Answers and Explanations 195
Answers
101–200
To find the limit, substitute –5 for x:
lim
( )
x
x
→−
= −
= −
5
1
5
1
5 5
1
25
191.
− 1
2
Note that substituting in the limiting value gives you an indeterminate form. For example, as x approaches 0 from the right, you have the indeterminate form ∞ – ∞, and as x
approaches 0 from the left, you have the indeterminate form –∞ + ∞.
Begin by getting common denominators:
lim
lim
lim
x
x
x
x
x x
x
x
x
x
x
x
→
→
→
+
−






=
+
−
+
+






=
− +
0
0
0
1
1
1
1
1
1
1
1 1
( (
)
+
x
x
1
Next, multiply the numerator and denominator of the fraction by the conjugate of the
numerator and simplify:
lim
lim
lim
x
x
x
x
x
x
x
x
x
x
x
→
→
→
− +
(
)
+
=
− +
(
)
+
(
)
+ +
(
)
+ +
(
)
=
0
0
0
1 1
1
1 1
1
1 1
1 1
1 1 1
1
1 1
1
1 1
1
1
1 1
0
0
− +
(
)
+
+ +
(
)
=
−
+
+ +
(
)
=
−
+
+ +
(
)
→
→
x
x
x
x
x
x
x
x
x
x
x
x
lim
lim
To find the limit, substitute 0 for x:
lim
( )
x
x
x
→
−
+
+ +
(
)
=
−
+
+ +
(
)
= −
= −
0
1
1
1 1
1
1 0 1 1 0
1
1 2
1
2