l'Hospital's+Rule+(including+logarithmic+method)

Property: L'Hôpital's rule

If And if Then

If the limit exists

This rule can be used to evaluate many indeterminate forms, including when the top and bottom evaluate to (before derivation):

(These pictures seem all squished, but I'm too tired to fix them)

This means that when finding the limit for a function approaching a value, henceforth called 'c', if simply inserting c into g(x) / h(x) returns one of the indeterminate forms listed above, L'Hospital's Rule should be utilized. This should be done by differentiating the top and bottom functions separately. This is explained because after differentiating separately, the value remains as if the original function had been used.

Also regardless of what Sara says, Lickitung is most definitely the lamest Pokemon. Honestly a GIF about something pretentious. If I could think of a retort I would post a retort GIF.

HEY LOOK A VIDEO IS UNDER HERE Here is a short, possibly uninformative video, about L'Hôpital's rule (double click): media type="file" key="L'Hospital Project.mp4" width="300" height="300" Song credit: Baba O'Riley - The Who

=LOGORITHMIC METHOD=

Use the logorithmic method when you have something that will simplify with logorithmic properties. Here's an example. It's similar to the one in the book, but not the same.



Yeah so Matt and Ben covered it pretty good, but I thought that I would throw in some examples:

For instance,

lim __lnx__ x->1 x-1

If this limit is evaluated as is, it comes out to the indeterminate form of 0/0.

So, L'hospital's rule is used by deriving the numerator and denominator of the function:

lim __1/x__ x->1 1

If this limit is evaluated, it comes out to 1/1 which is 1. Therefore, the limit is equal to 1.

L'hospital's rule can be used with other functions as well:

lim __-7x__ x->0 sinx

When this limit is evaluated without using l'hospital's rule, it is 0/0 which is indeterminate

lim __-7__ x->0 cosx

When evaluated, this limit comes out to -7/1 meaning that the limit equals -7.

L'hospitals rule can also be used multiple times to evaluate a limit. For example,

lim __5x ² - 1 __ x-> ∞ e^ x

This limit evaluates to the indeterminate form ∞/ ∞. So, l'hospital's rule must be used.

lim __10x__ x-> ∞ e^ x

However, this new limit still evaluates to the indeterminate form of ∞/ ∞. This limit can be evaluated by employing l'hospital's rule a second time:

lim __10__ x-> ∞ e^ x

This time, the limit goes to 10/ ∞ which is just 0. Limits can be checked by graphing. As shown below, the limit of 5x ² - 1 over e^x is 0 as x goes to infinity:

**__Sources:__** __Web__ http://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule //^ I only used it for examples of limits to use the Rule on, which I verified.// http://mathworld.wolfram.com/LHospitalsRule.html []

__Books__ //Calculus: Concepts and Applications//, by Paul Foerster.