L'Hopital's rule

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 in indeterminate form $\frac{0}{0}$ or $\frac{\infty }{\infty }$ with L’Hopital’s rule. Instead of applying different methods such as dominant term analysis, simply take the derivatives of the numerator and the denominator separately and re-evaluate. Repeat the process until direct substitution no longer results in either of those two indeterminate forms.

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, the derivatives of the numerator and the denominator must be found separately before re-evaluating a limit by direct substitution.

1. Form: $0/0$

Recall the special trig limit $\underset{x\to 0}{\lim }\frac{\sin (x)}{x}=1$. Other methods like the squeeze theorem can be used to prove this limit, but we’ll show how to use L’Hopital’s rule on it.

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

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

Therefore

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

Re-evaluating the limit with direct substitution results in $\boxed{1}$.

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 can only be applied to quotients, so when this indeterminate form results from an expression $f(x)\times g(x)$, rewrite it either as $\frac{f(x)}{\frac{1}{g(x)}}$ or $\frac{g(x)}{\frac{1}{f(x)}}$ depending on which form’s denominator is easier to differentiate or will lead somewhere. 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 1st option takes on the form $\frac{\infty }{\infty }$ while the 2nd is $\frac{0}{0}$. Let’s try applying L’Hopital’s rule on the 1st one.

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

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

Hm…this doesn’t seem like it will go anywhere because differentiating the denominator will just increase the exponent on $x$ and the $\frac{1}{0}$ will never go away. So let’s work with the 2nd 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 $0$, as it may seem.

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 results in $\frac{0}{0}$ so L’Hopital’s rule can be applied.

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

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

$\frac{0}{0}$ again, so using L’Hopital’s:

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

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

Sidenote

Infinity + infinity

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

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

When direct substitution results in any of these indeterminate forms, resolve it by using the power property of logarithms.

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

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 at this step results in $\infty \times 0$. Apply the previous technique of turning into a quotient.

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

Then L’Hopital’s rule can be applied because direct substitution results in $\frac{0}{0}$.

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

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 the other side throughout this process was $\ln (L)$ and the goal was to find just $L$ (which we defined as the limit problem). Solving for $L$,

$$\ln (L)=1$$

$$L=e$$

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

Key points

• Use L’Hopital’s rule only on quotients, when the indeterminate form is $\frac{0}{0}$ or $\pm \frac{\infty }{\infty }$.
• For the other indeterminate forms, a summary for what to do:

Form	What to do
$0\cdot \infty$	Rewrite as fraction
$\infty -\infty$	Combine
$0^0,{\infty }^0,1^{\infty }$	Use logarithms