Occasionally, one comes across a limit which results in or , which are called indeterminate limits. However, it is still possible to solve these in many cases due to L'Hôpital's rule. This rule also is vital in explaining how a number of other limits can be derived.
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 ,
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 .
can approach a finite value c, or .
Proof of the 0/0 case
Suppose that for real functions and , and and that exists. Thus and exist in an interval around , but maybe not at itself. This implies that both and are differentiable (and thus continuous) everywhere in except perhaps at . Thus, for any in , in any interval or , and are continuous and differentiable, with the possible exception of . Define
Note that , and that and are continuous in any interval or and differentiable in any interval or when is in . Cauchy's Mean Value Theorem tells us that for some in (if ) or (if ). Since , we have for and in . Note that is the same limit as since both and are being squeezed to . So taking the limit as of the last equation gives which is equivalent to .
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.
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 x once again gives one.
This time, plugging in for x gives you . You know the drill:
This time, though, there is no x term left! is the answer.
Sometimes, forms exist where it is not intuitively obvious how to solve them. One might think the value However, as was noted in the definition of an indeterminate form, this isn't possible to evaluate using the rules learned before now, and we need to use L'Hôpital's rule to solve.
Plugging the value of x into the limit yields
- (indeterminate form).
= = = (indeterminate form)
We now apply L'Hôpital's rule by taking the derivative of the top and bottom with respect to x.
Returning to the expression above
= = (indeterminate form)
We apply L'Hôpital's rule once again
Careful: this does not prove that because
Evaluate the following limits using L'Hôpital's rule: