# 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 start fraction, 0, divided by, 0, end fraction or start fraction, infinity, divided by, infinity, end fraction.
In other words, it helps us find $\displaystyle\lim_{x\to c}\dfrac{u(x)}{v(x)}$, where $\displaystyle\lim_{x\to c}u(x)=\lim_{x\to c}v(x)=0$ (or, alternatively, where both limits are plus minus, infinity).
The rule essentially says that if the limit $\displaystyle\lim_{x\to c}\dfrac{u'(x)}{v'(x)}$ exists, then the two limits are equal:
$\displaystyle\lim_{x\to c}\dfrac{u(x)}{v(x)}=\displaystyle\lim_{x\to c}\dfrac{u'(x)}{v'(x)}$

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

Let's find, for example, $\displaystyle\lim_{x\to 0}\dfrac{7x-\sin(x)}{x^2+\sin(3x)}$.
Substituting x, equals, 0 into start fraction, 7, x, minus, sine, left parenthesis, x, right parenthesis, divided by, x, start superscript, 2, end superscript, plus, sine, left parenthesis, 3, x, right parenthesis, end fraction results in the indeterminate form start fraction, 0, divided by, 0, end fraction. So let's use L’Hôpital’s rule.
\begin{aligned} &\phantom{=}\displaystyle\lim_{x\to 0}\dfrac{7x-\sin(x)}{x^2+\sin(3x)} \\\\ &=\displaystyle\lim_{x\to 0}\dfrac{\dfrac{d}{dx}[7x-\sin(x)]}{\dfrac{d}{dx}[x^2+\sin(3x)]}\qquad\gray{\text{L'Hopital's rule}} \\\\ &=\displaystyle\lim_{x\to 0}\dfrac{7-\cos(x)}{2x+3\cos(3x)} \\\\ &=\dfrac{7-\cos(0)}{2(0)+3\cos(3\cdot0)}\qquad\gray{\text{Substitution}} \\\\ &=2 \end{aligned}
Note that we were only able to use L’Hôpital’s rule because the limit $\displaystyle\lim_{x\to 0}\dfrac{\dfrac{d}{dx}[7x-\sin(x)]}{\dfrac{d}{dx}[x^2+\sin(3x)]}$ actually exists.
Problem 1.1
$\displaystyle\lim_{x\to 0}\dfrac{e^x-1}{2x}=?$

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, $\displaystyle\lim_{x\to 0}(1+2x)^{^{\LARGE\frac{1}{\sin(x)}}}$. Substituting x, equals, 0 into the expression results in the indeterminate form 1, start superscript, start superscript, infinity, end superscript, end superscript.
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, equals, left parenthesis, 1, plus, 2, x, right parenthesis, start superscript, start superscript, start fraction, 1, divided by, sine, left parenthesis, x, right parenthesis, end fraction, end superscript, end superscript, we will find $\displaystyle\lim_{x\to 0}\ln(y)$. Once we find it, we will be able to find $\displaystyle\lim_{x\to 0}y$.
natural log, left parenthesis, y, right parenthesis, equals, start fraction, natural log, left parenthesis, 1, plus, 2, x, right parenthesis, divided by, sine, left parenthesis, x, right parenthesis, end fraction
\begin{aligned} \ln(y)&=\ln\left((1+2x){^{\LARGE\frac{1}{\sin(x)}}}\right) \\\\ &=\dfrac{1}{\sin(x)}\ln(1+2x)\qquad\gray{\text{Logarithm properties}} \\\\ &=\dfrac{\ln(1+2x)}{\sin(x)} \end{aligned}
Substituting x, equals, 0 into start fraction, natural log, left parenthesis, 1, plus, 2, x, right parenthesis, divided by, sine, left parenthesis, x, right parenthesis, end fraction results in the indeterminate form start fraction, 0, divided by, 0, end fraction, so now it's L’Hôpital’s rule's turn to help us with our quest!
\begin{aligned} &\phantom{=}\displaystyle\lim_{x\to 0}\ln(y) \\\\ &=\displaystyle\lim_{x\to 0}\dfrac{\ln(1+2x)}{\sin(x)} \\\\ &=\displaystyle\lim_{x\to 0}\dfrac{\dfrac{d}{dx}[\ln(1+2x)]}{\dfrac{d}{dx}[\sin(x)]}\qquad\gray{\text{L'Hopital's rule}} \\\\ &=\displaystyle\lim_{x\to 0}\dfrac{\left(\dfrac{2}{1+2x}\right)}{\cos(x)} \\\\ &=\dfrac{\left(\dfrac{2}{1}\right)}{1}\qquad\gray{\text{Substitution}} \\\\ &=2 \end{aligned}
We found that $\displaystyle\lim_{x\to 0}\ln(y)=2$, which means $\displaystyle\lim_{x\to 0}y=e^2$.
Since natural log, left parenthesis, x, right parenthesis is continuous, we know that $\displaystyle\lim_{x\to 0}\ln(y)=\ln\left(\lim_{x\to 0}y\right)$, which means $\ln\left(\displaystyle\lim_{x\to 0}y\right)=2$:
\begin{aligned} \ln\left(\displaystyle\lim_{x\to 0}y\right)&=2 \\\\ e^{\ln\left(\displaystyle\lim_{x\to 0}y\right)}&=e^2\qquad\gray{\text{Raise }e\text{ by both sides}} \\\\ \displaystyle\lim_{x\to 0}y&=e^2 \end{aligned}
Problem 2.1
$\displaystyle\displaystyle\lim_{x\to0} [\cos\left(2\pi x \right)]^{^{\LARGE\frac 1 x}}=?$