Part II: The Answers
216
To find the limit as x → –∞, proceed in the same way:
lim
lim
lim
x
x
x
x
x
x
x
x
x
x
x
x
x
→−∞
→ −∞
→−∞
−
+
=
−
+
=
−
+
=
5
5
5
5
5 1
5 1
2
2
2
2
2
2
2
2
2
2
0 0 1
0 1
1
−
+
= −
Therefore, the only horizontal asymptote is y = –1.
250.
y = 1
3
In order to find any horizontal asymptotes of the function y
x
x
x
=
+
4
2
3
, you need
to consider the limit of the function as x → ∞ and as x → –∞. For the limit as x → ∞,
begin by multiplying the numerator and denominator by 1
2
x
:
lim
lim
lim
lim
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
→∞
→∞
→∞
→∞
+ =
+
( )
=
+
=
4
2
2
4
2
2
4
4
3
1
1 3
1
3
4 4
4
4
3
3
1 1
3
1 0
3
1
3
x
x
x
x
x
+
=
+
=
+
=
→∞
lim
In order to find the limit as x → –∞, proceed in the same way, noting that as x → –∞,
you still use 1
1
2
4
x
x
=
:
Answers
201–300