Suppose there are continuous functions and that are both zero at . Then, the limit cannot be found by substituting , since it yields which cannot be evaluated. We use as a notation for an expression known as an indeterminate form. In some cases, limits that lead to indeterminate forms may be evaluated by cancellation or rearrangement of terms (see Limits by Factoring for examples). However, this does not always work. For example, how would you evaluate Obviously, inserting will yield an indeterminate form of and, in this case, you can neither use algebraic manipulation nor rearrangement of terms to reduce this expression into a form that yields a valid limit.
Under certain circumstances, we can use a powerful theorem called L'Hôpital's rule to evaluate the limits that lead to indeterminate forms.
Suppose and are differentiable functions such that
- on an open interval containing
Directly applying leads the limit to an indeterminate form. Since both the term in the numerator and the term in the denominator are zero at and since and are both differentiable at , we can use L'Hôpital's rule:
Once again, directly applying produces an indeterminate form. Let the expression in the numerator be and the expression in the denominator . Since and are both zero and and are both differentiable at , we can use L'Hôpital's rule:
Since and as , we can use L'Hopital's rule:
Simplifying the expression, we obtain
So far we have looked at evaluating limits that reduce to the indeterminate form
But how would we evaluates limits that reduce to We can deal with such forms by writing them as quotients or rearranging them. The following identity is helpful:
Directly applying gives an indeterminate form of . This form of limit cannot be evaluated just like the case of . This seemingly formidable problem can be solved by introducing a variable substitution, . Using the substitution, the limit expression is changed as
In the case where application of L'Hôpital's rule yields an indeterminate form, if the resulting limit expression meets the conditions necessary to use L'Hôpital's rule, it can be used again. This can be quite confusing to understand. Let's look at the example below to see what this means.
Let and Since , and and are both differentiable at , we can use L'Hôpital's rule as follows:
The resulting expression yields an indeterminate form of . However, the term meets the criteria required to use L'Hôpital's rule:
Because the limit is in the form of , we can apply L'Hôpital's rule. The derivative of is , and thus by the chain rule, for some constant . Thus, the limit becomes
Note: We can also solve this limit by using the approximation for relatively small .
Evaluate the limit
Note that the limit is indeed in the form of However, we can simplify the given expression by noting that . So by taking their ratio, we are left with and its limit when approaches is simply
Evaluate the limit
Since the limit has the form of when , L'Hôpital's rule is applicable. Since we have
Given that and are finite constants such that evaluate .
We first note that the limit is of the form when , so L'Hôpital's rule is applicable:
If , the limit equals to which cannot be equal to a finite constant Hence is forced to take the value of . Continue by applying the rule a few more times to obtain
Similarly, as above, is forced to take a value of so . Then we have
Hence, or implying