Part II: The Answers
496
Answers
801–900
Substituting in the value x = 1 gives you the indeterminate form 0
0
, so use L’Hôpital’s rule:
lim
ln
cos
lim sin
x
x
x
x
x
x
x
→
→
− +
+
=
− +
−
( )
(
)( )
1
1
1
1
1 1
π
π
π
If you again substitute in the value x = 1, you still get the indeterminate form 0
0
. Using
L’Hôpital’s rule again gives you the limit:
lim sin
lim
cos
cos
x
x
x
x
x
x
→
→
− +
−
( )
(
)( )
=
−
−
( )
(
)( )
=
1
1
2
2
2
1 1
1
1
1
π
π
π
π
π
π
π
π
( )
= − 1
2
821.
e
−5
Here’s the given problem:
lim
x
x
x
→
−
(
)
0
1
1 5
Substituting in the value x = 0 gives you the indeterminate form 1
∞
. Create a new limit
by taking the natural logarithm of the original expression, y
x
x
= −
(
)
1 5
1 , so that
ln
ln
y
x
x
=
−
(
)
1 5
1 , or ln
ln
y
x
x
=
−
(
)
1 5 . Take the limit of the new expression:
lim ln
lim
ln
x
x
y
x
x
→
→
=
−
(
)
0
0
1 5
Then apply L’Hôpital’s rule:
lim
ln
lim
x
x
x
x
x
→
→
−
(
) =
−
−
= −
− ( )
= −
0
0
1 5
5
1 5
1
5
1 5 0
5
Because 1 5
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
x
→
→
→
−
−
(
) =
=
=
=
→
0
1
0
0
5
1 5
0
822.
1
Here’s the initial problem:
lim
x
x
x
→∞
1
2