Part II: The Answers
218
251.
y = 1 and y = –1
To find any horizontal asymptotes of the function y
x
x
=
+
2
2
, 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
x
:
lim
lim
lim
lim
x
x
x
x
x
x
x
x
x
x
x
x
x
x
→∞
→∞
→∞
→∞
+
=
( )
+
=
+
=
+
2
2
2
2
2
2
2
1
1
2
1
1
2
1
2
x x
x
x
2
2
1
1 2
1
1 0
1
=
+
=
+
=
→∞
lim
To find the limit as x → –∞, proceed in the same way, noting that as x → −∞, you need
to use the substitution 1
1
2
x
x
= −
:
lim
lim
lim
lim
x
x
x
x
x
x
x
x
x
x
x
x
→−∞
→ −∞
→−∞
→−∞
+
=
( )
+
=
−
+
=
−
2
2
2
2
2
1
1
2
1
1
2
1
x x
x
x
x
x
2
2
2
2
2
1
1 2
1
1 0
1
+
=
− +
= − +
= −
→−∞
lim
Therefore, the function has the horizontal asymptotes y = 1 and y = –1.
252.
removable discontinuity at x = –3, jump discontinuity at x = 3
The limit exists at x = –3 but isn’t equal to f
 
(–3), which corresponds to a removable
discontinuity.
At x = 3, the left-hand limit doesn’t equal the right-hand limit (both limits exist as finite
values); this corresponds to a jump discontinuity.
Answers
201–300