Part II: The Answers
192
Answers
101–200
185.
5
7
Note that substituting the limiting value, 5, into the function
x
x
x
2
2
25
3
16
5
−
−
+
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
→
→
→
−
−
+
=
−
+
−
−
=
+
5
2
2
5
5
25
3
16
5
5
5
5 3 1
5
3 3 1
x −
Then substitute 5 for x:
lim
( )
x
x
x
→
+
−
=
+
−
=
5
5
3 1
5 5
3 5 1
5
7
186.
− 1
6 6
Note that substituting in the limiting value gives you the indeterminate form 0
0
.
Begin by multiplying the numerator and denominator of the fraction by the conjugate
of the numerator:
lim
lim
(
)
lim
x
x
x
x
x
x
x
x
x x
x
x
x
x
→
→
+ −
−
=
+ −
(
)
−
+ +
(
)
+ +
(
)
=
3
2
3
3
2
3
3
2
3
3
2
3
2
x x
x
x
x
x x
x
x
x
x x
x
x
→
→
+ −
−
+ +
(
)
=
−
−
+ +
(
)
3
3
3 2
3
3
2
3
3
3
2
(
)
lim (
)
Next, factor –1 from the numerator and continue simplifying:
lim (
)
lim
(
)
(
)
lim
x
x
x
x
x x
x
x
x
x x
x
x
→
→
→
−
−
+ +
(
)
=
−
−
−
+ +
(
)
=
3
3
3
3
3
3
2
1
3
3
3
2
− −
+ +
(
)
1
3
2
x x
x