CK-12-Calculus

http://www.ck12.org Chapter 3. Applications of Derivatives

Evaluate limx→ 0 ex−x^1.
Solution:
Since limx→ 0 (ex− 1 ) =limx→ 0 x=0, L’Hospital’s Rule applies and we have

xlim→ 0 e

x− 1

x =limx→ 0

ex

1 =

1

1 =^1.

Example 4:
Evaluate limx→+∞xe^2 x
Solution:
Since limx→+∞x^2 =limx→+∞ex= +∞, L’Hospital’s Rule applies and we have

xlim→+∞x

2

ex=x→lim+∞

2 x
ex.
Here we observe that we still have the indeterminate form∞∞. So we apply L’Hospital’s Rule again to find the limit
as follows:

x→lim+∞x

2

ex=x→lim+∞

2 x

ex=x→lim+∞

2

ex=^0
L’Hospital’s Rule can be used repeatedly on functions like this. It is often useful because polynomial functions can
be reduced to a constant.
Let’s look at an example with trigonometric functions.
Example 5:
Evaluate limx→ 01 −xcos 2 x.
Solution:
Since limx→ 0 ( 1 −cosx) =limx→ 0 x^2 =0, L’Hospital’s Rule applies and we have

xlim→ 01 −xcos^2 x=xlim→ 0 sin 2 xx=limx→ 0 cos 2 x=^12.

Lesson Summary

1. We learned to examine end behavior of functions on infinite intervals.


2. We determined horizontal asymptotes of rational functions.


3. We examined indeterminate forms of limits of rational functions.


4. We applied L’Hospital’s Rule to find limits of rational functions.

Multimedia Links

For an introduction to L’Hopital’s Rule(8.0), see Khan Academy, L’Hopital’s Rule (8:51).