a) Initially, there are 25 kg of undissolved salt, so A(0) = A 0 = 25. To<br />
find the number k in A(t) = A 0 e kt , we use the fact that at t = 10, then A(10) = 15.<br />
That is,<br />
A(10) = 15 = 25e 10k A(t) = A 0e kt , A 0 = 25; A(10) = 15<br />
e 10k = 3 5<br />
10k = ln 3 5<br />
k = 1 ln 0.6<br />
10<br />
So, A(t) = 25e ( 1 10 ln 0.6)t . The amount of undissolved salt that remains after one day<br />
(24 h) is<br />
A(24) = 25e ( 1<br />
10 ln 0.6)24 ≈ 7.337 kilograms<br />
(b) We want to find t so that A(t) = 25e ( 1 10 ln 0.6)t = 1 kg. Then<br />
2<br />
1<br />
2 = 25e( 1<br />
e ( 1<br />
10 ln 0.6)t = 1 50<br />
( ) 1<br />
10 ln 0.6 t = ln 1<br />
50<br />
t ≈ 76.582<br />
10 ln 0.6)t<br />
common error<br />
Sometimes students think that a function<br />
cannot cross a horizontal asymptote. A<br />
function cannot cross a vertical asymptote;<br />
however, horizontal asymptotes describe<br />
the function’s behavior as x approaches<br />
positive or negative infinity. Horizontal<br />
asymptotes give no information about<br />
the function’s behavior anywhere else. A<br />
function may cross a horizontal asymptote<br />
for values of x that are not near infinity;<br />
however, as x becomes infinitely large (or<br />
small), the function will approach, but will<br />
not cross, a limiting value. That limiting<br />
value defines the horizontal asymptote.<br />
Here is a graph of the function<br />
x<br />
fx=− + 2<br />
( )<br />
x + 1 . This function has a<br />
2<br />
horizontal asymptote at y = 0.<br />
The function crosses that asymptote at<br />
x = −2, then slowly approaches the<br />
horizontal asymptote from above as x<br />
approaches negative infinity.<br />
y<br />
After approximately 76.6 h, 1 kg of undissolved salt will remain.<br />
2<br />
(c) Since 1<br />
25<br />
ln 0.6 ≈−0.051, we have lim A(t) = lim<br />
10 t→∞ t→∞ (25e−0.051t ) = lim = 0<br />
t→∞ e 0.051t<br />
(d) As t becomes unbounded, the amount of undissolved salt in the water approaches<br />
0 kg. Eventually, there will be no undissolved salt. ■<br />
1<br />
28<br />
26<br />
24<br />
22<br />
21<br />
2<br />
x<br />
NOW WORK Problem 79.<br />
22<br />
4 Find the Horizontal Asymptotes of a Graph<br />
Limits at infinity have an important geometric interpretation. When lim f (x) = M,<br />
x→∞<br />
it means that as x becomes unbounded in the positive direction, the value of f can<br />
be made as close as we please to a number M. That is, the graph of y = f (x) is<br />
as close as we please to the horizontal line y = M by choosing x sufficiently large.<br />
Similarly, lim f (x) = L means that the graph of y = f (x) is as close as we please<br />
x→−∞<br />
to the horizontal line y = L for x unbounded in the negative direction. These lines are<br />
horizontal asymptotes of the graph of f.<br />
TRM Section 1.5: Worksheet 3<br />
This worksheet contains 4 rational<br />
functions and their corresponding graphs<br />
to help the students identify the horizontal<br />
asymptotes of each of the functions.<br />
common error<br />
Functions may have infinitely many vertical<br />
asymptotes. Students may incorrectly believe<br />
that functions can have infinitely many horizontal<br />
asymptotes as well. Remind them that functions may<br />
have 0, 1, or 2 horizontal asymptotes. A function will<br />
not have more than 2 horizontal asymptotes, but a<br />
function can have infinitely many vertical asymptotes.<br />
Why can a function have no more than 2 horizontal<br />
asymptotes? Recall that a horizontal asymptote<br />
is found by finding the limit of the function as x<br />
approaches positive infinity and as x approaches<br />
negative infinity. If both of those limits yield unique<br />
constants, the function has 2 horizontal asymptotes.<br />
If one of those two limits is constant, or if they<br />
have the same value, the function has 1 horizontal<br />
asymptote. If neither of those limits is constant, the<br />
function has 0 horizontal asymptotes.<br />
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138 Chapter 1 • Limits and Continuity<br />
Teaching Tip<br />
Examples of functions that have 0, 1, or 2<br />
horizontal asymptotes:<br />
Zero Horizontal Asymptotes<br />
y = x 2<br />
y = sinx<br />
One Horizontal Asymptote<br />
y = e x<br />
This function has the horizontal asymptote<br />
y = 0.<br />
2x<br />
+ 1<br />
y =<br />
x<br />
This function has the horizontal asymptote<br />
y = 2.<br />
Two Horizontal Asymptotes<br />
−<br />
y = tan 1 x<br />
This function has horizontal asymptotes at<br />
π<br />
y = ± . 2<br />
| x|<br />
y =<br />
x<br />
This function has horizontal asymptotes at<br />
y = ± 1.<br />
Teaching Tip<br />
Functions may have 0, 1, or 2 horizontal<br />
asymptotes. Use this fact to remind<br />
students that they have to find the limit<br />
of the function as x approaches positive<br />
infinity and as x approaches negative<br />
infinity.<br />
lim f(x) L<br />
x→ ∞<br />
Horizontal<br />
asymptote<br />
y M<br />
Figure 62<br />
y f (x)<br />
y<br />
y<br />
4<br />
2<br />
2<br />
Horizontal<br />
asymptote<br />
y L<br />
x<br />
lim f(x) M<br />
x→ ∞<br />
3<br />
y <br />
4<br />
4 2<br />
2 4<br />
Figure 63 f (x) = 3x − 2<br />
4x − 1<br />
x<br />
DEFINITION Horizontal Asymptote<br />
The line y = L is a horizontal asymptote of the graph of a function f for x unbounded<br />
in the negative direction if lim f (x) = L.<br />
x→−∞<br />
The line y = M is a horizontal asymptote of the graph of a function f for x<br />
unbounded in the positive direction if lim f (x) = M.<br />
x→∞<br />
In Figure 62, y = L is a horizontal asymptote as x → −∞ because lim f (x) = L.<br />
x→−∞<br />
The line y = M is a horizontal asymptote as x →∞because lim f (x) = M. To<br />
x→∞<br />
identify horizontal asymptotes, we find the limits of f at infinity.<br />
EXAMPLE 11 Finding the Horizontal Asymptotes of a Graph<br />
Find the horizontal asymptotes, if any, of the graph of f (x) = 3x − 2<br />
4x − 1 .<br />
3x − 2 3x − 2<br />
Solution We examine the two limits at infinity: lim and lim<br />
x→−∞ 4x − 1 x→∞ 4x − 1 .<br />
3x − 2<br />
Since lim<br />
x→−∞ 4x − 1 = 3 4 , the line y = 3 is a horizontal asymptote of the graph<br />
4<br />
of f for x unbounded in the negative direction.<br />
3x − 2<br />
Since lim<br />
x→∞ 4x − 1 = 3 4 , the line y = 3 is a horizontal asymptote of the graph<br />
4<br />
of f for x unbounded in the positive direction. ■<br />
Figure 63 shows the graph of f (x) = 3x − 2<br />
4x − 1 and the horizontal asymptote y = 3 4 .<br />
NOW WORK Problem 63 (find any horizontal asymptotes)<br />
and AP®Practice Problems 1, 4, and 7.<br />
5 Find the Asymptotes of the Graph of a Rational Function<br />
In the next example we find the horizontal asymptotes and vertical asymptotes, if any,<br />
of the graph of a rational function.<br />
EXAMPLE 12<br />
Finding the Asymptotes of the Graph of a<br />
Rational Function<br />
Find any asymptotes of the graph of the rational function R(x) = 3x 2 − 12<br />
2x 2 − 9x + 10 .<br />
Solution We begin by factoring R.<br />
R(x) = 3x 2 − 12 3(x − 2)(x + 2)<br />
=<br />
2x 2 − 9x + 10 (2x − 5)(x − 2)<br />
The domain of R is<br />
{x| x = 5 }<br />
2 and x = 2 . Since R is a rational function, it is<br />
continuous on its domain, that is, all real numbers except x = 5 and x = 2.<br />
2<br />
To check for vertical asymptotes, we find the limits as x approaches 5 and 2. First<br />
2<br />
we consider lim<br />
x→ 5 2<br />
lim<br />
x→ 5 −<br />
2<br />
R(x) = lim<br />
x→ 5 −<br />
2<br />
− R(x).<br />
[ 3(x − 2)(x + 2)<br />
(2x − 5)(x − 2)<br />
]<br />
= lim<br />
x→ 5 −<br />
2<br />
[ ] 3(x + 2)<br />
(2x − 5)<br />
= 3 lim<br />
x→ 5 −<br />
2<br />
x + 2<br />
2x − 5 = −∞<br />
That is, as x approaches 5 from the left, R becomes unbounded in the negative direction.<br />
2<br />
The graph of R has a vertical asymptote on the left at x = 5 2 .<br />
AP® CaLC skill builder<br />
for example 12<br />
Finding the Asymptotes of the Graph of a<br />
Rational Function<br />
2x<br />
+ 1<br />
Find any asymptotes of the function fx ( ) =<br />
x − 2 .<br />
Solution<br />
Vertical Asymptote(s)<br />
The function is undefined at x = 2, so we will<br />
check the one-sided limits at x = 2.<br />
x +<br />
lim 2 1<br />
− ≈ a number close to 5<br />
−<br />
→ x 2 very small negative and<br />
x 2<br />
x +<br />
thus lim 2 1<br />
−<br />
≈−∞ .<br />
−<br />
x→2<br />
x 2<br />
We do not have to do further verification. This<br />
function has a vertical asymptote at x = 2.<br />
Horizontal Asymptote(s)<br />
We must find the limit of the function as x<br />
approaches positive and negative infinity.<br />
x +<br />
− = x<br />
lim 2 1 lim 2 = lim 2 = 2<br />
x →∞ x 2 x →∞ x x →∞<br />
The function has a horizontal asymptote at y = 2.<br />
x +<br />
− = x<br />
lim 2 1 lim 2 = lim 2 = 2<br />
x →−∞ x 2 x →−∞ x x →−∞<br />
Since these two limits are the same, this function<br />
has only one horizontal asymptote: y = 2.<br />
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Section 1.5 • Assess Your Understanding 139<br />
3<br />
y <br />
2<br />
y<br />
8<br />
4<br />
8 4<br />
4 8 12<br />
4<br />
8<br />
12<br />
5<br />
x <br />
2<br />
Figure 64 R(x) = 3x2 − 12<br />
2x 2 − 9x + 10<br />
1.5 Assess Your Understanding<br />
Concepts and Vocabulary<br />
1. True or False ∞ is a number.<br />
1<br />
2. (a) lim =<br />
x→0 − x<br />
1<br />
; (b) lim<br />
x→0 + x = ;<br />
(c) lim x→0 +<br />
x<br />
To determine the behavior to the right of x = 5 , we find the right-hand limit.<br />
2<br />
lim<br />
x→ 5 +<br />
2<br />
R(x) = lim<br />
x→ 5 +<br />
2<br />
[ ] 3(x + 2)<br />
(2x − 5)<br />
= 3 lim<br />
x→ 5 +<br />
2<br />
x + 2<br />
2x − 5 =∞<br />
As x approaches 5 from the right, R becomes unbounded in the positive direction. The<br />
2<br />
graph of R has a vertical asymptote on the right at x = 5 2 .<br />
Next we consider lim<br />
x→2<br />
R(x).<br />
lim R(x) = lim<br />
x→2 x→2<br />
3. True or False The graph of a rational function has a vertical<br />
asymptote at every number x at which the function is not defined.<br />
4. If lim f (x) =∞, then the line x = 4 is a(n)<br />
x→4<br />
asymptote<br />
of the graph of f.<br />
1<br />
1<br />
5. (a) lim = ; (b) lim<br />
x→∞ x x→∞ x 2 = ;<br />
(c) lim x→∞<br />
6. True or False lim x→−∞<br />
7. (a) lim<br />
x→−∞ ex = ; (b) lim<br />
x→∞ ex = ; (c) lim<br />
x→∞ e−x =<br />
8. True or False The graph of a function can have at most two<br />
horizontal asymptotes.<br />
3(x − 2)(x + 2)<br />
(2x − 5)(x − 2) = lim<br />
x→2<br />
3(x + 2)<br />
2x − 5<br />
=<br />
3(2 + 2)<br />
2 · 2 − 5 = 12<br />
−1 =−12<br />
Since the limit is not infinite, the function R does not have a vertical asymptote at 2.<br />
Since 2 is not in the domain of R, the graph of R has a hole at the point (2, −12).<br />
To check for horizontal asymptotes, we find the limits at infinity.<br />
3x 2 − 12<br />
lim R(x) = lim<br />
x→∞ x→∞ 2x 2 − 9x + 10 = lim 3x 2<br />
x→∞ 2x = lim 3<br />
2 x→∞ 2 = 3 ↑<br />
2<br />
(2)<br />
lim R(x) = lim 3x 2 − 12<br />
x→−∞ x→−∞ 2x 2 − 9x + 10 = lim 3x 2<br />
x→−∞ 2x = lim 3<br />
2 x→−∞ 2 = 3 ↑<br />
2<br />
(2)<br />
The line y = 3 is a horizontal asymptote of the graph of R for x unbounded in the<br />
2<br />
negative direction and for x unbounded in the positive direction. ■<br />
The graph of R and its asymptotes are shown in Figure 64. Notice the hole in the<br />
graph at the point (2, −12).<br />
NOW WORK Problem 69 and AP® Practice Problems 8 and 10.<br />
Skill Building<br />
In Problems 9–16, use the accompanying graph of y = f (x).<br />
9. Find lim f (x).<br />
x→∞<br />
10. Find lim f (x).<br />
x→−∞<br />
PAGE<br />
129 11. Find lim f (x).<br />
x→−1 −<br />
12. Find lim f (x).<br />
x→−1 +<br />
13. Find lim f (x).<br />
x→3 −<br />
14. Find lim f (x).<br />
x→3 +<br />
PAGE<br />
131 15. Identify all<br />
vertical<br />
asymptotes.<br />
16. Identify all<br />
horizontal<br />
asymptotes.<br />
y<br />
12<br />
8<br />
4<br />
4 4 8 12<br />
x 1 x 3<br />
y 2<br />
x<br />
Must-Do Exercises for<br />
Exam Readiness<br />
AB: 2–26, 27–59 odd, 67–71 odd, AP ®<br />
Practice Problems<br />
BC: 17–26, 30–35, 45, 53, 55, 63, 73, 78,<br />
all AP ® Practice Problems<br />
TRM Full Solutions to Section<br />
1.5 Problems and AP® Practice<br />
Problems<br />
Answers to Section 1.5<br />
Problems<br />
1. False.<br />
2. (a) −∞ (b) ∞ (c) − ∞<br />
3. False.<br />
4. Vertical.<br />
5. (a) 0 (b) 0 (c) ∞<br />
6. False.<br />
7. (a) 0 (b) ∞ (c) 0<br />
8. True.<br />
9. 2<br />
10. 0<br />
11. ∞<br />
12. ∞<br />
13. ∞<br />
14. ∞<br />
15. x = −1, x = 3<br />
16. y = 0, y = 2<br />
Section 1.5 • Assess Your Understanding<br />
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140 Chapter 1 • Limits and Continuity<br />
17. −3 18. 0<br />
19. ∞ 20. − ∞<br />
21. 0 22. ∞<br />
23. ∞ 24. ∞<br />
25. x = −3, x = 0, x = 4<br />
26. y = 0, y =−3<br />
27. − ∞ 28. − ∞ 29. ∞<br />
30. − ∞ 31. ∞ 32. ∞<br />
33. ∞ 34. ∞ 35. − ∞<br />
36. ∞ 37. − ∞ 38. ∞<br />
39. − ∞ 40. − ∞ 41. − ∞<br />
42. − ∞ 43. 0 44. 0<br />
45. 2 5<br />
46. 1 47. 1<br />
48. 2 5<br />
49. 0 50. 0<br />
51. 5 4<br />
52. − 1 3<br />
53. 0<br />
54. 0 55. 0 56. 0<br />
57. 3 58. 4 59. −∞<br />
60. −∞<br />
61. x = 0 is a vertical asymptote. y = 3 is<br />
a horizontal asymptote.<br />
62. x = 0 is a vertical asymptote. y = 2 is<br />
a horizontal asymptote.<br />
63. x = −1 and x = 1 are vertical<br />
asymptotes. y = 1 is a horizontal<br />
asymptote.<br />
64. x = −1 and x = 1 are vertical<br />
asymptotes. y = 2 is a horizontal<br />
asymptote.<br />
3<br />
65. x = is a vertical asymptote.<br />
2<br />
2<br />
2<br />
y =− and y = are<br />
2<br />
2<br />
horizontal asymptotes.<br />
66. x = 6 is a vertical asymptote. y = 0 is<br />
a horizontal asymptote.<br />
67. (a) { x| x ≠−2, x ≠ 0}<br />
(b) y = 0 is a horizontal asymptote.<br />
(c) x = −2 and x = 0 are vertical<br />
asymptotes.<br />
(d) See TSM.<br />
68. (a) { x| x ≠± 1}<br />
(b) y = 0 is a horizontal asymptote.<br />
(c) x = −1 and x = 1 are vertical<br />
asymptotes.<br />
(d) See TSM.<br />
⎧ 3 ⎫<br />
69. (a) ⎨xx<br />
| ≠ x ≠ ⎬<br />
⎩ 2 , 2<br />
⎭<br />
1<br />
(b) y = is a horizontal asymptote.<br />
2<br />
3<br />
(c) x = is a vertical asymptote.<br />
2<br />
(d) See TSM.<br />
In Problems 17–26, use the graph below of y = f (x).<br />
17. Find lim f (x). 18. Find lim f (x).<br />
x→∞ x→−∞<br />
19. Find lim f (x). 20. Find lim f (x).<br />
x→−3− x→−3 +<br />
21. Find lim<br />
x→0 − f (x).<br />
23. Find lim<br />
x→4 − f (x).<br />
25. Identify all vertical asymptotes.<br />
26. Identify all horizontal asymptotes.<br />
y 3<br />
y<br />
22. Find lim f (x).<br />
x→0 +<br />
24. Find lim f (x).<br />
x→4 +<br />
y 0 4 2<br />
2 4 6<br />
4<br />
2<br />
2<br />
4<br />
In Problems 27–42, find each limit.<br />
PAGE<br />
130 27. lim<br />
x→2 −<br />
3x<br />
x − 2<br />
x→2 + x 2 − 4<br />
29. lim<br />
5<br />
31.<br />
5x + 3<br />
lim<br />
x→−1 + x(x + 1)<br />
x→−3 − x 2 − 9<br />
33. lim<br />
1<br />
35. lim<br />
1 − x<br />
x→3<br />
x 3 x 0 x 4<br />
(3 − x) 2 36. lim<br />
2x + 1<br />
28. lim<br />
x→−4 + x + 4<br />
2x<br />
30. lim<br />
x→1 − x 3 − 1<br />
5x + 3<br />
32. lim<br />
x→0 − 5x(x − 1)<br />
x<br />
34. lim<br />
x→2 + x 2 − 4<br />
x + 2<br />
x→−1 (x + 1) 2<br />
37. lim cot x 38. lim tan x<br />
x→π− x→−π/2 −<br />
39. lim csc(2x)<br />
x→π/2 +<br />
40. lim x→−π/2 −<br />
41. lim ln(x + 1)<br />
x→−1 +<br />
42. lim ln(x − 1)<br />
x→1 +<br />
In Problems 43–60, find each limit.<br />
5<br />
43. lim<br />
x→∞ x 2 + 4<br />
PAGE<br />
133 45.<br />
2x + 4<br />
lim<br />
x→∞ 5x<br />
PAGE<br />
134 47.<br />
x 3 + x 2 + 2x − 1<br />
lim<br />
x→∞ x 3 + x + 1<br />
PAGE<br />
134 49.<br />
x 2 + 1<br />
lim<br />
x→−∞ x 3 − 1<br />
51.<br />
[ ]<br />
3x<br />
lim<br />
x→∞ 2x + 5 − x2 + 1<br />
4x 2 + 8<br />
1<br />
44. lim<br />
x→−∞ x 2 − 9<br />
x + 1<br />
46. lim<br />
x→∞ x<br />
2x 2 − 5x + 2<br />
48. lim<br />
x→∞ 5x 2 + 7x − 1<br />
x 2 − 2x + 1<br />
50. lim<br />
x→∞ x 3 + 5x + 4<br />
[<br />
52. lim<br />
x→∞<br />
70. (a) { x| x ≠−3}<br />
(b) y = 1 is a horizontal asymptote.<br />
(c) x =−3 is a vertical asymptote.<br />
(d) See TSM.<br />
71. (a) { x| x ≠0, x ≠1}<br />
(b) There are no horizontal asymptotes.<br />
(c) x = 0 is a vertical asymptote.<br />
(d) See TSM.<br />
72. (a) { x| x ≠ 1}<br />
(b) There are no horizontal asymptotes.<br />
(c) x = 1 is a vertical asymptote.<br />
(d) See TSM.<br />
x<br />
PAGE<br />
131<br />
1<br />
x 2 + x + 4 − x + 1<br />
3x − 1<br />
]<br />
[ ( )] 5x + 1<br />
53. lim 2e x<br />
x→−∞ 3x<br />
√ x + 2<br />
55. lim<br />
PAGE<br />
134 57. lim<br />
x→∞<br />
x→∞ 3x − 4<br />
√<br />
3x 2 − 1<br />
x 2 + 4<br />
PAGE<br />
135 59.<br />
5x 3<br />
lim<br />
x→−∞ x 2 + 1<br />
54. lim<br />
x→−∞<br />
[<br />
( )]<br />
x<br />
e x 2 + x − 3<br />
2x 3 − x 2<br />
√<br />
3x<br />
56. lim<br />
3 + 2<br />
x→∞ x 2 + 6<br />
( ) 16x 3 2/3<br />
+ 2x + 1<br />
58. lim<br />
x→∞ 2x 3 + 3x<br />
x 4<br />
60. lim<br />
x→−∞ x − 2<br />
In Problems 61–66, find any horizontal or vertical asymptotes of the<br />
graph of f .<br />
61. f (x) = 3 + 1 x<br />
PAGE<br />
138 63. f (x) = x2<br />
x 2 − 1<br />
65. f (x) =<br />
√<br />
2x 2 − x + 10<br />
2x − 3<br />
62. f (x) = 2 − 1 x 2<br />
64. f (x) = 2x2 − 1<br />
x 2 − 1<br />
√<br />
3<br />
x<br />
66. f (x) =<br />
2 + 5x<br />
x − 6<br />
In Problems 67–72, for each rational function R:<br />
(a) Find the domain of R.<br />
(b) Find any horizontal asymptotes of R.<br />
(c) Find any vertical asymptotes of R<br />
(d) Discuss the behavior of the graph at numbers where R is not<br />
defined.<br />
67. R(x) = −2x2 + 1<br />
2x 3 + 4x 2 68. R(x) = x3<br />
x 4 − 1<br />
PAGE<br />
139 69. R(x) = x2 + 3x − 10<br />
2x 2 − 7x + 6<br />
70. R(x) =<br />
x(x − 1)2<br />
(x + 3) 3<br />
71. R(x) = x3 − 1<br />
x − x 2 72. R(x) = 4x5<br />
x 3 − 1<br />
Applications and Extensions<br />
In Problems 73 and 74:<br />
(a) Sketch a graph of a function f that has the given properties.<br />
(b) Define a function that describes the graph.<br />
73. f (3) = 0, lim f (x) = 1, lim<br />
x→∞<br />
lim f (x) =∞, lim<br />
x→1− 74. f (2) = 0, lim<br />
x→∞<br />
f (x) =∞, lim<br />
lim<br />
x→0<br />
f (x) = 1,<br />
x→−∞<br />
f (x) = −∞<br />
x→1 +<br />
f (x) = 0, lim<br />
f (x) = −∞, lim<br />
x→5− f (x) = 0,<br />
x→−∞<br />
f (x) =∞<br />
x→5 +<br />
75. Newton’s Law of Cooling Suppose an object is heated to a<br />
temperature u 0. Then at time t = 0, the object is put into a<br />
medium with a constant lower temperature T causing the object to<br />
cool. Newton’s Law of Cooling states that the temperature u of<br />
the object at time t is given by u = u(t) = (u 0 − T )e kt + T ,<br />
where k < 0 is a constant.<br />
(a) Find lim u(t). Is this the value you expected? Explain why<br />
t→∞<br />
or why not.<br />
(b) Find lim u(t). Is this the value you expected? Explain why<br />
t→0 +<br />
or why not.<br />
Source: Submitted by the students of Millikin University.<br />
73. Answers will vary. Sample answer:<br />
(a)<br />
26 24 22<br />
y<br />
6<br />
4<br />
2<br />
24<br />
26<br />
x<br />
(b) fx= − 3<br />
( )<br />
x −1<br />
2 4 6<br />
x<br />
Answers continue on p. 141<br />
140<br />
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76. Environment A utility company burns coal to generate<br />
electricity. The cost C, in dollars, of removing p% of the<br />
pollutants emitted into the air is<br />
Find the cost of removing:<br />
C = 70,000p<br />
100 − p , 0 ≤ p < 100<br />
(a) 45% of the pollutants.<br />
(b) 90% of the pollutants.<br />
(c) Find lim C.<br />
p→100 −<br />
(d) Interpret the answer found in (c).<br />
77. Pollution Control The cost C, in thousands of dollars, to<br />
remove a pollutant from a lake is<br />
C(x) =<br />
5x<br />
100 − x , 0 ≤ x < 100<br />
where x is the percent of pollutant removed. Find lim C(x).<br />
x→100 −<br />
Interpret your answer.<br />
78. Population Model A rare species of insect was discovered in<br />
the Amazon Rain Forest. To protect the species, entomologists<br />
declared the insect endangered and transferred 25 insects to a<br />
protected area. The population P of the new colony t days after<br />
the transfer is<br />
50(1 + 0.5t)<br />
P(t) =<br />
2 + 0.01t<br />
(a) What is the projected size of the colony after 1 year<br />
(365 days)?<br />
(b) What is the largest population that the protected area can<br />
sustain? That is, find lim<br />
t→∞ P(t).<br />
(c) Graph the population P as a function of time t.<br />
(d) Use the graph from (c) to describe the regeneration of the<br />
insect population. Does the graph support the answer to (b)?<br />
PAGE<br />
137 79. Population of an Endangered Species Often environmentalists<br />
capture several members of an endangered species and transport<br />
them to a controlled environment where they can produce<br />
offspring and regenerate their population. Suppose six American<br />
bald eagles are captured, tagged, transported to Montana, and set<br />
free. Based on past experience, the environmentalists expect the<br />
population to grow according to the model<br />
where t is measured in years.<br />
500<br />
P(t) =<br />
1 + 82.3e −0.162t<br />
(a) If the model is correct, how many bald eagles can the<br />
environment sustain? That is, find lim<br />
t→∞ P(t).<br />
(b) Graph the population P as a function of time t.<br />
(c) Use the graph from (b) to describe the growth of the<br />
bald eagle population. Does the graph support the answer<br />
to (a)?<br />
80. Hailstones Hailstones typically originate at an altitude of about<br />
3000 meters (m). If a hailstone falls from 3000 m with no air<br />
resistance, its speed when it hits the ground would be about<br />
240 meters/second (m/s), which is 540 miles/hour (mi/h)! That<br />
would be deadly! But air resistance slows the hailstone<br />
considerably. Using a simple model of air resistance, the speed<br />
v = v(t) of a hailstone of mass m as a function of time t is given<br />
by v(t) = mg<br />
k (1 − e−kt/m ) m/s, where g = 9.8m/s 2 and k is a<br />
constant that depends on the size of the hailstone, its mass, and the<br />
conditions of the air. For a hailstone with a diameter<br />
d = 1 centimeter (cm) and mass m = 4.8 × 10 −4 kg, k has been<br />
measured to be 3.4 × 10 −4 kg/s.<br />
(a) Determine the limiting speed of the hailstone by finding<br />
lim v(t). Express your answer in meters per second and<br />
t→∞<br />
miles per hour, using the fact that 1 mi/h ≈ 0.447 m/s. This<br />
speed is called the terminal speed of the hailstone.<br />
(b) Graph v = v(t). Does the graph support the answer to (a)?<br />
81. Damped Harmonic Motion The motion of a spring is given by<br />
the function<br />
x(t) = 1.2e −t/2 cos t + 2.4e −t/2 sin t<br />
where x is the distance in meters from the the equilibrium position<br />
and t is the time in seconds.<br />
(a) Graph y = x(t). What is lim x(t), as suggested by the<br />
t→∞<br />
graph?<br />
(b) Find lim x(t).<br />
t→∞<br />
(c) Compare the results of (a) and (b). Is the answer to (b)<br />
supported by the graph in (a)?<br />
82. Decomposition of Chlorine in a Pool Under certain water<br />
conditions, the free chlorine (hypochlorous acid, HOCl) in a<br />
swimming pool decomposes according to the law of uninhibited<br />
decay, C = C(t) = C(0)e kt , where C = C(t) is the amount<br />
(in parts per million, ppm) of free chlorine present at time t<br />
(in hours) and k is a negative number that represents the rate<br />
of decomposition. After shocking his pool, Ben immediately<br />
tested the water and found the concentration of free chlorine to<br />
be C 0 = C(0) = 2.5 ppm. Twenty-four hours later, Ben<br />
tested the water again and found the amount of free chlorine<br />
to be 2.2 ppm.<br />
(a) What amount of free chlorine will be left after 72 hours?<br />
(b) When the free chlorine reaches 1.0 ppm, the pool should be<br />
shocked again. How long can Ben go before he must shock<br />
the pool again?<br />
(c) Find lim<br />
t→∞ C(t).<br />
(d) Interpret the answer found in (c).<br />
83. Decomposition of Sucrose Reacting with water in an<br />
acidic solution at 35 ◦ C, the amount A of sucrose (C 12H 22O 11)<br />
decomposes into glucose (C 6H 12O 6) and fructose (C 6H 12O 6)<br />
according to the law of uninhibited decay A = A(t) = A(0)e kt ,<br />
where A = A(t) is the amount (in moles) of sucrose present at<br />
time t (in minutes) and k is a negative number that represents the<br />
rate of decomposition. An initial amount A 0 = A(0) = 0.40 mole<br />
of sucrose decomposes to 0.36 mole in 30 minutes.<br />
(a) How much sucrose will remain after 2 hours?<br />
(b) How long will it take until 0.10 mole of sucrose remains?<br />
(c) Find lim<br />
t→∞ A(t).<br />
(d) Interpret the answer found in (c).<br />
84. Macrophotography A camera lens can be approximated by a<br />
thin lens. A thin lens of focal length f obeys the thin-lens<br />
equation 1 f<br />
78. (a) 1624 insects<br />
(b) 2500 insects<br />
(c)<br />
y<br />
2500<br />
2000<br />
1500<br />
1000<br />
t 5 1 year<br />
500<br />
1000 2000 3000 4000 5000 x<br />
Time (days)<br />
(d) The graph supports the answer<br />
to (b).<br />
79. (a) 500 bald eagles<br />
(b)<br />
y<br />
500<br />
400<br />
300<br />
200<br />
100<br />
20 40 60 80 100 x<br />
Time (years)<br />
(c) Answers will vary. See TSM for<br />
sample answer. The graph supports the<br />
answer to (a).<br />
80. (a) 13.84 m/s or 30.96 mi/h<br />
(b)<br />
y<br />
13.84<br />
12<br />
10<br />
8<br />
6<br />
4<br />
2<br />
2 4 6 8 10 x<br />
Time (seconds)<br />
The graph supports the answer to (a).<br />
81. (a)<br />
x<br />
2.0<br />
1.5<br />
= 1 p + 1 q , where p > f is the distance from the 1.0<br />
Number of insects<br />
Number of bald eagles<br />
Speed (meters/second)<br />
0.5<br />
74. Answers will vary. Sample answer:<br />
(a)<br />
76. (a) $57,272.73<br />
(b) $630,000<br />
y<br />
4<br />
(c) ∞<br />
2<br />
24 22 2 4 6 8 x<br />
22<br />
24<br />
x − 2<br />
(b) fx ( ) =<br />
2<br />
x ( x−<br />
5)<br />
(b) u 0<br />
75. (a) T<br />
(d) Interpretations will vary. See TSM for<br />
sample interpretation.<br />
77. ∞. Interpretations will vary. Sample<br />
interpretation: As the percentage of the<br />
pollutants removed from the air increases<br />
toward 100%, the cost of removing those<br />
pollutants increases without bound.<br />
0.5<br />
1 2 3 4 5 6 7 8 9<br />
The graph suggests that lim xt () = 0.<br />
t→∞<br />
(b) 0<br />
(c) The graph in (a) supports the<br />
answer in (b).<br />
82. (a) Approx. 1.7 ppm<br />
(b) Roughly 172 h<br />
(c) 0 ppm<br />
(d) In the long run, all of the free<br />
chlorine in the pool will decompose.<br />
t<br />
Answers continue on p. 142<br />
Section 1.5 • Assess Your Understanding<br />
141<br />
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142 Chapter 1 • Limits and Continuity<br />
83. (a) 0.26 moles<br />
(b) 395 min<br />
(c) 0 moles<br />
(d) In the long run, all of the sucrose<br />
will decompose. See TSM.<br />
84. (a) Not continuous.<br />
(b) A camera cannot focus on an object<br />
placed close to its focal length because<br />
the distance of the image from the lens<br />
becomes unbounded.<br />
85. a<br />
≠ 0; b can be any real number;<br />
c = 0; and d ≠ 0. See TSM.<br />
86. a ≠ 0; b can be any real number;<br />
c = 0; and d ≠ 0.<br />
87. See TSM.<br />
88. See TSM.<br />
89. See TSM.<br />
90. See TSM.<br />
91. (a) For table, see TSM.<br />
(b) e ≈ 2.718281828<br />
(c) Answers will vary. Sample answer:<br />
The results from (a) and (b) agree to<br />
five decimal places.<br />
92. The property requires the exponent to<br />
be a constant, independent of x, but in<br />
x<br />
⎛ 1 ⎞<br />
lim +<br />
→∞⎝<br />
⎜1<br />
x ⎠<br />
⎟ the exponent is x.<br />
x<br />
93. (a) ∞<br />
(b) The result suggests that it is not<br />
possible to reach the speed of light.<br />
lens to the object being photographed and q is the distance from<br />
the lens to the image formed by the lens. See the figure below.<br />
To photograph an object, the object’s image must be formed on<br />
the photo sensors of the camera, which can only occur if q is<br />
positive.<br />
Object<br />
p<br />
Lens<br />
q<br />
Photo<br />
sensors<br />
(a) Is the distance q of the image from the lens continuous as the<br />
distance of the object being photographed approaches the<br />
focal length f of the lens? (Hint: First solve the thin-lens<br />
equation for q and then find lim q.) p→ f +<br />
(b) Use the result from (a) to explain why a camera (or any lens)<br />
cannot focus on an object placed close to its focal length.<br />
In Problems 85 and 86, find conditions on a, b, c, and d so that the<br />
graph of f has no horizontal or vertical asymptotes.<br />
85. f (x) = ax3 + b<br />
cx 4 + d<br />
86. f (x) = ax + b<br />
cx + d<br />
87. Explain why the following properties are true. Give an example of<br />
each.<br />
1<br />
(a) If n is an even positive integer, then lim x→c (x − c) n =∞.<br />
1<br />
(b) If n is an odd positive integer, then lim<br />
x→c − (x − c) n = −∞.<br />
1<br />
(c) If n is an odd positive integer, then lim<br />
x→c + (x − c) n =∞.<br />
88. Explain why a rational function, whose numerator and<br />
denominator have no common zeros, will have vertical<br />
asymptotes at each point of discontinuity.<br />
89. Explain why a polynomial function of degree 1 or higher cannot<br />
have any asymptotes.<br />
AP® Practice Problems<br />
PAGE<br />
138 1. For x > 0, the line y = 1 is an asymptote of the graph of a<br />
function f . Which of the following statements must be true?<br />
(A) f (x) = 1 for x > 0. (B) lim f (x) =∞<br />
x→1<br />
(C)<br />
lim<br />
x→∞<br />
PAGE<br />
3x 3 + 4x 2 − x + 10<br />
135 2. lim<br />
x→∞ 2x 4 − x 3 + 2x 2 − 2 =<br />
3<br />
(A) –5 (B) 0 (C)<br />
2<br />
PAGE<br />
5x 3 − x<br />
135 3. lim<br />
x→∞ 8 − x 3 =<br />
5<br />
(A) –5 (B) (C) 5<br />
8<br />
f (x) = 1 (D) lim<br />
x→−∞ f (x) = 1<br />
(D) ∞<br />
(D) ∞<br />
90. If P and Q are polynomials of degree m and n, respectively,<br />
P(x)<br />
discuss lim<br />
x→∞ Q(x) when:<br />
(a) m > n (b) m = n (c) m < n<br />
<br />
91. (a) Use a table to investigate lim 1 + 1 x<br />
.<br />
x→∞ x<br />
<br />
CAS (b) Find lim 1 + 1 x<br />
.<br />
x→∞ x<br />
(c) Compare the results from (a) and (b). Explain the possible<br />
causes of any discrepancy.<br />
Challenge Problems<br />
<br />
92. lim 1 + 1 <br />
x→∞ x<br />
= 1, but lim<br />
x→∞<br />
<br />
1 + 1 x<br />
> 1. Discuss why the<br />
x<br />
<br />
lim<br />
x→∞ f (x) n<br />
cannot be used to find the<br />
property lim [ f x→∞ (x)]n =<br />
second limit.<br />
93. Kinetic Energy At low speeds the kinetic energy K , that is, the<br />
energy due to the motion of an object of mass m and speed v, is<br />
given by the formula K = K (v) = 1 2 mv2 . But this formula is<br />
only an approximation to the general formula, and works only for<br />
speeds much less than the speed of light, c. The general formula,<br />
which holds for all speeds, is<br />
⎡<br />
⎤<br />
Kgen(v) = mc 2 ⎢<br />
1<br />
⎣<br />
− 1⎥<br />
⎦<br />
1 − v2<br />
c 2<br />
(a) As an object is accelerated closer and closer to the speed of<br />
light, what does its kinetic energy Kgen approach?<br />
(b) What does the result suggest about the possibility of reaching<br />
the speed of light?<br />
PAGE<br />
138 4. Find all the horizontal asymptotes of the graph of y = 2 + 3x<br />
4 − 3 x .<br />
(A) y = –1 only<br />
(B) y = 1 2 only<br />
(C) y = –1 and y = 0 (D) y = –1 and y = 1 2<br />
PAGE<br />
131 5. Find all the vertical asymptotes of the graph of<br />
r(x) = x2 + 5x + 6<br />
.<br />
x 3 − 4x<br />
(A) x = 0 and x = –2 (B) x = 0 and x = 2<br />
(C) x = –2 and x = 2 (D) x = 0, x = –2 and x = 2<br />
PAGE<br />
134 6. lim<br />
x→−∞<br />
√<br />
8x 2 − 4x<br />
x + 2<br />
=<br />
(A) –∞ (B) −2 √ 2 (C) 4 (D) 2 √ 2<br />
Answers to AP® Practice Problems<br />
1. C<br />
2. B<br />
3. A<br />
4. D<br />
5. B<br />
6. B<br />
142<br />
Chapter 1 • Limits and Continuity<br />
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Section 1.6 • The ε-δ Definition of a Limit 143<br />
PAGE<br />
138 7. The graph of which of the following functions has an<br />
asymptote of y = 1?<br />
(A) y = cos x (B) y = x − 1<br />
x<br />
(C) y = e −x (D) y = ln x<br />
PAGE<br />
139 8. If the graph of f (x) = ax − b has a vertical asymptote<br />
x + c<br />
x = –5 and horizontal asymptote y = –3, then a + c =<br />
3<br />
(A) –8 (B) –2 (C) (D) 2<br />
5<br />
PAGE<br />
x<br />
130 9. lim<br />
x→1 − ln x is<br />
(A) –∞ (B) –1 (C) 1 (D) ∞<br />
PAGE<br />
139 10. The function f (x) = 2x<br />
|x|−1 has<br />
(A) no vertical asymptote and one horizontal asymptote.<br />
(B) one vertical asymptote and one horizontal asymptote.<br />
(C) two vertical asymptotes and one horizontal asymptote.<br />
(D) two vertical asymptotes and two horizontal asymptotes.<br />
PAGE<br />
5x + 1<br />
130 11. lim<br />
x→2 − 2x − 4 =<br />
(A) –∞ (B) − 5 2<br />
(C)<br />
5<br />
2<br />
(D) ∞<br />
7. B<br />
8. D<br />
9. A<br />
10. D<br />
11. A<br />
AP® EXAM INSIGHT The ε-δ definition<br />
of a limit is not assessed on the AP®<br />
Calculus AB or BC Exam. However, your<br />
teacher may include this topic in the<br />
course if time permits. The ε-δ definition<br />
is an essential piece of calculus. The limit<br />
rules and theorems we have used in this<br />
chapter are proved using the ε-δ<br />
definition, which makes the ε-δ<br />
definition part of the foundation of<br />
calculus.<br />
y<br />
10<br />
8<br />
7.3<br />
6.7<br />
6<br />
4<br />
(2, 10)<br />
1.6 The ε-δ Definition of a Limit<br />
OBJECTIVES When you finish this section, you should be able to:<br />
1 Use the ε-δ definition of a limit (p. 145)<br />
Throughout the chapter, we stated that we could be sure a limit was correct only if it<br />
was based on the ε-δ definition of a limit. In this section, we examine this definition and<br />
how to use it to prove a limit exists, to verify the value of a limit, and to show that a limit<br />
does not exist.<br />
Consider the function f defined by<br />
{<br />
3x + 1 if x = 2<br />
f (x) =<br />
10 if x = 2<br />
whose graph is given in Figure 65.<br />
As x gets closer to 2, the value f (x) gets closer to 7. If in fact, by taking x close<br />
enough to 2, we can make f (x) as close to 7 as we please, then lim f (x) = 7.<br />
x→2<br />
Suppose we want f (x) to differ from 7 by less than 0.3; that is,<br />
−0.3 < f (x) − 7 < 0.3<br />
6.7 < f (x)
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<strong>Sullivan</strong><br />
144 Chapter 1 • Limits and Continuity<br />
That is, whenever x = 2 and x differs from 2 by less than 0.1, then f (x) differs from 7<br />
by less than 0.3.<br />
Now, generalizing the question, we ask, for x = 2, how close must x be to 2 to<br />
guarantee that f (x) differs from 7 by less than any given positive number ε? (ε might<br />
be extremely small.) The statement “ f (x) differs from 7 by less than ε” means<br />
−ε < f (x) − 7
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Section 1.6 • The ε-δ Definition of a Limit 145<br />
DEFINITION Limit of a Function<br />
Let f be a function defined everywhere in an open interval containing c, except possibly<br />
at c. Then the limit as x approaches c of f(x) is L, written<br />
lim f (x) = L<br />
x→c<br />
if, given any number ε>0, there is a number δ>0 so that<br />
whenever 0 < |x − c| 0 so that<br />
whenever 0 < |x − (−1)| =|x + 1|
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<strong>Sullivan</strong><br />
146 Chapter 1 • Limits and Continuity<br />
EXAMPLE 2<br />
Using the ε-δ Definition of a Limit<br />
Use the ε-δ definition of a limit to prove that:<br />
(a) lim<br />
x→c<br />
A = A, where A and c are real numbers<br />
(b) lim<br />
x→c<br />
x = c, where c is a real number<br />
Solution (a) f (x) = A is the constant function whose graph is a horizontal line. Given<br />
any ε>0, we must find δ>0 so that whenever 0 < |x − c| < δ, then | f (x) − A|
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Section 1.6 • The ε-δ Definition of a Limit 147<br />
To ensure that both inequalities are satisfied, we select δ to be the smaller of the<br />
numbers 1 and ε {<br />
5 , abbreviated as δ = min 1, ε }<br />
. Now,<br />
5<br />
{<br />
whenever |x − 2| 0, we need to find a positive number δ so that whenever 0 < |x −c| 0, then x > c 2 > 0, and 1 x < 2 c . Now<br />
1<br />
∣ x − 1 |x − c|<br />
c ∣ =<br />
c|x|<br />
< 2 1<br />
·|x − c| Substitute<br />
c2 |x| < 2 c .<br />
We can make<br />
1<br />
∣ x − 1 c2<br />
c ∣
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<strong>Sullivan</strong><br />
148 Chapter 1 • Limits and Continuity<br />
( ) c<br />
So, given any ε>0, we choose δ = min<br />
2 , c2<br />
2 · ε . Then whenever 0 < |x−c| 0. ■ NOW WORK Problem 25.<br />
The ε-δ definition of a limit can be used to show that a limit does not exist, or that a<br />
limit is not equal to a specific number. Example 5 illustrates how the ε-δ definition of a<br />
limit is used to show that a limit is not equal to a specific number.<br />
NOTE In a proof by contradiction, we<br />
assume that the conclusion is not true<br />
and then show this leads to a<br />
contradiction.<br />
EXAMPLE 5<br />
Showing a Limit Is Not Equal to a Specific Number<br />
Use the ε-δ definition of a limit to prove the statement lim<br />
x→3<br />
(4x − 5) = 10.<br />
Solution We use a proof by contradiction. Assume lim<br />
x→3<br />
(4x −5) = 10 and choose ε = 1.<br />
(Any smaller positive number ε will also work.) Then there is a number δ>0, so that<br />
whenever 0 < |x − 3|
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Section 1.6 • The ε-δ Definition of a Limit 149<br />
Suppose x 1 is a rational number satisfying 0 < |x 1 − c| 1 2 , and from the right inequality, we have L < 1 2 .<br />
Since it is impossible for both inequalities to be satisfied, we conclude that lim<br />
x→c<br />
f (x)<br />
does not exist. ■<br />
EXAMPLE 7<br />
Using the ε-δ Definition of a Limit<br />
Prove that if lim<br />
x→c<br />
f (x) >0, then there is an open interval around c, for which f (x) >0<br />
everywhere in the interval except possibly at c.<br />
Solution Suppose lim<br />
x→c<br />
f (x) = L > 0. Then given any ε>0, there is a δ>0 so that<br />
whenever 0 < |x − c|
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150 Chapter 1 • Limits and Continuity<br />
Figures 68 and 69 illustrate limits at infinity.<br />
y<br />
y 5 f (x)<br />
y<br />
y 5 f (x)<br />
L 1 <br />
L<br />
L 2 <br />
L 1 <br />
L<br />
L 2 <br />
M<br />
x . M<br />
x<br />
x , N<br />
N<br />
x<br />
lim f(x) 5 L<br />
x→∞<br />
Figure 68 For any ε>0, there is a positive<br />
number M so that whenever x > M, then<br />
| f (x) − L| 0, there is a<br />
negative number N so that whenever x < N,<br />
then | f (x)−L| 0, suppose there is a δ>0<br />
so that whenever 0 < |x − c| < δ, then | f (x) − L|
<strong>Sullivan</strong> AP˙<strong>Sullivan</strong>˙Chapter01 October 8, 2016 17:4<br />
Section 1.6 • Assess Your Understanding 151<br />
6. True or False A function f has a limit L at infinity, if for<br />
any given ε>0, there is a positive number M so that<br />
whenever x > M, then | f (x) − L| > ε.<br />
Skill Building<br />
In Problems 7–12, for each limit, find the largest δ that “works” for<br />
the given ε.<br />
7. lim x→1<br />
(2x) = 2, ε = 0.01 8. lim x→2<br />
(−3x) =−6, ε = 0.01<br />
9. lim(6x − 1) = 11 10. lim (2 − 3x) = 11<br />
x→2 x→−3<br />
ε = 1 ε = 1 2<br />
3<br />
(<br />
11. lim − 1 ) (<br />
x→2 2 x + 5 = 4 12. lim 3x + 1 )<br />
= 3<br />
x→ 5 2<br />
6<br />
ε = 0.01 ε = 0.3<br />
13. For the function f (x) = 4x − 1, we have lim x→3<br />
f (x) = 11.<br />
For each ε>0, find a δ>0 so that<br />
whenever 0 < |x − 3| 0, find a δ>0 so that<br />
whenever 0 < |x + 2| 0, find a δ>0 so that<br />
∣ whenever 0 < |x + 3| 0, find a δ>0 so that<br />
∣ whenever 0 < |x − 2| 0, let δ ≤ . See TSM for<br />
(c) d ≤ 0.00025<br />
complete proof. 2<br />
ε<br />
(d) δ ≤ ε<br />
4<br />
20. Given any e > 0, let δ ≤ . See TSM for<br />
complete proof. 3<br />
}<br />
.<br />
ε<br />
21. Given any e > 0, let δ ≤ . See<br />
TSM for complete proof. 5<br />
ε<br />
22. Given any e > 0, let δ ≤ . See<br />
TSM for complete proof. 2<br />
⎧ ε ⎫<br />
23. Given any e > 0, let δ ≤ min⎨1, ⎬ .<br />
See TSM for complete proof. ⎩ 3 ⎭<br />
⎧ ε ⎫<br />
24. Given any e > 0, let δ ≤ min⎨1, ⎬ .<br />
See TSM for complete proof. ⎩ 4 ⎭<br />
⎧ ε ⎫<br />
25. Given any e > 0, let δ ≤ min⎨1, 2 ⎬ .<br />
See TSM for complete proof. ⎩ 7 ⎭<br />
⎧ ε ⎫<br />
26. Given any e > 0, let δ ≤ min⎨1, 15 ⎬ .<br />
See TSM for complete proof. ⎩ 4 ⎭<br />
27. Given any e > 0, let δ ≤ ε<br />
3 . See TSM<br />
for complete proof.<br />
28. Given any e > 0, let d ≤ e. See TSM<br />
for complete proof.<br />
⎧ ε ⎫<br />
29. Given any e > 0, let δ ≤ min⎨1, ⎬ .<br />
See TSM for complete proof. ⎩ 3 ⎭<br />
⎧ ε ⎫<br />
30. Given any e > 0, let δ ≤ min⎨1, ⎬ .<br />
See TSM for complete proof. ⎩ 19 ⎭<br />
31. Given any e > 0, let d ≤ min{1, 6e}.<br />
See TSM for complete proof.<br />
⎧ ε ⎫<br />
32. Given any e > 0, let δ ≤ min⎨1, 4 ⎬ .<br />
See TSM for complete proof. ⎩ 5 ⎭<br />
33. See TSM.<br />
34. See TSM.<br />
35. Given any e > 0, let<br />
⎧ ε ⎫<br />
δ ≤ min⎨1, 234 ⎬.<br />
⎩ 7 ⎭<br />
See TSM for complete proof.<br />
⎧ ε ⎫<br />
36. Given any e > 0, let δ ≤ min⎨1, ⎬ .<br />
⎩ 5<br />
See TSM for complete proof. ⎭<br />
37. Given any e > 0, let d ≤ min{1, 26e}.<br />
See TSM for complete proof.<br />
38. See TSM.<br />
ε<br />
39. Given any e > 0, let δ ≤ . See<br />
TSM for complete proof. 1 + | m|<br />
40. Given any e > 0, let<br />
⎧ ε ⎫<br />
δ ≤ min⎨<br />
1 ⎬<br />
⎩ 3 , 3 . See TSM.<br />
13 ⎭<br />
41. x must be within 0.05 of 3.<br />
42. x must be within approx. 0.087 of 3.<br />
43. See TSM.<br />
1<br />
44. Given any e > 0, let N ≤− . See<br />
TSM for complete proof. ε<br />
1<br />
45. Given any e > 0, let M ≥<br />
2<br />
ε<br />
. See<br />
TSM for complete proof.<br />
46. N =− 10<br />
47. See TSM.<br />
Section 1.6 • Assess Your Understanding<br />
151<br />
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152 Chapter 1 • Limits and Continuity<br />
48. See TSM.<br />
49. See TSM.<br />
50. See TSM.<br />
51. lim fx ( ) = 0;<br />
x→0<br />
lim fx ( ) does not exist.<br />
x→1<br />
See TSM for discussion and proof.<br />
52. lim fx ( ) = 0. See TSM for discussion<br />
x→0<br />
and proof.<br />
⎧ ε ⎫<br />
53. Given any e > 0, let δ ≤ min⎨1, ⎬.<br />
⎩ 47<br />
See TSM for complete proof. ⎭<br />
54. See TSM.<br />
55. M = 101.<br />
1<br />
56. Given any e > 0, let δ ≤<br />
1 + | a | . See<br />
TSM for complete proof.<br />
57. See TSM.<br />
58. K = 12.<br />
48. Explain why in the ε-δ definition of a limit, the inequality<br />
0 < |x − c| 0 if n is even<br />
x→c x→c<br />
(p. 95)<br />
[ ] m/n<br />
• lim[ f (x)] m/n = lim f (x) , provided [ f (x)] m/n is<br />
x→c x→c<br />
defined for positive integers m and n (p. 95)<br />
[ ] f (x)<br />
lim f (x)<br />
x→c<br />
• lim = , provided lim g(x) = 0 (p. 96)<br />
x→c g(x) lim g(x) x→c<br />
x→c<br />
• If P is a polynomial function, then lim P(x) = P(c). (p. 96)<br />
x→c<br />
• If R is a rational function and if c is in the domain of R,<br />
then lim R(x) = R(c). (p. 96)<br />
x→c<br />
1.3 Continuity<br />
Definitions<br />
• Continuity at a number (p. 103)<br />
• Removable discontinuity (p. 105)<br />
• One-sided continuity at a number (p. 105)<br />
• Continuity on an interval (p. 106)<br />
• Continuity on a domain (p. 107)<br />
152<br />
Chapter 1 • Limits and Continuity<br />
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Chapter 1 • Chapter Review 153<br />
Properties of Continuity<br />
• A polynomial function is continuous on its domain, all real<br />
numbers. (p. 107)<br />
• A rational function is continuous on its domain. (p. 107)<br />
• If the functions f and g are continuous at a number c, and if k<br />
