Assignment #2 Solutions

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