L'Hôpital's rule

In computing limits, we met limits of the form $\dfrac{0}{0}$. We solved them with algebra: factoring, conjugates, or combining fractions. But algebra does not always work. For example, no algebra trick simplifies

$\displaystyle\lim_{x\to 1}\dfrac{\ln x}{x-1}$.

L'Hôpital's rule uses derivatives to find limits like this one. (The name is French. It is pronounced "low-pee-TAHL.")

Indeterminate forms

If both the numerator and the denominator approach $0$, the limit has the indeterminate form $\dfrac{0}{0}$. "Indeterminate" means that the form alone does not tell us the answer. For example, all of these have the form $\dfrac{0}{0}$ at $x=0$:

$\dfrac{2x}{x}\to 2$,   $\dfrac{x^2}{x}\to 0$,   $\dfrac{x}{x^3}\to\infty$.

The same is true when both approach $\infty$ (or $-\infty$). That is the indeterminate form $\dfrac{\infty}{\infty}$.

The rule

L'Hôpital's Rule: Suppose $f$ and $g$ are differentiable near $a$, and $g'(x)\ne 0$ near $a$ (except possibly at $a$). If $\dfrac{f(x)}{g(x)}$ has the form $\dfrac{0}{0}$ or $\dfrac{\infty}{\infty}$ as $x\to a$, then

$\displaystyle\lim_{x\to a}\dfrac{f(x)}{g(x)}=\lim_{x\to a}\dfrac{f'(x)}{g'(x)}$,

if the limit on the right exists (or is $\infty$ or $-\infty$).

The rule also works for one-sided limits and for $x\to\infty$ or $x\to-\infty$.

Two warnings:

Optional Why the rule works (a simple case)

Suppose $f(a)=g(a)=0$, and $f'$ and $g'$ are continuous with $g'(a)\ne 0$. Since $f(a)$ and $g(a)$ are $0$, we can subtract them:

$\begin{align*}&\dfrac{f(x)}{g(x)}\\&=\dfrac{f(x)-f(a)}{g(x)-g(a)}\end{align*}$

Now divide the numerator and the denominator by $x-a$:

$\begin{align*}&=\dfrac{\dfrac{f(x)-f(a)}{x-a}}{\dfrac{g(x)-g(a)}{x-a}}\end{align*}$

As $x\to a$, the top approaches $f'(a)$, and the bottom approaches $g'(a)$. So the limit is $\dfrac{f'(a)}{g'(a)}$.

Near $a$, both graphs look like their tangent lines. So their quotient is close to the quotient of the slopes.

The form $\dfrac{0}{0}$

Example 1: Find

$\displaystyle\lim_{x\to 1}\dfrac{\ln x}{x-1}$.

Solution:

At $x=1$: $\ln 1=0$ and $1-1=0$. So the form is $\dfrac{0}{0}$.

Use L'Hôpital's rule. The derivative of $\ln x$ is $\dfrac{1}{x}$, and the derivative of $x-1$ is $1$:

$\begin{align*}&\lim_{x\to 1}\dfrac{\ln x}{x-1}\\&=\lim_{x\to 1}\dfrac{1/x}{1}\\&=1\end{align*}$

Example 2: Find

$\displaystyle\lim_{x\to 0}\dfrac{1-\cos x}{x^2}$.

Solution:

At $x=0$, the form is $\dfrac{0}{0}$. Use the rule:

$\displaystyle\lim_{x\to 0}\dfrac{1-\cos x}{x^2}=\lim_{x\to 0}\dfrac{\sin x}{2x}$

This is still $\dfrac{0}{0}$. So use the rule again:

$\displaystyle\lim_{x\to 0}\dfrac{\sin x}{2x}=\lim_{x\to 0}\dfrac{\cos x}{2}=\dfrac{1}{2}$

You may use the rule as many times as you need, as long as the form is still indeterminate each time.

A note about $\dfrac{\sin x}{x}$: L'Hôpital's rule gives $\displaystyle\lim_{x\to 0}\dfrac{\sin x}{x}=\lim_{x\to 0}\dfrac{\cos x}{1}=1$. This is a good check. But it is not a proof, because we used this same limit to find the derivative of $\sin x$. The real proof is in the Squeeze Theorem.

The form $\dfrac{\infty}{\infty}$

L'Hôpital's rule can compare how fast two functions grow.

Example 3: Find

$\displaystyle\lim_{x\to\infty}\dfrac{e^x}{x^2}$.

Solution:

Both $e^x$ and $x^2$ approach $\infty$. Use the rule twice:

$\begin{align*}&\lim_{x\to\infty}\dfrac{e^x}{x^2}\\&=\lim_{x\to\infty}\dfrac{e^x}{2x}\\&=\lim_{x\to\infty}\dfrac{e^x}{2}\\&=\infty\end{align*}$

So $e^x$ grows much faster than $x^2$. The same steps show that $e^x$ grows faster than any power of $x$.

Example 4: Find

$\displaystyle\lim_{x\to\infty}\dfrac{\ln x}{\sqrt{x}}$.

Solution:

Both approach $\infty$. The derivative of $\sqrt{x}$ is $\dfrac{1}{2\sqrt{x}}$:

$\begin{align*}&\lim_{x\to\infty}\dfrac{\ln x}{\sqrt{x}}\\&=\lim_{x\to\infty}\dfrac{1/x}{1/(2\sqrt{x})}\\&=\lim_{x\to\infty}\dfrac{2\sqrt{x}}{x}\\&=\lim_{x\to\infty}\dfrac{2}{\sqrt{x}}\\&=0\end{align*}$

So $\ln x$ grows more slowly than $\sqrt{x}$. In fact, $\ln x$ grows more slowly than any positive power of $x$.

Other indeterminate forms

Some limits are indeterminate in other ways. Rewrite them as $\dfrac{0}{0}$ or $\dfrac{\infty}{\infty}$ first.

Products: $0\cdot\infty$

If one factor approaches $0$ and the other approaches $\infty$, write the product as a quotient. Divide by the reciprocal of one factor.

Example 5: Find

$\displaystyle\lim_{x\to 0^+}x\ln x$.

Solution:

As $x\to 0^+$, $x\to 0$ and $\ln x\to-\infty$. Write $x$ as $\dfrac{1}{1/x}$:

$x\ln x=\dfrac{\ln x}{1/x}$

Now the form is $\dfrac{-\infty}{\infty}$. The derivative of $\dfrac{1}{x}$ is $-\dfrac{1}{x^2}$:

$\begin{align*}&\lim_{x\to 0^+}\dfrac{\ln x}{1/x}\\&=\lim_{x\to 0^+}\dfrac{1/x}{-1/x^2}\\&=\lim_{x\to 0^+}(-x)\\&=0\end{align*}$

Differences: $\infty-\infty$

If both terms approach $\infty$, combine them into one fraction. For example, as $x\to 0^+$, both terms below approach $\infty$:

$\dfrac{1}{x}-\dfrac{1}{\sin x}=\dfrac{\sin x-x}{x\sin x}$

The right side has the form $\dfrac{0}{0}$, so L'Hôpital's rule can be used.

Powers: $1^\infty$, $0^0$, and $\infty^0$

For a limit of $y=f(x)^{g(x)}$, take the natural logarithm first:

$\ln y=g(x)\ln f(x)$

Find the limit $L$ of $\ln y$. Then the limit of $y$ is $e^L$.

Example 6: Find

$\displaystyle\lim_{x\to 0^+}x^x$.

Solution:

This has the form $0^0$. Let $y=x^x$. Then

$\ln y=x\ln x$.

By Example 5, $x\ln x\to 0$. So $\ln y\to 0$, and

$\displaystyle\lim_{x\to 0^+}x^x=e^0=1$.

Example 7: Find

$\displaystyle\lim_{x\to\infty}\left(1+\dfrac{1}{x}\right)^x$.

Solution:

This has the form $1^\infty$. Let

$y=\left(1+\dfrac{1}{x}\right)^x$.

Then

$\ln y=x\ln\left(1+\dfrac{1}{x}\right)=\dfrac{\ln(1+1/x)}{1/x}$.

This has the form $\dfrac{0}{0}$. Use L'Hôpital's rule. The numerator needs the chain rule:

$\begin{align*}&\lim_{x\to\infty}\ln y\\&=\lim_{x\to\infty}\dfrac{\dfrac{1}{1+1/x}\cdot\left(-\dfrac{1}{x^2}\right)}{-\dfrac{1}{x^2}}\end{align*}$

The factors ${-\dfrac{1}{x^2}}$ cancel:

$\begin{align*}&=\lim_{x\to\infty}\dfrac{1}{1+1/x}\end{align*}$

As $x\to\infty$, $\dfrac{1}{x}\to 0$:

$\begin{align*}&\lim_{x\to\infty}\ln y=\dfrac{1}{1+0}=1\end{align*}$

Since $\ln y\to 1$, $y$ approaches $e^1$:

$\displaystyle\lim_{x\to\infty}\left(1+\dfrac{1}{x}\right)^x=e^1=e$.

This limit appears in compound interest. It is another way to define the number $e$ (see derivatives of exponential and logarithmic functions).

Check the form first

Example 8: What is wrong with this work?

$\displaystyle\lim_{x\to 0}\dfrac{x+1}{x+2}=\lim_{x\to 0}\dfrac{1}{1}=1$

Solution:

The form is not indeterminate. At $x=0$, the quotient is $\dfrac{1}{2}$. So direct substitution gives the correct answer:

$\displaystyle\lim_{x\to 0}\dfrac{x+1}{x+2}=\dfrac{1}{2}$

L'Hôpital's rule does not apply here, and using it gave a wrong answer.

Summary

FormWhat to do
$\dfrac{0}{0}$ or $\dfrac{\infty}{\infty}$Use L'Hôpital's rule.
$0\cdot\infty$Rewrite the product as a quotient.
$\infty-\infty$Combine into one fraction.
$1^\infty$, $0^0$, $\infty^0$Take $\ln$, find the limit of $\ln y$, then use $e$.

← Optimization problemsAntiderivatives →