Assignment #2 Solutions

Question 1:
Part of the graph of y = f(x) =
Find:
lim f(x) f(−2) lim f(x) lim
x→−2− x→−2 +
Assignment #2 Solutions
x 2 − 1 if x ≤−1orx ≥ 1
2 − x 2 if − 1

2
Question 2: (Show your reasoning.)
Note: It is not necessary to give specific Limit Laws with their #’s, but beware of using ”Laws” that don’t exist! The
graphs are not a necessary part of the solutions.
x
(a) Find lim
x→1
2 y
3
− 1
2
x − 1
x
Solution: lim
x→1
2 − 1 (x − 1)(x + 1)
= lim
=(by Limit Law #12)
x − 1 x→1 x − 1
lim(x
+ 1) =(by Limit Law #1)
x→1
lim x + lim 1 =(by Limit Laws #8 and #7)
x→1 x→1
1 + 1 = 2
1
-2
0
-1 0
-1
1 2
-2
x
Universitas
DEO PAT-
ET RIÆ
Saskatchewanensis
©2003 Doug MacLean

|x − 3|
(b) Find lim
x→3 + 3 − x
|x − 3|
Solution: Since |x − 3| =x − 3 when x>3, we have lim = lim
x→3 + 3 − x x→3 +
x − 3
3 − x =
= lim −1 =(by Limit Law #7)
x→3 +
−1
|x − 3|
(c) Find lim
x→3− 3 − x
|x − 3|
Solution: Since |x − 3| =(−x − 3) = 3 − x when x>3, we have lim = lim
x→3− 3 − x x→3− 3 − x
3 − x =
= lim 1 =(by Limit Law #7)
x→3− 1
y
2
1
-2
3
0
0
-1
1 2 3 4 5
Universitas
DEO PAT-
ET RIÆ
Saskatchewanensis
©2003 Doug MacLean
x

4
2
(d) Find lim
x→2− x − 2
Solution: Since x − 2 < 0 when x 0 when x>2,
x
(f) Find lim
x→−2
2 + 4x + 4
x2 + 3x + 2
2
x − 2
2
x − 2
is negative, so we have lim
x→2− 2
= −∞
x − 2
is positive, so we have lim
x→2 +
2
= ∞
x − 2
x
Solution: lim
x→−2
2 + 4x + 4
x2 (x + 2)
= lim
+ 3x + 2 x→−2
2
=(by Limit Law #12)
(x + 2)(x + 1)
x + 2
lim =(by Limit Law #5)
x→−2 x + 1
lim x + 2
x→−2
=(by Limit Law #1)
lim x + 1
x→−2
lim x + lim
x→−2 x→−2 2
=(by Limit Laws #7 and #8)
lim x + lim 1
x→−2 x→−2
−2 + 2 0
= = 0
−2 + 1 −1
y
5
4
3
2
1
0
-5 -4 -3 -2 -1
-1
0 1 2 3 4
-2
-3
-4
-5
y
5
4
3
2
1
0
-2 -1
-1
0 1 2 3 4
-2
-3
-4
-5
x
Universitas
DEO PAT-
ET RIÆ
x
Saskatchewanensis
©2003 Doug MacLean