Calculus/L'Hôpital's Rule

← Differentiation/Basics of Differentiation/Exercises Calculus Extrema and Points of Inflection →
L'Hôpital's Rule

L'Hôpital's RuleEdit

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.

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   ,


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   ,   or   .

Proof of the   caseEdit

Suppose that for real functions   ,   and that   exists. Thus   and   exist in an interval   around   , but maybe not at   itself. This implies that both   are differentiable (and thus continuous) everywhere in   except perhaps at  . Thus, for any   , in any interval   or   ,   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 tells us that   for some   or   . Since   , we have   for   .

Note that since   or   , by the squeeze theorem


This implies


So taking the limit as   of the last equation gives   which is equivalent to   .


Example 1Edit


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

Example 2Edit


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 1.

Example 3Edit


This time, plugging in   for x gives you   . You know the drill:


This time, though, there is no x term left!   is the answer.

Example 4Edit

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   .


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:












← Differentiation/Basics of Differentiation/Exercises Calculus Extrema and Points of Inflection →
L'Hôpital's Rule