What you’ll learn:

• How to apply L’Hopital’s rule on limits with indeterminate form $\frac{0}{0}$ or $\frac{\infty }{\infty }$
• How to handle other indeterminate forms by rewriting or using logarithms

Another application of derivatives is evaluating limits that produce the indeterminate forms $\frac{0}{0}$ or $\frac{\infty }{\infty }$. In these cases, L’Hopital’s rule lets you replace the original quotient with a new one formed by differentiating the numerator and denominator separately, then trying direct substitution again. You can repeat this process until direct substitution no longer gives $\frac{0}{0}$ or $\frac{\infty }{\infty }$.

L’Hopital’s rule

If $\underset{x\to a}{\lim }f(x)=\underset{x\to a}{\lim }g(x)=0$ or $\underset{x\to a}{\lim }f(x)=\underset{x\to a}{\lim }g(x)=\infty$, then

$$\underset{x\to a}{\lim }\frac{f(x)}{g(x)}=\underset{x\to a}{\lim }\frac{f^{\prime }(x)}{g^{\prime }(x)}$$

Sidenote

Not the quotient rule

Don’t use the quotient rule. With L’Hopital’s rule, you differentiate the numerator and denominator separately, then form the new quotient and re-check the limit.

1. Form: $0/0$

Recall the special trig limit $\underset{x\to 0}{\lim }\frac{\sin (x)}{x}=1$. You can prove this limit using other tools (like the squeeze theorem), but here we’ll evaluate it using L’Hopital’s rule.

Since direct substitution gives $\frac{0}{0}$, apply L’Hopital’s rule.

• The derivative of the numerator, $\sin (x)$, is $\cos (x)$.
• The derivative of the denominator, $x$, is $1$.

Therefore

$$\underset{x\to 0}{\lim }\frac{\sin (x)}{x}=\underset{x\to 0}{\lim }\frac{\cos (x)}{1}$$

Now direct substitution works, so the limit is $\boxed{1}$.

Here are a few more examples where direct substitution results in $\frac{0}{0}$.

$$\underset{x\to 0}{\lim }\frac{e^x-1}{x}$$

Solution

$$\underset{x\to 0}{\lim }\frac{\sqrt{1+x}-1}{x}$$

Solution

2. Form: $\infty /\infty$

L’Hopital’s rule can also be applied when direct substitution results in the indeterminate form $\frac{\infty }{\infty }$.

$$\underset{x\to \infty }{\lim }\frac{x^3+5}{e^x}$$

Solution

$$\underset{x\to \infty }{\lim }\frac{\ln (e^x+x)}{x}$$

Solution

3. Form $0\times \infty$

L’Hopital’s rule applies only to quotients. If you have an indeterminate product $f(x)\times g(x)$, rewrite it as a fraction:

• $\frac{f(x)}{\frac{1}{g(x)}}$, or
• $\frac{g(x)}{\frac{1}{f(x)}}$

Choose the version whose denominator is easier to differentiate or leads to a useful form. For example,

$$\underset{x\to 0^{+}}{\lim }x\cot (x)$$

can be written either as

$$\underset{x\to 0^{+}}{\lim }\frac{\cot (x)}{1/x}$$

or

$$\underset{x\to 0^{+}}{\lim }\frac{x}{1/\cot (x)}$$

$$=\underset{x\to 0^{+}}{\lim }\frac{x}{\tan (x)}$$

The first option has the form $\frac{\infty }{\infty }$, while the second has the form $\frac{0}{0}$. Try applying L’Hopital’s rule to the first one:

$$\underset{x\to 0^{+}}{\lim }\frac{\cot (x)}{1/x}$$

$$\underset{x\to 0^{+}}{\lim }\frac{-{\csc }^2(x)}{-1/x^2}$$

This approach doesn’t simplify the indeterminate behavior in a helpful way, because differentiating $1/x$ keeps producing powers of $x$ in the denominator. So switch to the second option:

$$=\underset{x\to 0^{+}}{\lim }\frac{x}{\tan (x)}$$

$$=\underset{x\to 0^{+}}{\lim }\frac{1}{{\sec }^2(x)}$$

$$=\boxed{1}$$

4. Form $\infty -\infty$

$\infty -\infty$ is an indeterminate form because the limit is not necessarily $0$.

Find

$$\underset{x\to 1^{+}}{\lim }\frac{x}{x-1}-\frac{1}{\ln (x)}$$

Combine into one rational expression.

$$\underset{x\to 1^{+}}{\lim }\frac{x}{x-1}\cdot \frac{\ln (x)}{\ln (x)}-\frac{1}{\ln (x)}\cdot \frac{x-1}{x-1}$$

$$=\underset{x\to 1^{+}}{\lim }\frac{x\ln (x)-(x-1)}{(x-1)\ln (x)}$$

Direct substitution gives $\frac{0}{0}$, so apply L’Hopital’s rule.

$$=\underset{x\to 1^{+}}{\lim }\frac{x(1/x)+\ln (x)-1}{(x-1)(1/x)+\ln (x)}$$

$$=\underset{x\to 1^{+}}{\lim }\frac{\ln (x)}{1-(1/x)+\ln (x)}$$

Direct substitution still gives $\frac{0}{0}$, so apply L’Hopital’s rule again:

$$=\underset{x\to 1^{+}}{\lim }\frac{(1/x)}{(1/x^2)+(1/x)}$$

$$=\boxed{\frac{1}{2}}$$

Sidenote

Infinity + infinity

$\infty +\infty$ is not an indeterminate form - the limit is $\infty$. Similarly, $-\infty -\infty$ becomes $-\infty$.

5. Form: $1^{\infty },0^0,{\infty }^0$

When direct substitution produces one of these indeterminate exponential forms, rewrite the expression using logarithms and the power property.

$$\underset{x\to \infty }{\lim }(1+\frac{1}{x}{)}^x$$

Let

$$\begin{array}{rl}L& =\underset{x\to \infty }{\lim }(1+\frac{1}{x}{)}^x\\ \ln (L)& =\underset{x\to \infty }{\lim }\ln \left[\right(1+\frac{1}{x}{)}^x]\\ & =\underset{x\to \infty }{\lim }x\cdot \ln (1+\frac{1}{x})\end{array}$$

Direct substitution here gives $\infty \times 0$. Rewrite it as a quotient:

$$\underset{x\to \infty }{\lim }\frac{\ln (1+\frac{1}{x})}{\frac{1}{x}}$$

Now direct substitution gives $\frac{0}{0}$, so apply L’Hopital’s rule.

$$\underset{x\to \infty }{\lim }\frac{\frac{x}{x+1}\cdot (-\frac{1}{x^2})}{-\frac{1}{x^2}}$$

After canceling $-\frac{1}{x^2}$, the limit becomes

$$\underset{x\to \infty }{\lim }\frac{x+1}{x}=\boxed{1}$$

(using either dominant term analysis or applying L’Hopital’s rule again).

Remember that this value is for $\ln (L)$, and the goal is to find $L$.

$$\ln (L)=1$$

$$L=e$$

Therefore

$$\boxed{\underset{x\to \infty }{\lim }(1+\frac{1}{x}{)}^x=e}$$

AP tip:

Remember this limit definition of $e$.

Examples

$$\underset{n\to 0}{\lim }(1+n{)}^{1/n}$$

Solution

$$\underset{x\to \infty }{\lim }(1-\frac{1}{x}{)}^{\cos (x)}$$

Solution

$$\underset{x\to 0^{+}}{\lim }{x}^x$$

Solution