Answers and Explanations 493
Answers
801–900
813.
0
Here’s the initial problem:
lim
tan
x
x
x
→∞
−
−
−
1
4
1
π
Make sure you have an indeterminate form before applying L’Hospital’s rule, or you
can get incorrect results! Notice that as x → ∞ in the numerator, tan
−
( )−
(
)
1
4
x π →
π π π
2 4 4
− = , and as x → ∞ in the denominator, (x – 1) → ∞. Therefore, the solution
becomes
lim
tan
x
x
x
→∞
−
( )−
−
=
1
4
1
0
π
814.
–∞
Here’s the given problem:
lim
tan
x
x
x
→∞
−
−
+
1
3
2
1
1
π
Applying the limit gives you the indeterminate form 0
0
. Applying L’Hôpital’s rule gives
you the following:
lim
tan
lim
lim
x
x
x
x
x
x
x
x
x
→∞
−
→∞
→∞
−
+
=
+
−
+
(
)
=
+
(
)
−
1
3
2
2
3
2
3
2
2
1
1
1
1
3
1
1
π
3 3
1
1 2
3
3
2
2
3
6
2
4
x
x
x
x
x
x
x
+
(
)
= −
+
+
+
→∞
lim
This limit is still an indeterminate form, and you can continue using L’Hôpital’s rule
four more times to arrive at the solution. However, because the degree of the numerator is larger than the degree of the denominator, you can find the limit by noting that
−
+
+
+
= −
= −
= −∞
→∞
→∞
→∞
lim
lim
lim
x
x
x
x
x
x
x
x
x
x
1 2
3
3
3
3
3
6
2
4
6
4
2
815.
0
The given problem is
lim sec
tan
x
x
x
→
−
−
(
)
π
2