Answers and Explanations 499
Answers
801–900
826.
1
Here’s the given problem:
lim tan
x
x
x
→
+
(
)
0
3
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
= (
)
tan 3
, so that
ln
ln tan
y
x
x
= (
)
3
, or ln
ln tan
y x
x
=
(
)
3 . Next, take the limit of the new expression:
lim ln
lim ln tan
lim
ln(tan x)
x
x
x
y
x
x
x
→
→
→
−
+
+
+
=
(
)=
0
0
0
1
3
3
Apply L’Hôpital’s rule:
lim
ln tan
lim
sec
tan
x
x
x
x
x
x
x
→
−
→
−
+
+
(
) =
( )
−
0
1
0
2
2
3
3
3
3
Rewrite the expression and simplify:
lim
sec
tan
lim
cos
cos
x
x
x
x
x
x
x
→
−
→
+
+
( )
−
= −
( )
( )


 


 
0
2
2
0
2
2
3
3
3
3
3
3 3
3
1
3
3
3
0
x
x
x
x
x
x
( )
( )


 


 
= −
( )


 


 
( )


 
→
+
sin
lim cos
s in
 

  ( )
= −( )( )( ) =
x
1 1 0 0
Because tan 3x
y
x
(
) = and y e
y
=
ln , it follows that
lim tan
lim
lim
ln
lim ln
x
x
x
x
y
y
x
y
e
e
e
x
→
→
→
+
+
+
→ +
(
) =
=
=
=
=
0
0
0
0
1
3
0
827.
0
The given problem is
lim tan ln
x
x x
→
+
0
Applying the limit gives you the indeterminate form 0(–∞). Rewrite the product as a
quotient and use L’Hôpital’s rule to get the limit:
lim tan ln
lim ln
cot
lim csc
lim
sin
x
x
x
x
x x
x
x
x
x
x
→
→
→
→
+
+
+
+
=
=
−
=
−
0
0
0
2
0
1
x x
x
x
x
x
x
x
sin
lim sin
lim sin
(
)
= −
( ) ⋅ ( )
= − ⋅
=
→
→
+
+
0
0
1 0
0