Part II: The Answers
500
Answers
801–900
828.
e
Here’s the initial problem:
lim
x
x
x
→∞
+
( )
1 1
Applying the limit gives you the indeterminate form 1
∞
. Create a new limit by taking the
natural logarithm of the original expression, y
x
x
= +
( )
1 1 , so that ln
ln
y
x
x
=
+
( )
1 1 , or
ln
ln
y x
x
=
+
( )
1 1 . Noting that lim ln
x
x
x
→∞
+
( )
1 1 is an indeterminate product, you can
rewrite this limit as
lim ln
lim
ln
lim
ln
x
x
x
x
x
x
x
x
x
→∞
→∞
−
→∞
−
−
+
( ) =
+
( ) =
+
(
)
1 1
1 1
1
1
1
1
Next, apply L’Hôpital’s rule:
lim
ln
lim
lim
x
x
x
x
x
x
x
x
x
→∞
−
−
→∞
−
−
−
→∞
+
(
) =
+





 −
( )
−
=
+
1
1
1
1
1
1
1
1
2
2
− −
→∞
=
+
= +
=
1
1
1 1
1
1 0
1
lim
x
x
Because 1 1
+
( ) =
x
y
x
and y e
y
=
ln
, it follows that
lim
lim
lim
ln
lim ln
x
x
x
x
y
y
x
y
e
e
e e
x
→∞
→∞
→∞
+
( ) = =
=
=
→∞
1 1
1
829.
e
4
The given problem is
lim
x
x
x
e
x
→
+
+
(
)
0
2
Applying the limit gives you the indeterminate form 1
∞
. Create a new limit by taking the
natural logarithm of the original expression, y
e
x
x
x
=
+
(
)
2 , so that ln
ln
y
e
x
x
x
=
+
(
)
2 ,
or ln
ln
y
e
x
x
x
=
+
(
)
2
. Then use L’Hôpital’s rule on the new limit: