Answers and Explanations 187
Answers
101–200
174.
− 7
4
Note that substituting the limiting value, 2, into the function x
x
x
x
2
2
3
10
8
12
+
−
−
+
gives you the
indeterminate form 0
0
.
To find the limit, first factor the numerator and denominator and simplify:
lim
lim
(
)(
)
(
)(
)
lim
(
x
x
x
x
x
x
x
x
x
x
x
x
→
→
→
+
−
−
+
=
−
+
−
−
=
2
2
2
2
2
3 10
8
12
2
5
2
6
+ +
−
5
6
)
(
)
x
Then substitute 2 for x:
lim
x
x
x
→
+
(
)
−
(
)
= +
−
= −
2
5
6
2 5
2 6
7
4
175.
1
2
Note that substituting the limiting value, –5, into the function x
x
x
2
2
5
25
+
−
gives you the
indeterminate form 0
0
.
To find the limit, first factor the numerator and denominator and simplify:
lim
lim
(
)
(
)(
)
lim
x
x
x
x
x
x
x x
x
x
x
x
→−
→−
→−
+
−
=
+
−
+
=
−
5
2
2
5
5
5
25
5
5
5
5
Then substitute –5 for x:
lim
x
x
x
→−
−
= −
− −
=
5
5
5
5 5
1
2