is a real number, then the functions f + g, f − g, f · g, and<br />
kf are also continuous at c. If g(c) = 0, the function f g is<br />
continuous at c. (p. 108)<br />
• If a function g is continuous at c and a function f is<br />
continuous at g(c), then the composite function<br />
( f ◦ g)(x) = f (g(x)) is continuous at c. (p. 109)<br />
• If f is a one-to-one function that is continuous on its domain,<br />
then its inverse function f −1 is also continuous on its domain.<br />
(p. 110)<br />
The Intermediate Value Theorem Let f be a function that is<br />
continuous on a closed interval [a, b] with f (a) = f (b). If N is any<br />
number between f (a) and f (b), then there is at least one number c in<br />
the open interval (a, b) for which f (c) = N. (p. 110)<br />
1.4 Limits and Continuity of Trigonometric, Exponential,<br />
and Logarithmic Functions<br />
Basic Limits<br />
sin θ<br />
• lim = 1 θ→0 θ<br />
(p. 119)<br />
cos θ − 1<br />
• lim = 0 θ→0 θ<br />
(p. 121)<br />
• lim sin x = sin c x→c<br />
(p. 122)<br />
• lim cos x = cos c x→c<br />
(p. 122)<br />
• lim a x = a c ; x→c<br />
a > 0, a = 1 (p. 124)<br />
• lim log a x = log a c; x→c<br />
a > 0, a = 1, and c > 0 (p. 124)<br />
Squeeze Theorem If the functions f , g, and h have the property that<br />
for all x in an open interval containing c, except possibly at c,<br />
f (x) ≤ g(x) ≤ h(x), and if lim f (x) = lim h(x) = L,<br />
x→c x→c<br />
then lim g(x) = L. (p. 118)<br />
x→c<br />
Properties of Continuity<br />
• The six trigonometric functions are continuous on their<br />
domains. (pp. 122–123)<br />
• The six inverse trigonometric functions are continuous on their<br />
domains. (p. 123)<br />
• An exponential function is continuous on its domain, all real<br />
numbers. (p. 124)<br />
• A logarithmic function is continuous on its domain, all<br />
positive real numbers. (p. 124)<br />
1.5 Infinite Limits; Limits at Infinity; Asymptotes<br />
Basic Limits<br />
1<br />
• lim<br />
x→0 − x = −∞ lim 1<br />
=∞<br />
x→0 + x<br />
(p. 128)<br />
1<br />
• lim =∞ x→0 x2 (p. 128)<br />
• lim ln x = −∞<br />
x→0 +<br />
(p. 129)<br />
1<br />
• lim<br />
x→∞ x = 0 lim 1<br />
= 0<br />
x→−∞ x<br />
(p. 132)<br />
• lim ln x =∞<br />
x→∞<br />
(p. 136)<br />
• lim<br />
x→−∞ ex = 0<br />
lim<br />
x→∞ ex =∞ (p. 136)<br />
Definitions<br />
• Vertical asymptote (p. 131)<br />
• Horizontal asymptote (p. 138)<br />
Properties of Limits at Infinity (p. 132): If k is a real number,<br />
n ≥ 2 is an integer, and the functions f and g approach real<br />
numbers as x →∞, then:<br />
• lim A = A, where A is a constant<br />
x→∞<br />
• lim [kf(x)] = k lim f (x)<br />
x→∞ x→∞<br />
• lim [ f (x) ± g(x)] = lim f (x) ± lim g(x)<br />
x→∞ x→∞ x→∞<br />
[ ][ ]<br />
• lim [ f (x)g(x)] = lim f (x) lim g(x) x→∞ x→∞ x→∞<br />
f (x)<br />
lim f (x)<br />
• lim<br />
x→∞ g(x) = x→∞<br />
provided lim g(x) = 0<br />
lim g(x) x→∞<br />
x→∞<br />
[ ] n<br />
• lim [ f x→∞ (x)]n = lim f (x) x→∞<br />
√ √<br />
n<br />
• lim f (x) = n lim f (x), where f (x) >0 if n is even<br />
x→∞<br />
x→∞<br />
1.6 The ε-δ Definition of a Limit<br />
Definitions<br />
• Limit of a Function (p. 145)<br />
• Limit at Infinity (p. 149)<br />
• Infinite Limit (p. 150)<br />
• Infinite Limit at Infinity (p. 150)<br />
Properties of limits<br />
• If lim f (x) >0, then there is an open interval around c, for<br />
x→c<br />
which f (x) >0 everywhere in the interval, except possibly<br />
at c. (p. 149)<br />
• If lim f (x)
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<strong>Sullivan</strong><br />
154 Chapter 1 • Limits and Continuity<br />
TRM Full Solutions to Chapter 1<br />
Review Exercises and AP® Review<br />
Problems<br />
Answers to Chapter 1 Review<br />
Exercises<br />
− x<br />
1. lim 1 cos = 0. For table,<br />
x→0<br />
1+<br />
cos x<br />
see TSM.<br />
2. lim fx ( ) =−3.<br />
3.<br />
x→1<br />
y<br />
22 21<br />
22<br />
24<br />
26<br />
28<br />
1 2 x<br />
lim fx ( ) does not exist.<br />
x→2<br />
y<br />
7<br />
6<br />
5<br />
4<br />
3<br />
2<br />
4. (a) 3<br />
(b) 2<br />
5. −<br />
x<br />
3<br />
2<br />
1<br />
6. 6x + 2<br />
7. 1<br />
8. 7 2<br />
9. −π<br />
10. 2<br />
11. 0<br />
12. 0<br />
13. 27<br />
14. 3<br />
15. 4<br />
16. 1 2<br />
17. 2 3<br />
18. −1<br />
19. − 1 6<br />
20. − 1 4<br />
1 2 3 4<br />
x<br />
21. 6<br />
22. 0<br />
23. 1<br />
24. 1<br />
25. −1<br />
26. 3<br />
27. 8<br />
28. 0<br />
OBJECTIVES<br />
AP® Review<br />
Section You should be able to . . . Example Review Exercises Problems<br />
1.1 1 Discuss the slope of a tangent line to a graph (p. 78) 1 4<br />
2 Investigate a limit using a table (p. 80) 2–4 1 5<br />
3 Investigate a limit using a graph (p. 82) 5–8 2, 3<br />
1.2 1 Find the limit of a sum, a difference, and a product (p. 91) 1–6 8, 10, 12, 14, 22,<br />
26, 29, 30, 47, 48<br />
2 Find the limit of a power and the limit of a root (p. 94) 7–9 11, 18, 28, 55<br />
3 Find the limit of a polynomial (p. 95) 10 10, 22<br />
4 Find the limit of a quotient (p. 96) 11–14 13–17, 19–21, 3<br />
23–25, 27, 56<br />
5 Find the limit of an average rate of change (p. 98) 15 37<br />
6 Find the limit of a difference quotient (p. 99) 16 5, 6, 49<br />
1.3 1 Determine whether a function is continuous at a number (p. 103) 1–4 31–36 11<br />
2 Determine intervals on which a function is continuous (p. 106) 5, 6 39–42 8<br />
3 Use properties of continuity (p. 108) 7, 8 39–42<br />
4 Use the Intermediate Value Theorem (p. 110) 9, 10 38, 44–46 6<br />
1.4 1 Use the Squeeze Theorem to find a limit (p. 117) 1 7, 69<br />
2 Find limits involving trigonometric functions (p. 119) 2, 3 9, 51–55 4, 10<br />
3 Determine where the trigonometric functions are continuous (p. 122) 4 63–65<br />
4 Determine where an exponential or a logarithmic 5 43<br />
function is continuous (p. 124)<br />
1.5 1 Investigate infinite limits (p. 128) 1–3 57, 58<br />
2 Find the vertical asymptotes of a graph (p. 131) 4 61, 62 1<br />
3 Investigate limits at infinity (p. 131) 5–10 59, 60 2<br />
4 Find the horizontal asymptotes of a graph (p. 137) 11 61, 62 1<br />
5 Find the asymptotes of the graph of a rational function (p. 138) 12 67, 68 7<br />
1.6 1 Use the ε-δ definition of a limit (p. 145) 1–7 50, 66<br />
REVIEW EXERCISES<br />
1 − cos x<br />
1. Use a table of numbers to investigate lim x→0 1 + cos x .<br />
In Problems 2 and 3, use a graph to investigate lim f (x).<br />
{ x→c<br />
2x − 5 if x < 1<br />
2. f (x) =<br />
at c = 1<br />
6 − 9x if x ≥ 1<br />
{<br />
x<br />
3. f (x) =<br />
2 + 2 if x < 2<br />
at c = 2<br />
2x + 1 if x ≥ 2<br />
4. For f (x) = x 2 − 3:<br />
(a) Find the slope of the secant line joining (1, −2) and (2, 1).<br />
(b) Find the slope of the tangent line to the graph<br />
of f at (1, −2).<br />
In Problems 5 and 6, for each function find the limit of the difference<br />
f (x + h) − f (x)<br />
quotient lim<br />
.<br />
h→0 h<br />
5. f (x) = 3 6. f (x) = 3x 2 + 2x<br />
x<br />
7. Find lim f (x) if 1 + sin x ≤ f (x) ≤|x|+1<br />
x→0<br />
In Problems 8–22, find each limit.<br />
8. lim<br />
(2x − 1 )<br />
x→2 x<br />
9. lim x→π<br />
(x cos x)<br />
29. lim fx ( ) = 7; lim fx ( ) = 7; lim fx ( ) = 7.<br />
x→2<br />
− x→ 2<br />
+ x→2<br />
(<br />
10. lim x 3 + 3x 2 − x − 1 ) √ 3<br />
11. lim x(x + 2) 3<br />
x→−1<br />
x→0<br />
12. lim [(2x + 3)(x 5 x 3 − 27<br />
+ 5x)] 13. lim<br />
x→0 x→3 x − 3<br />
( x 2<br />
14. lim x→3 x − 3 − 3x )<br />
x 2 − 4<br />
15. lim<br />
x − 3<br />
x→2 x − 2<br />
x 2 + 3x + 2<br />
x 3 + 5x 2 + 6x<br />
16. lim<br />
x→−1 x 2 17. lim<br />
+ 4x + 3<br />
x→−2 x 2 + x − 2<br />
18. lim<br />
(x 2 − 3x + 1 ) 15<br />
√<br />
3 − x<br />
19. lim<br />
2 + 5<br />
x→1 x<br />
x→2 x 2 − 4<br />
20. lim x→0<br />
{ 1<br />
x<br />
[<br />
1<br />
(2 + x) 2 − 1 4<br />
]}<br />
21. lim x→0<br />
(x + 3) 2 − 9<br />
x<br />
22. lim[(x 3 − 3x 2 + 3x − 1)(x + 1) 2 ]<br />
x→1<br />
In Problems 23–28, find each one-sided limit, if it exists.<br />
x 2 + 5x + 6<br />
23. lim<br />
x→−2 + x + 2<br />
|x − 5|<br />
24. lim<br />
x→5 + x − 5<br />
x 2 − 16<br />
26. lim<br />
27. lim<br />
x→ 3/2 +2x x→4 − x − 4<br />
In Problems 29 and 30, find lim f (x) and lim<br />
x→c− x→c<br />
given c. Determine whether lim f (x) exists.<br />
{ x→c<br />
2x + 3 if x < 2<br />
29. f (x) =<br />
at c = 2<br />
9 − x if x ≥ 2<br />
|x − 1|<br />
25. lim<br />
x→1 − x − 1<br />
28. lim<br />
x→1 + √<br />
x − 1<br />
+<br />
f (x) for the<br />
154<br />
Chapter 1 • Limits and Continuity<br />
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⎧<br />
⎨ 3x + 1 if x < 3<br />
30. f (x) = 10 if x = 3<br />
⎩<br />
4x − 2 if x > 3<br />
at c = 3<br />
In Problems 31–36, determine whether f is continuous at c.<br />
⎧<br />
⎨ 5x − 2 if x < 1<br />
31. f (x) = 5 if x = 1 at c = 1<br />
⎩<br />
2x + 1 if x > 1<br />
⎧<br />
⎨ x 2 if x < −1<br />
32. f (x) = 2 if x =−1 at c =−1<br />
⎩<br />
−3x − 2 if x > −1<br />
⎧<br />
⎪⎨<br />
4 − 3x 2 if x < 0<br />
33. f (x) = 4 if x = 0 at c = 0<br />
⎪⎩<br />
<br />
16 − x 2 if 0 < x ≤ 4<br />
⎧ √<br />
⎨ 4 + x if − 4 ≤ x ≤ 4<br />
<br />
34. f (x) =<br />
⎩ x 2 − 16<br />
at c = 4<br />
if x > 4<br />
x − 4<br />
35. f (x) =2x at c = 1 36. f (x) =|x − 5 | at c = 5<br />
2<br />
37. (a) Find the average rate of change of f (x) = 2x 2 − 5x from 1<br />
to x.<br />
(b) Find the limit as x approaches 1 of the average rate of change<br />
found in (a).<br />
38. A function f is defined on the interval [−1, 1] with the following<br />
properties: f is continuous on [−1, 1] except at 0, negative at −1,<br />
positive at 1, but with no zeros. Does this contradict the<br />
Intermediate Value Theorem?<br />
In Problems 39–43 find all numbers x for which f is continuous.<br />
39. f (x) =<br />
x<br />
x 3 − 27<br />
41. f (x) =<br />
2x + 1<br />
x 3 + 4x 2 + 4x<br />
43. f (x) = 2 −x<br />
40. f (x) =<br />
x 2 − 3<br />
x 2 + 5x + 6<br />
42. f (x) = √ x − 1<br />
44. Use the Intermediate Value Theorem to determine whether<br />
2x 3 + 3x 2 − 23x − 42 = 0 has a zero in the<br />
interval [3, 4].<br />
In Problems 45 and 46, use the Intermediate Value Theorem to<br />
approximate the zero correct to three decimal places.<br />
45. f (x) = 8x 4 − 2x 2 + 5x − 1 on the interval [0, 1] .<br />
46. f (x) = 3x 3 − 10x + 9; zero between −3 and −2.<br />
| x |<br />
| x |<br />
47. Find lim (1 − x) and lim (1 − x).<br />
x→0 + x x→0 − x<br />
|x|<br />
What can you say about lim (1 − x)?<br />
x→0 x<br />
x 2<br />
48. Find lim x→2 x − 2 − 2x <br />
. Then comment on the statement<br />
x − 2<br />
x 2<br />
that this limit is given by lim x→2 x − 2 − lim 2x<br />
x→2 x − 2 .<br />
Chapter 1 • Review Exercises 155<br />
f (x + h) − f (x)<br />
49. Find lim<br />
for f (x) = √ x.<br />
h→0 h<br />
50. For lim(2x + 1) = 7, find the largest possible δ that<br />
x→3<br />
“works” for ε = 0.01.<br />
In Problems 51–60, find each limit.<br />
sin x<br />
51. lim cos(tan x) 52. lim<br />
4<br />
x→0 x→0 x<br />
53. lim x→0<br />
tan (3x)<br />
tan(4x)<br />
55. lim x→0<br />
cos x − 1<br />
x<br />
54. lim x→0<br />
cos x 3 − 1<br />
x<br />
10<br />
56. lim x→0<br />
e 4x − 1<br />
e x −1<br />
2 + x<br />
57. lim tan x 58. lim<br />
x→π/2 + x→−3 (x + 3) 2<br />
3x 3 − 2x + 1<br />
59. lim<br />
x→∞ x 3 − 8<br />
3x 4 + x<br />
60. lim<br />
x→∞ 2x 2<br />
In Problems 61 and 62, find any vertical and horizontal<br />
asymptotes of f .<br />
61. f (x) = 4x − 2<br />
62. f (x) = 2x<br />
x + 3<br />
x 2 − 4<br />
⎧<br />
tan x ⎪⎨ if x = 0<br />
2x<br />
63. Let f (x) =<br />
. Is f continuous at 0?<br />
⎪⎩ 1<br />
if x = 0<br />
2<br />
⎧<br />
⎨ sin(3x)<br />
if x = 0<br />
64. Let f (x) = x<br />
. Is f continuous at 0?<br />
⎩<br />
1 if x = 0<br />
<br />
cos π x + π <br />
65. The function f (x) =<br />
2<br />
is not defined at 0.<br />
x<br />
Decide how to define f (0) so that f is continuous at 0.<br />
66. Use the ε-δ definition of a limit to prove lim<br />
x→−3 (x2 − 9) = 18.<br />
67. (a) Sketch a graph of a function f that has the following<br />
properties:<br />
f (−1) = 0<br />
lim f (x) = −∞ lim<br />
x→4− lim<br />
x→∞ f (x) = 2<br />
f (x) =∞<br />
x→4 +<br />
lim f (x) = 2<br />
x→−∞<br />
(b) Define a function that describes your graph.<br />
68. (a) Find the domain and the intercepts (if any) of<br />
R(x) = 2x2 − 5x + 2<br />
5x 2 − x − 2 .<br />
(b) Discuss the behavior of the graph of R at numbers<br />
where R is not defined.<br />
(c) Find any vertical or horizontal asymptotes of the function R.<br />
69. If 1 − x 2 ≤ f (x) ≤ cos x for all x in the interval − π 2 < x < π 2 ,<br />
show that lim x→0<br />
f (x) = 1.<br />
49.<br />
2<br />
1<br />
x<br />
50. 0.005<br />
51. 1<br />
52. 1 4<br />
53. 3 4<br />
54. 0<br />
55. 0<br />
56. 4<br />
57. −∞<br />
58. −∞<br />
59. 3<br />
60. ∞<br />
61. x = −3 is a vertical asymptote. y = 4<br />
is a horizontal asymptote.<br />
62. x = −2 and x = 2 are vertical<br />
asymptotes. y = 0 is a horizontal<br />
asymptote.<br />
63. Yes<br />
64. No<br />
65. f (0) = −π<br />
66. See TSM.<br />
67. Answers will vary. One possibility is<br />
(a)<br />
y<br />
8<br />
6<br />
4<br />
2<br />
8 6 4 2<br />
2<br />
2 4 6 8 10 12 14 x<br />
4<br />
6<br />
30. lim fx ( ) = 10; lim fx ( ) = 10; lim fx ( ) = 10.<br />
x→3<br />
− x→ 3<br />
+ x→3<br />
31. f is discontinuous at c = 1.<br />
32. f is discontinuous at c =−1.<br />
33. f is continuous at c = 0.<br />
34. f is continuous at c = 4.<br />
1<br />
35. f is discontinuous at c =<br />
2 .<br />
36. f is continuous at c = 5.<br />
37. (a) 2x − 3 (b) −1<br />
38. Does not contradict IVT, because f is not<br />
continuous on the given interval.<br />
39. f is continuous on { xx | ≠ 3}.<br />
40. f is continuous on { xx | ≠−3, x ≠−2}.<br />
41. f is continuous on { xx | ≠−2, x ≠ 0}.<br />
42. f is continuous on { xx | ≥ 1}.<br />
43. f is continuous on the set of all real numbers.<br />
44. f is continuous on [3, 4], f (3) < 0, and<br />
f (4) > 0, so f has a zero on (3, 4).<br />
45. 0.215<br />
46. −2.171<br />
x<br />
x<br />
lim | | (1 − x) = 1, lim | | (1 − x) =−1,<br />
+ −<br />
x→0 x<br />
x→0<br />
x<br />
47.<br />
x<br />
lim | | (1 − x) does not exist.<br />
x→0<br />
x<br />
⎛<br />
2<br />
x<br />
48. ⎜<br />
⎝ − − 2x<br />
⎞<br />
lim<br />
−<br />
⎟<br />
⎠<br />
= 2. See TSM for<br />
x→2<br />
x 2 x 2<br />
comment.<br />
2x<br />
+ 2<br />
(b) f( x)=<br />
x − 4<br />
⎧<br />
⎪ 1<br />
68. (a) ⎨<br />
≠ ± 41<br />
⎫<br />
⎪<br />
xx ⎬ , x-intercepts<br />
10<br />
⎩⎪<br />
⎭⎪<br />
1<br />
and 2, y-intercept −1<br />
2<br />
(b) See TSM.<br />
1<br />
(c) Vertical asymptotes x = − 41<br />
10<br />
1<br />
and x = + 41<br />
, horizontal<br />
10<br />
2<br />
asymptote y =<br />
5<br />
69. See TSM.<br />
Chapter 1 • Chapter Review<br />
155<br />
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<strong>Sullivan</strong><br />
156 Chapter 1 • Limits and Continuity<br />
Answers to Chapter 1 AP®<br />
Review Problems<br />
1. B<br />
2. B<br />
3. C<br />
4. C<br />
5. C<br />
6. B<br />
7. C<br />
8. C<br />
9. D<br />
10. B<br />
11. D<br />
TRM Full Solutions for the<br />
Chapter 1 Project<br />
CHAPTER 1 AP® REVIEW PROBLEMS<br />
u 1. Which line is an asymptote to the graph of f (x) = e 2x ?<br />
u 2.<br />
(A) x = 0 (B) y = 0 (C) y = 2 (D) y = x<br />
3x 3 + 4x<br />
lim<br />
x→∞ 5 − 2x 4 =<br />
(A) − 3 2<br />
(B) 0<br />
(C)<br />
3<br />
5<br />
(D)<br />
3<br />
2<br />
u 3. If f (x) = 5x 3 − 1, then lim x→0<br />
f (x) − f (0)<br />
x 3 =<br />
(A) 0 (B) 1 (C) 5 (D) The limit does not exist.<br />
θ 2<br />
u 4. lim θ→0 1 − cos θ =<br />
(A) 0 (B) 1 (C) 2 (D) 4<br />
u 5. The table gives values of three functions:<br />
x –0.15 –0.1 –0.05 0 0.05 0.1 0.15<br />
f(x) 0.075 0.05 0.025 –4 0.025 0.05 0.075<br />
g(x) –8.3 –8.2 –8.1 undefined –7.9 –7.8 –7.7<br />
h(x) 1.997 1.99 1.9975 1 1.005 1.02 1.045<br />
For which of these functions does the table suggest that the<br />
limit as x approaches 0 exists?<br />
(A) f only (B) h only<br />
(C) f and g only (D) f and h only<br />
u 6. If a function f is continuous on the closed interval [1, 4] and if<br />
f (1) = 6 and f (4) = –1, then which of the following must be<br />
true?<br />
(A) f (c) = 0 for some number c in the open interval (–1, 6).<br />
(B) f (c) = 1 for some number c in the open interval (1, 4).<br />
(C) f (c) = 1 for some number c in the open interval (–1, 6).<br />
(D) f (c) = –2 for any number c in the open interval (1, 4).<br />
u 7. Which are the equations of the asymptotes of the graph of the<br />
x<br />
function f (x) =<br />
x(x 2 − 9) ?<br />
(A) x = –3, x = 0, x = 3, y = 0<br />
(B) x = –3, x = 0, x = 3, y = 1<br />
(C) x = –3, x = 3, y = 0<br />
(D) x = –3, x = 3, y = 1<br />
u 8. Find the value of k that makes the function<br />
{<br />
x<br />
f (x) =<br />
2 + 2 if x ≤−1<br />
kx + 4 if x > −1<br />
continuous for all real numbers.<br />
(A) –3 (B) –1 (C) 1 (D) 3<br />
u 9. An odd function f is continuous for all real numbers.<br />
If lim f (x) =−2, then which of the statements<br />
x→∞<br />
must be true?<br />
I. f has no vertical asymptotes.<br />
II. lim f (x) = 0<br />
x→0<br />
III. The horizontal asymptotes of the graph of f are y = –2<br />
and y = 2.<br />
(A) I only (B) III only<br />
(C) I and III only (D) I, II, and III<br />
u 10. lim x→0<br />
sin(2x)<br />
tan(3x) =<br />
(A) 0<br />
(B)<br />
2<br />
3<br />
(C) 1<br />
(D)<br />
u 11. If a function f is continuous for all real numbers, and if<br />
f (x) = x3 + 8<br />
when x = −2, then f (–2) =<br />
x + 2<br />
(A) 0 (B) 4 (C) 8 (D) 12<br />
3<br />
2<br />
CHAPTER 1 PROJECT<br />
Pollution in Clear Lake<br />
AP Photo<br />
The Toxic Waste Disposal<br />
Company (TWDC) specializes in<br />
the disposal of a particularly<br />
dangerous pollutant, Agent Yellow<br />
(AY). Unfortunately, instead of<br />
safely disposing of this pollutant,<br />
the company simply dumped AY<br />
in (formerly) Clear Lake.<br />
Fortunately, they have been caught and are now defending<br />
themselves in court.<br />
The facts below are not in dispute. As a result of TWDC’s<br />
activity, the current concentration of AY in Clear Lake is now<br />
10 ppm (parts per million). Clear Lake is part of a chain of rivers<br />
and lakes. Fresh water flows into Clear Lake and the<br />
contaminated water flows downstream from it. The Department<br />
of Environmental Protection estimates that the level of<br />
contamination in Clear Lake will fall by 20% each year. These<br />
facts can be modeled as<br />
p(0) = 10 p(t + 1) = 0.80p(t)<br />
where p = p(t), measured in ppm, is the concentration of<br />
pollutants in the lake at time t, in years.<br />
1. Explain how the above equations model the facts.<br />
2. Create a table showing the values of t for t = 0, 1, 2,...,20.<br />
3. Show that p(t) = 10(0.8) t .<br />
4. Use technology to graph p = p(t).<br />
5. What is lim p(t)?<br />
t→∞<br />
Lawyers for TWDC looked at the results in 1–5 above and argued<br />
that their client has not done any real damage. They concluded<br />
that Clear Lake would eventually return to its former clear and<br />
unpolluted state. They even called in a mathematician, who wrote<br />
the following on a blackboard:<br />
lim p(t) = 0<br />
t→∞<br />
and explained that this bit of mathematics means, descriptively,<br />
that after many years the concentration of AY will, indeed, be<br />
close to zero.<br />
156<br />
Chapter 1 • Limits and Continuity<br />
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Chapter 1 Project • Pollution in Clear Lake 157<br />
Concerned citizens booed the mathematician’s testimony. Fortunately,<br />
one of them has taken calculus and knows a little bit about limits. She<br />
noted that, although “after many years the concentration of AY will<br />
approach zero,” the townspeople like to swim in Clear Lake and state<br />
regulations prohibit swimming unless the concentration of AY is<br />
below 2 ppm. She proposed a fine of $100,000 per year for each full<br />
year that the lake is unsafe for swimming. She also questioned the<br />
mathematician, saying, “Your testimony was correct as far as it went,<br />
but I remember from studying calculus that talking about the eventual<br />
concentration of AY after many, many years is only a small part of the<br />
story. The more precise meaning of your statement lim p(t) = 0 is<br />
t→∞<br />
that given some tolerance T for the concentration of AY, there is some<br />
time N (which may be very far in the future) so that for all t > N,<br />
p(t) N,<br />
then p(t) N,<br />
then p(t)
<strong>Sullivan</strong> AP˙<strong>Sullivan</strong>˙Chapter01 October 8, 2016 17:4<br />
<strong>Sullivan</strong><br />
158 Chapter 1 • Limits and Continuity<br />
TRM Full Solutions for AP®<br />
Practice Exam, Big Idea 1: Limits<br />
Answers to AP® Practice Exam,<br />
Big Idea 1: Limits<br />
1. A<br />
2. B<br />
3. C<br />
4. C<br />
5. C<br />
6. B<br />
7. A<br />
8. D<br />
9. C<br />
10. A<br />
11. A<br />
12. D<br />
13. B<br />
14. C<br />
15. C<br />
AP ® Practice Exam, Big Idea 1: Limits<br />
Section 1: Multiple Choice<br />
3x 4 − 6x 3 + 3x 2<br />
1. lim x→1 x 3 − 3x 2 =<br />
+ 2x<br />
(A) 0 (B) 1 (C) 2 (D) nonexistent<br />
2. lim x→4<br />
x − 4<br />
4 − x =<br />
(A) –4 (B) –1 (C) 0 (D) nonexistent<br />
<br />
3x + sin x<br />
3. lim<br />
=<br />
x→0 2x<br />
(A) 0 (B) 1 (C) 2 (D) nonexistent<br />
4. Suppose a, b, and c are constants. Find<br />
4ax 4 − 2bx 3 + cx − 1<br />
lim<br />
x→∞ x 4 + b − 2<br />
(A) a (B) 2 (C) 4a (D) nonexistent<br />
x 2 − 16<br />
5. lim √ =<br />
x→4 x − 2<br />
(A) 2 (B) 4 (C) 32 (D) nonexistent<br />
6. lim x→0<br />
e x sin x tan x<br />
x 2 =<br />
(A) 0 (B) 1 (C) e (D) nonexistent<br />
7. Let h be defined by<br />
h(x) =<br />
<br />
f (x) · g(x) x ≤ 1<br />
k + x x > 1<br />
If lim f (x) = 2 and lim g(x) =−2, then for what value<br />
x→1 x→1<br />
of k is h continuous?<br />
(A) –5 (B) –4 (C) –2 (D) 2<br />
Use the figure below for Questions 8 and 9.<br />
y<br />
a b c d x<br />
Graph of f<br />
8. The graph of f is shown in the figure above. Which of the<br />
following statements is false?<br />
(A) lim f (x) exists. x→a<br />
(B) lim f (x) exists.<br />
x→b<br />
(C) lim f (x) exists.<br />
x→c− (D) lim f (x) does not exist.<br />
x→d− 9. The graph of f is shown in the figure above. If f is defined at k,<br />
but lim f (x) does not exist, then k =<br />
x→k<br />
(A) a (B) b (C) c (D) 0<br />
10. Which of the following functions has the horizontal asymptotes<br />
y = 1 and y = –1?<br />
(A) f (x) = 2 π tan−1 (x) (B) f (x) = e −x + 1<br />
(C) f (x) = 1 − x2<br />
1 + x 2 (D) f (x) = 2x2 − 1<br />
2x 2 + x<br />
11. Let f be a continuous function for which f (–2) = 1 and<br />
f (5) = –3. The function g is also continuous and g(x) = f −1 (x)<br />
for all x. The Intermediate Value Theorem guarantees that<br />
(A) g(c) = 2 for at least one c between –3 and 1.<br />
(B) g(c) = 0 for at least one c between –2 and 5.<br />
(C) f (c) = 0 for at least one c between –3 and 1.<br />
(D) f (c) = 2 for at least one c between –2 and 5.<br />
12. The line x = c is a vertical asymptote of the graph of the<br />
function f . Which of the following statements cannot be true?<br />
(A) lim f (x) =∞ (B) lim f (x) = c<br />
x→c− x→∞<br />
(C) f (c) is undefined (D) f is continuous at x = c.<br />
x 2 − 9<br />
13. The graph of the function f (x) =<br />
2x 2 has a vertical<br />
− 5x − 3<br />
asymptote at x = a and a horizontal asymptote at y = b. What<br />
are the values of the constants a and b?<br />
(A) a = –3, b = 2 (B) a =− 1 2 , b = 1 2<br />
(C) a = 1 2 , b = 1 (D) a = 3, b = 2<br />
2<br />
14. The graph of y = 2x2 + 2x + 3<br />
4x 2 has<br />
− 4x<br />
(A) a horizontal asymptote at y = 1 , but no vertical asymptote.<br />
2<br />
(B) no horizontal asymptote, but vertical asymptotes at x = 0<br />
and x = 1.<br />
(C) a horizontal asymptote at y = 1 and vertical asymptotes at<br />
2<br />
x = 0 and x = 1.<br />
(D) a horizontal asymptote at x = 2, but no vertical asymptote.<br />
⎧<br />
⎨ x 2 + x<br />
if x = 0<br />
15. Let f (x) = x<br />
⎩<br />
1 if x = 0<br />
Which of the following statements is true?<br />
I. f (0) exists<br />
II. lim f (x) exists<br />
x→0<br />
III. f is continuous at x = 0<br />
(A) I only (B) II only (C) I, II, III (D) None<br />
158<br />
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Chapter 1 • AP ® Practice Exam, Big Idea 1: Limits 159<br />
16. Let f be the function defined by<br />
⎧<br />
5x + 17 if −5 ≤ x < −3<br />
⎪⎨<br />
4 if x =−3<br />
f (x) =<br />
⎪⎩<br />
8 − x 2 if −3 < x < 3<br />
x − 4 if x ≥ 3<br />
For what values of x is f NOT continuous?<br />
(A) none (B) –3 only (C) 3 only (D) –3 and 3<br />
k<br />
17. If lim = 0, which of the following combinations of p and k<br />
x→∞ x p<br />
is not possible?<br />
(A) p = –1, k = 1 (B) p = 0, k = 0<br />
(C) p = 2, k = –1 (D) p = 3, k = –2<br />
18. Which of the following functions does not satisfy<br />
lim f (x) =∞?<br />
x→∞<br />
(A) f (x) = e 2x (B) f (x) = ln(x − 5)<br />
(C) f (x) = 3e −x (D) f (x) = √ x + 4<br />
19. If f (x) = x2 − 9<br />
which of the following is false?<br />
x − 3<br />
(A) lim f (x) = 6<br />
x→3<br />
(B) f is continuous for all real numbers x.<br />
(C) lim f (x) =∞<br />
x→∞<br />
(D) The domain of f is {x|x = 3}<br />
20. Let f be the function given by f (x) = x2 − 3x<br />
. For what value<br />
x − a<br />
of a is f continuous for all real numbers x?<br />
(A) None (B) a = –3 (C) a = –1 (D) a = 3<br />
Section 2: Free Response<br />
Show all of your work. Be sure to indicate clearly the methods you use<br />
to obtain your results.<br />
1. The tax on gross income in a small country is computed using<br />
the table below.<br />
Gross Income, i<br />
Tax, T<br />
$0 up to but not<br />
$0<br />
including $30,000<br />
$30,000 up to but not 10% of gross income<br />
including $50,000<br />
$50,000 and higher $5000 + 20% of gross income<br />
in excess of $50,000<br />
(a) Use a piecewise-defined function to represent T as a<br />
function of i.<br />
(b) Find lim<br />
i→0 + T (i).<br />
(c) Determine the numbers, if any, at which T is discontinuous.<br />
(d) Use the definition of continuity to explain your answer to (c).<br />
16. B<br />
17. A<br />
18. C<br />
19. B<br />
20. A<br />
Free Response<br />
1. (a)<br />
⎧0 if 0≤ i < 30,000<br />
⎪<br />
Ti () = ⎨0.1i<br />
if 30,000 ≤ i < 50,00<br />
⎪<br />
⎩⎪<br />
0.2i− 5000 if i > 50,00<br />
(b) 0<br />
(c) T is discontinuous at 30,000.<br />
(d) Answers will vary.<br />
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