Answers and Explanations 501
Answers
801–900
lim ln
lim
ln
lim
lim
x
x
x
x
x
x
x
y
e
x
x
e
x
e
→
→
→
→
+
+
+
+
=
+
(
)
=
+
(
)
+
(
)
=
0
0
0
0
2
2
1
1
2 2
1
2
1
0
2 2
1
4
0
0
e
e
x
e
e
x
x
+
(
)
+
(
)
=
+
(
)
+
=
=
( )
Because e
x
y
x
x
+
(
) =
2
and y e
y
=
ln , it follows that
lim
lim
lim
ln
lim ln
x
x
x
x
x
y
y
e
x
y
e
e
e
x
→
→
→
+
+
+
→ +
+
(
) =
=
=
=
0
2
0
0
4
0
830.
–∞
Here’s the given problem:
lim
ln
x
x
x
x
→
+
−
−






1
2
2
1
1
Applying the limit gives you the indeterminate form ∞ – ∞, so write the expression as a
single fraction by getting common denominators and then apply L’Hôpital’s rule:
lim
ln
ln
lim
ln
x
x
x
x x
x x
x
x x x
x
x
x
→
→
+
+
( )− +
( ) −
(
)
=
( )+ ( ) −
1
2
2
2
1
2
1
1
2
1 2
1 2 2 1
2
−
(
) + ( )( )
ln x
x
Multiply the numerator and denominator by x:
lim
ln
ln
x
x
x x
x
x
x
→
+
( )−
− +
1
2
2
2
2
2
1 2
Notice that as x → 1
+
in the numerator, 2x
2 
(ln x) – x
2 → –1, and as x → 1
+
in the denominator, x
2
– 1 + 2x
2 
ln x → 0
+
. Therefore, you get the following answer:
lim
ln
ln
x
x
x x
x
x
x
→
+
−
− +
= −∞
1
2
2
2
2
2
1 2
831.
1
2
Here’s the given problem:
lim ln
x
x x
→
+
− −
(
)
1
1
1
1
Evaluating the limit gives you the indeterminate form ∞ – ∞, so write the expression as
a single fraction by getting a common denominator and then use L’Hôpital’s rule: