Part II: The Answers
498
Answers
801–900
or ln
ln
y
x
x
x
=
−
(
)
−
+
(
)
1
3 1
3
4
. Next, rewrite the expression as a fraction and use properties
of logarithms to expand the natural logarithm (this step will make finding the
derivative a bit easier):
lim ln
lim
ln
lim
ln
ln
x
x
x
y
x
x
x
x
x
→∞
→∞
−
→∞
=
−
+
(
)
−
(
)
=
−
(
)−
+
(
3 1
3
4
1
3 1
3
4
1
) )
−
(
)
−
x 1
1
Apply L’Hôpital’s rule:
lim
ln
ln
lim
x
x
x
x
x
x
x
x
→∞
−
→∞
−
−
(
)−
+
(
)
−
(
)
=
−
−
+
− −
(
)
= −
3 1
3
4
1
3
3 1
3
3
4
1
1
2
l lim
lim
x
x
x
x
x
x
x
x
→∞
→∞
−
(
)
−
(
) +
(
)


 


 
= −
−
+
+
1
15
3 1 3
4
15
30 15
9
2
2
2
9 9
4
15
9
x −
= −
Because 3 1
3
4
1
x
x
y
x
−
+
( ) =
−
and y e
y
=
ln , it follows that
lim
lim
lim
ln
lim ln
x
x
x
x
y
y
x
x
y
e
e
e
x
→∞
−
→∞
→∞
−
−
+
( ) = =
=
=
→∞
3 1
3
4
1
15 9
825.
1
Here’s the given problem:
lim
x
x
x
→
+
( )
0
2
2
Substituting in the value x = 0 gives you the indeterminate form 0
0
. Create a new limit
by taking the natural logarithm of the original expression, y
x
x
= ( )
2
2
, so that
ln
ln
y
x
x
=
( )
( )
2
2
, or ln
ln
y x
x
=
( )
(
)
2
2
. Then take the limit of the new expression:
lim ln
lim
ln
lim
ln
x
x
x
y
x
x
x
x
→
→
→
−
+
+
+
=
( )=
( )
0
0
2
0
2
2
2
Apply L’Hôpital’s rule:
lim
ln
lim
lim
x
x
x
x
x
x
x
x
→
−
→
−
→
+
+
+
( ) =
( )
−
=
−
=
0
2
0
3
0
2
2
1
2
2
2
2
0
Because 2
2
x
y
x
( ) = and y e
y
=
ln , it follows that
y
x
y
e
e
e
x
x
x
x
y
y
x
=
( ) =
=
=
=
=
→
→
→
+
+
+
→ +
lim
lim
lim
ln
lim ln
0
0
0
0
1
2
2
0