Ans: 1 c lim f(x) - Berkeley City College

1. 

The graph of a function f is drawn above, answer the questions: 

a f(−4) =? Ans: 2 b lim f(x) = ? Ans: 1 c lim f(x) =? Ans: 1 d lim f(x) =? Ans: 1 

x→−4 − x→−4 + x→−4 

e f(−2) =? Ans: 4 f lim f(x) = ? Ans: 5 g lim f(x) =? Ans: 4 h lim f(x) =? Ans: 

x→−2 − x→−2 + x→−2 

Does not exist 

i f(0) =? Ans: Undefined. j lim f(x) = ? Ans: 4 k 

x→0− Does not exist 

lim f(x) =? Ans: 2 l 

x→0 + 

lim f(x) =? Ans: 

x→0 

m f(2) =? Ans: 0 n lim f(x) = ? Ans: 0 o lim f(x) =? Ans: -2 p 

x→2− x→2 + 

not exist 

lim f(x) =? Ans: Does 

x→2 

q f(4) =? Ans: Undefined 

Does not exist 

r lim 

x→4 − f(x) = ? Ans: -1 

s lim 

x→4 + f(x) =? Ans: ∞ 

t lim 

x→4 

f(x) =? Ans: 

2. Evaluate the limit: 

a. lim 

x→1 

4x + 3 

Ans. lim 

x→1 

4x + 3 = 7 

b. lim 

x→1 

x 2 − 1 

x − 1 


x→1 

x 2 − 1 

x − 1 = 2 

c. lim 

x→1 

x 3 − 1 

x − 1 


x→1 

x 3 − 1 

x − 1 = 3

d. lim 

x→1 

x − 1 

x 2 − 1 


x→1 

x − 1 

x 2 − 1 = 1 2 

e. lim 

x→1 

x − 1 

x 2 − 2x + 1 


x→1 

x − 1 

x 2 − 2x + 1 = DNE 

x 2 − 5x 

f. lim 

x→5 x 2 + 2x − 35 

x 2 − 5x 


x→5 x 2 + 2x − 35 = 5 

12 

g. lim 

x→7 

√ x + 2 − 3 

x − 7 

√ x + 2 − 3 

Ans: lim 

= 1 

x→7 x − 7 6 

h. lim 

x→0 

sin x 

x 


x→0 

sin x 

x = 1 

i. lim 

x→2 

f(x) − f(2) 

x − 2 

f(x) − f(2) 


= 3 

x→2 x − 2 

j. lim 

x→3 

f(x) − f(3) 

x − 3 

f(x) − f(3) 


= 6 

x→3 x − 3 

k. lim 

x→c 

f(x) − f(c) 

x − c 

f(x) − f(c) 


= 4c 

x→c x − c 

l. lim 

h→0 

f(x + h) − f(x) 

h 

if f(x) = 3x + 1 

if f(x) = x 2 − 1 

if f(x) = 2x 2 

f(x + h) − f(x) 


= −4 

h→0 h 

m. lim 

h→0 

f(x + h) − f(x) 

h 

if f(x) = −4x − 2 

if f(x) = sin x 

f(x + h) − f(x) 


= cos x 

h→0 h 

n. lim 

h→0 

f(x + h) − f(x) 

h 

if f(x) = cos x 

f(x + h) − f(x) 


= − sin x 

h→0 h 

o. lim 

x→2 − |x − 2| 

x − 2 

Ans: −1

p. lim 

x→2 + |x − 2| 

x − 2 

Ans: 1 

x 

q. lim 

x→−1 − x + 1 

Ans: ∞ 

x 

r. lim 

x→−1 + x + 1 

Ans: −∞ 

3. Let 

⎧ 

⎨ x 2 + 1 if x < −3 

f(x) = x − 2 if − 3 ≤ x < 9 

⎩ √ x if 9 ≤ x 

Evaluate: 

a. lim 

x→−3 − f(x) 


b. lim 

x→−3 + f(x) 

Ans: −5 

c. lim 

x→−3 f(x) 

Ans: Limit Does Not Exist. 

d. lim 

x→−1 − f(x) 


e. lim 

x→−1 + f(x) 


f. lim 

x→−1 f(x) 


g. lim 

x→0 − f(x) 


h. lim 

x→0 + f(x) 


i. lim 

x→0 

f(x) 


j. lim 

x→2 − f(x) 


k. lim 

x→2 + f(x) 


l. lim 

x→2 

f(x)


m. lim 

x→9 − f(x) 


n. lim 

x→9 + f(x) 


o. lim 

x→9 

f(x) 

Ans: Limit Does Not Exist 

p. lim 

x→16 − f(x) 


q. lim 

x→16 + f(x) 


r. lim 

x→16 f(x) 


4. Think about: 

For any real number x, let f be defined by: 

What is lim 

x→0 

f(x)? 

Ans: Limit Does Not Exist. 

f(x) = 

5. For any real number x, let f be defined by: 

What is lim 

x→0 

f(x)? 


⎧ 

⎪⎨ 

f(x) = 

⎪⎩ 

{ 0 if x is irrational 

1 if x is rational 

0 if x is irrational 

0 if x = 0 

1 

n 

if x ≠ 0, x = m , m, n are co-prime, n > 0 

n 

6. For the following given function f, find the point(s) where f is discontinuous. Catagorize the point(s) you 

found as removable or unremovable discontinuity. If the point is a removable discontinuity, explain how you can 

redefine the function at that point to make the function continuous at that point. 

a. f(x) = 5 

Ans: f is continuous on whole real line. 

b. f(x) = x 3 + 2x + 7 


c. f(x) = |x| 


d. f(x) = sin x 

x

Ans: f is discontinuous at x = 0. Define f(x) = 1 if x = 0 and f will be continuous at 0, so x = 0 is a removable 

discontinuity. 

e. f(x) = x2 − 3x + 2 

x 2 − 4 

Ans: f is discountinuous at x = −2 and x = 2. Define f(x) = 1 if x = 2, and f will be continuous at 2, so x = 2 

4 

is a removable discontinuity. lim f(x) does not exist, so x = −2 is a non-removable discontinuity. 

x→−2 

f. f(x) = 

|x + 4| 

x + 4 

Ans: f is discontinuous at x = −4. This is a non-removable discontinuity. 

g. f(x) = e −1/x2 

Ans: f is discontinuous at x = 0. Define f(x) = 0 if x = 0 will make f continuous at 0, so x = 0 is a removable 

discontinuity. 

⎧ 

⎨ x + 3 if x < −1 

h. f(x) = x 2 + 1 if − 1 ≤ x < 4 

⎩ 

−2x + 10 if 4 ≤ x 

Ans: f has a non-removable discontinuity at x = 4. 

⎧ 

⎨ 2 if x ≤ 3 

i. f(x) = x − 1 if 3 ≤ x < 5 

⎩ 

2x − 3 if 5 ≤ x 

Ans: f has a non-removable discontinuity at x = 5. 

7. Using the same graph of the function at problem (1), answer the following question: 

a. For which value(s) of x does f have a removable discontinuity? 

Ans: x = −4 

b. For which value(s) of x does f have a non-removable discontiuity? 

Ans: x = −2, x = 0, x = 2, x = 4 

c. For which value(s) of x is f discontinuous but has a limit? 

Ans: x = −4 

d. For which value(s) of x is f discontinuous and not have a limit, but has a left and/or right hand limit? 

Ans: x = −2, x = 0, x = 2, x = 4 

e. For which value(s) of x does f not have any left or right hand limit? 

Ans: None. 

8. Find the slope of the tangent line to the given function at the give value of x, using the limit definition. 

a. f(x) = 3, x = −2 

Ans: m = 0 

b. f(x) = 3x + 1, x = 1 


c. f(x) = x 2 − 1, x = 3 


d. f(x) = √ x, x = 9 

Ans: m = 1 6 

9. For the functions in (8), find the equation of the tangent line to the given function at the give value of x.

a. Ans: y = 3 

b. Ans: y = 3x + 1 

c. Ans: y = 6x − 10 

d. Ans: y = 1 6 x + 3 2

Ans: 1 c lim f(x) - Berkeley City College

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