Occasionally, one comes across a limit which results in or , which are called indeterminate limits. However, it is still possible to solve these by using L'Hôpital's rule. This rule is vital in explaining how other limits can be derived.
Definition: Indeterminate Limit
If exists, where or , the limit is said to be indeterminate.
All of the following expressions are indeterminate forms.
These expressions are called indeterminate because you cannot determine their exact value in the indeterminate form. Depending on the situation, each indeterminate form could evaluate to a variety of values.
If is indeterminate of type or ,
then , where .
In other words, if the limit of the function is indeterminate, the limit equals the derivative of the top over the derivative of the bottom. If that is indeterminate, L'Hôpital's rule can be used again until the limit isn't or .
Suppose that for real functions and , and that exists. Thus and exist in an interval around , but maybe not at itself. Thus, for any , in any interval or , and are continuous and differentiable, with the possible exception of . Define
Note that , , and that are continuous in any interval or and differentiable in any interval or when .
Cauchy's Mean Value Theorem (see 3.9) tells us that for some or . Since , we have for .
Since or , by the squeeze theorem
So taking the limit as of the last equation gives , which is equivalent to the more commonly used form .
Since plugging in 0 for x results in , use L'Hôpital's rule to take the derivative of the top and bottom, giving:
Plugging in 0 for x gives 1 here.
Note that it is logically incorrect to prove this limit by using L'Hôpital's rule, as the same limit is required to prove that the derivative of the sine function exists: it would be a form of begging the question, or circular reasoning. An alternative way to prove this limit equal one is using squeeze theorem.
First, you need to rewrite the function into an indeterminate limit fraction:
Now it's indeterminate. Take the derivative of the top and bottom:
Plugging in 0 for once again gives 1.
This time, plugging in for x gives you . So using L'Hôpital's rule gives:
Therefore, is the answer.
Plugging the value of x into the limit yields
- (indeterminate form).
We now apply L'Hôpital's rule by taking the derivative of the top and bottom with respect to .
We apply L'Hôpital's rule once again
Similarly, this limit also yields the same result
This does not prove that because using the same method,
Evaluate the following limits using L'Hôpital's rule: