If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

## Differential Calculus

### Course: Differential Calculus>Unit 4

Lesson 8: L’Hôpital’s rule: composite exponential functions

# L'Hôpital's rule review

L'Hôpital's rule helps us find many limits where direct substitution ends with the indeterminate forms 0/0 or ∞/∞. Review how (and when) it's applied.

## What is L'Hôpital's rule?

L'Hôpital's rule helps us evaluate indeterminate limits of the form $\frac{0}{0}$ or $\frac{\mathrm{\infty }}{\mathrm{\infty }}$.
In other words, it helps us find $\underset{x\to c}{lim}\frac{u\left(x\right)}{v\left(x\right)}$, where $\underset{x\to c}{lim}u\left(x\right)=\underset{x\to c}{lim}v\left(x\right)=0$ (or, alternatively, where both limits are $±\mathrm{\infty }$).
The rule essentially says that if the limit $\underset{x\to c}{lim}\frac{{u}^{\prime }\left(x\right)}{{v}^{\prime }\left(x\right)}$ exists, then the two limits are equal:
$\underset{x\to c}{lim}\frac{u\left(x\right)}{v\left(x\right)}=\underset{x\to c}{lim}\frac{{u}^{\prime }\left(x\right)}{{v}^{\prime }\left(x\right)}$

## Using L'Hôpital's rule to find limits of quotients

Let's find, for example, $\underset{x\to 0}{lim}\frac{7x-\mathrm{sin}\left(x\right)}{{x}^{2}+\mathrm{sin}\left(3x\right)}$.
Substituting $x=0$ into $\frac{7x-\mathrm{sin}\left(x\right)}{{x}^{2}+\mathrm{sin}\left(3x\right)}$ results in the indeterminate form $\frac{0}{0}$. So let's use L’Hôpital’s rule.
$\begin{array}{rl}& \phantom{=}\underset{x\to 0}{lim}\frac{7x-\mathrm{sin}\left(x\right)}{{x}^{2}+\mathrm{sin}\left(3x\right)}\\ \\ & =\underset{x\to 0}{lim}\frac{\frac{d}{dx}\left[7x-\mathrm{sin}\left(x\right)\right]}{\frac{d}{dx}\left[{x}^{2}+\mathrm{sin}\left(3x\right)\right]}\phantom{\rule{2em}{0ex}}\text{L’Hopital’s rule}\\ \\ & =\underset{x\to 0}{lim}\frac{7-\mathrm{cos}\left(x\right)}{2x+3\mathrm{cos}\left(3x\right)}\\ \\ & =\frac{7-\mathrm{cos}\left(0\right)}{2\left(0\right)+3\mathrm{cos}\left(3\cdot 0\right)}\phantom{\rule{2em}{0ex}}\text{Substitution}\\ \\ & =2\end{array}$
Note that we were only able to use L’Hôpital’s rule because the limit $\underset{x\to 0}{lim}\frac{\frac{d}{dx}\left[7x-\mathrm{sin}\left(x\right)\right]}{\frac{d}{dx}\left[{x}^{2}+\mathrm{sin}\left(3x\right)\right]}$ actually exists.
Problem 1.1
$\underset{x\to 0}{lim}\frac{{e}^{x}-1}{2x}=?$

Want to try more problems like this? Check out this exercise.

## Using L'Hôpital's rule to find limits of exponents

Let's find, for example, $\underset{x\to 0}{lim}\left(1+2x{\right)}^{{}^{\frac{1}{\mathrm{sin}\left(x\right)}}}$. Substituting $x=0$ into the expression results in the indeterminate form ${1}^{{}^{\mathrm{\infty }}}$.
To make the expression easier to analyze, let's take its natural log (this is a common trick when dealing with composite exponential functions). In other words, letting $y=\left(1+2x{\right)}^{{}^{\frac{1}{\mathrm{sin}\left(x\right)}}}$, we will find $\underset{x\to 0}{lim}\mathrm{ln}\left(y\right)$. Once we find it, we will be able to find $\underset{x\to 0}{lim}y$.
$\mathrm{ln}\left(y\right)=\frac{\mathrm{ln}\left(1+2x\right)}{\mathrm{sin}\left(x\right)}$
Substituting $x=0$ into $\frac{\mathrm{ln}\left(1+2x\right)}{\mathrm{sin}\left(x\right)}$ results in the indeterminate form $\frac{0}{0}$, so now it's L’Hôpital’s rule's turn to help us with our quest!
$\begin{array}{rl}& \phantom{=}\underset{x\to 0}{lim}\mathrm{ln}\left(y\right)\\ \\ & =\underset{x\to 0}{lim}\frac{\mathrm{ln}\left(1+2x\right)}{\mathrm{sin}\left(x\right)}\\ \\ & =\underset{x\to 0}{lim}\frac{\frac{d}{dx}\left[\mathrm{ln}\left(1+2x\right)\right]}{\frac{d}{dx}\left[\mathrm{sin}\left(x\right)\right]}\phantom{\rule{2em}{0ex}}\text{L’Hopital’s rule}\\ \\ & =\underset{x\to 0}{lim}\frac{\left(\frac{2}{1+2x}\right)}{\mathrm{cos}\left(x\right)}\\ \\ & =\frac{\left(\frac{2}{1}\right)}{1}\phantom{\rule{2em}{0ex}}\text{Substitution}\\ \\ & =2\end{array}$
We found that $\underset{x\to 0}{lim}\mathrm{ln}\left(y\right)=2$, which means $\underset{x\to 0}{lim}y={e}^{2}$.
Problem 2.1
$\underset{x\to 0}{lim}\left[\mathrm{cos}\left(2\pi x\right){\right]}^{{}^{\frac{1}{x}}}=?$