Answers and Explanations 495
Answers
801–900
If you again take the limit, you get the indeterminate form − ∞
∞
. Use L’Hôpital’s rule one
more time to get the final answer:
lim
lim
x
x
x
x
x
e
e
→−∞ −
→−∞
−
=
−
=
6
6
0
818.
1
The given problem is
lim
x
x
xe
x
→∞
−
(
)
1
Applying the limit gives you the indeterminate form ∞ – ∞. Begin by factoring out x;
then rewrite the expression as a fraction and use L’Hôpital’s rule:
lim
lim
lim
li
x
x
x
x
x
x
x e
e
x
e
x
x
→∞
→∞
−
→∞
−
(
) =
−
=
( ) −






−
=
1
1
1
1
2
2
1
1
1
1
m m
x
x
e
e
→∞
=
=
1
0
1
819.
∞
The given problem is
lim
x
x
e
x
→∞
+
3 1
2
Substituting in the value x = ∞ gives you the indeterminate form ∞
∞
, so use L’Hôpital’s
rule:
lim
lim
x
x
x
x
e
x
e
x
→∞
+
→∞
+
=
( )( )
3 1
2
3 1
3
2
If you again substitute in the value x = ∞, you still get the indeterminate form ∞
∞
.
Applying L’Hôpital’s rule one more time gives you the limit:
lim
lim
x
x
x
x
e
x
e
→∞
+
→∞
+
( )( ) =
( )( ) = ∞
3 1
3 1
3
2
3
3
2
820.
− 1
2
π
Here’s the initial problem:
lim
ln
cos
x
x
x
x
→
− +
+
1
1
1
π