# L'Hôpital's rule

L'Hôpital's rule is a theorem used to find the limit of certain types of indeterminate forms; indeterminate forms are expressions that result from attempting to compute a limit through use of substitution. For example, rational functions whose limits evaluate to 0/0 or ∞/∞ are referred to as indeterminate forms, since the expression does not provide enough information to evaluate the limit. Other indeterminate forms include 0 · ±∞, ∞ - ∞, 1^{∞}, 0^{0}, and ∞^{0}.

Indeterminate forms can be thought of as a contest between the terms in an expression where there are competing rules that make it unclear which term is dominant; this makes it unclear what the limit of the function is without further examination. There are a number of different ways to evaluate limits (refer to the limit page for more information). L'Hôpital's rule is only used in cases where a limit is of the indeterminate form 0/0 or ∞/∞, and the limit cannot be computed using other methods.

L'Hôpital's rule is as follows: If f and g are differentiable and g'(x) ≠ 0 on an open interval containing a (except possibly at a) and one of the following is true,

then:

In other words, L'Hôpital's rule states that for indeterminate forms of the appropriate type (0/0 or ∞/∞), the limit can be found by differentiating both expressions, which often results in a simplified expression whose limit can be computed through substitution. It is worth noting that L'Hôpital's rule can only be used for these specific cases, and cannot be used to compute all limits. Furthermore, in certain cases, L'Hôpital's rule must be applied multiple times before the limit can be computed through substitution; in other cases, no matter how many times L'Hôpital's rule is applied, the result will still be an indeterminate form, and the limit may not exist, or may need to be evaluated using different methods. Below is an example of how to use L'Hôpital's rule.

Example

Find .

Substituting 1 into the limit results in the indeterminate form 0/0:

Since the above expression fits the necessary criteria, we can use L'Hôpital's rule to determine the limit:

As mentioned above, in some cases we need to apply L'Hôpital's rule several times before we can determine a limit. Below is such an example.

Example

Find .

As x approaches ∞, both the numerator and denominator approach infinity, resulting in the indeterminate form ∞/∞. Thus, we can use L'Hôpital's rule, and differentiate the expression:

Notice that the resulting expression has the indeterminate form ∞/∞, so we still cannot determine the limit, and instead differentiate the expression once more:

After differentiating the expression a second time, we get an expression that is no longer an indeterminate form, so we can compute the limit and find that:

## Other indeterminate forms

Although L'Hôpital's rule can only be applied to limits of the indeterminate forms 0/0 and ∞/∞, it is sometimes possible to algebraically re-arrange expressions such that they meet the criteria necessary to use L'Hôpital's rule. The examples below illustrate this for different cases.

### Indeterminate form: 0 · ∞

Limits of the indeterminate form 0 · ±∞ can be converted into the indeterminate form 0/0 or ∞/∞ by rewriting the product as a quotient; given and , we can rewrite f(x) · g(x) as the quotient,

,

which results in an expression of the indeterminate form 0/0 or ∞/∞, allowing us to apply L'Hôpital's rule.

Example

Find .

Plugging π/2 in results in an expression of the indeterminate form 0 · ∞. In order to make use of L'Hôpital's rule, re-write the limit as a quotient such that substitution results in an expression of the indeterminate form 0/0:

Then, we use L'Hôpital's rule and differentiate the expression to find the limit:

### Indeterminate forms: 1^{∞}, 0^{0}, ∞^{0}

Limits of the indeterminate forms 1^{∞}, 0^{0}, or ∞^{0} may be converted into an expression of the indeterminate form 0/0 or ∞/∞ by using the natural logarithm function.

Example

Find

Plugging π/2 into the limit yields an expression of the indeterminate form 1^{∞}, so we apply the natural logarithm to the limit in an attempt to get the expression into a form that allows us to use L'Hôpital's rule:

Since sin(π/2) = 1, we can drop the sin(x) term:

After dropping the sin(x) term, the expression is now of the form 0/0, so we can apply L'Hôpital's rule:

Remember that we took the natural logarithm of the original limit, so we need to exponentiate our result to find the original limit. Thus, e^{0} = 1, and the limit is 1.

Limits of the form 0^{0} and ∞^{0} can be computed in a similar manner to this example.

### Indeterminate form: ∞ - ∞

Limits of the indeterminate form ∞ - ∞ can be converted to a limit of the form 0/0 or ∞/∞ by exponentiating the limit and using logarithm rules. In cases where have f and g that are fractions, we can simply combine them into a single quotient using the least common denominator, then use L'Hôpital's rule.

Examples

Find the following limits:

1. As x → 1^{+}, both terms in the expression approach ∞, so the limit is of the form ∞ - ∞. Since both terms are fractions, we can combine the terms,

resulting in an expression of the form 0/0, allowing us to apply L'Hôpital's rule:

This expression is again of the indeterminate form 0/0, so we apply L'Hôpital's rule once more:

Thus, the limit is 1.

2. As x → ∞, both terms approach ∞, so the limit is of the form ∞ - ∞. In this case, we cannot use the same method as before, since the terms of the expression are not fractions. Instead, we exponentiate both sides of the equation, assuming some real number L is the limit:

The resulting expression is of the indeterminate form ∞/∞, so we can apply L'Hôpital's rule:

Thus, L = ∞, which is an infinite limit, so we can say that the limit does not exist.

## When not to use L'Hôpital's rule

It is important to note that L'Hôpital's rule cannot be used for just any limit. To reiterate, it can only be used for limits that result in (or can be converted to) the indeterminate forms 0/0 or ∞/∞ after having evaluated the limit through substitution. Using it in situations where this is not the case will result in an incorrect outcome.

Example

Find .

This limit can be evaluated through substitution. Plugging x = 0 into the limit,

we find that the limit is 2. However, if we had applied L'Hôpital's rule instead of computing the limit as we did above, we would find an incorrect limit:

Thus, it is important to first confirm that we have a limit of the appropriate form by using substitution before applying L'Hôpital's rule to a limit.

It is also important to note that even in cases where a limit is of the indeterminate form 0/0 or ∞/∞, this does not guarantee that the limit exists, or that it can be determined using L'Hôpital's rule. It is entirely possible that repeated applications of L'Hôpital's rule always results in an indeterminate form. In such cases, the limit may not exist. However, just because a limit cannot be determined using L'Hôpital's rule does not necessarily mean that the limit does not exist; sometimes, it may be possible to use other methods to compute the limit.

Example

Find .

As x → ∞, both terms approach ∞, so we have a limit of the indeterminate form ∞/∞, and can apply L'Hôpital's rule:

The resulting limit is still of the form ∞/∞, and if we were to apply L'Hôpital's rule once more, we would end up with the same limit we began with. Regardless how many times we apply L'Hôpital's rule, we would continue to alternate between the same two results. Thus, this limit cannot be computed using L'Hôpital's rule. However, it can be computed by simplifying the expression algebraically. To do so, we first rewrite x + 1 as . We also complete the square of the denominator to find that x^{2} + 2x + 3 = (x + 1)^{2} + 2, allowing us to simplify the expression: