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.

# 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 limit, start subscript, x, \to, c, end subscript, start fraction, u, left parenthesis, x, right parenthesis, divided by, v, left parenthesis, x, right parenthesis, end fraction, where limit, start subscript, x, \to, c, end subscript, u, left parenthesis, x, right parenthesis, equals, limit, start subscript, x, \to, c, end subscript, v, left parenthesis, x, right parenthesis, equals, 0 (or, alternatively, where both limits are plus minus, infinity).
The rule essentially says that if the limit limit, start subscript, x, \to, c, end subscript, start fraction, u, prime, left parenthesis, x, right parenthesis, divided by, v, prime, left parenthesis, x, right parenthesis, end fraction exists, then the two limits are equal:
limit, start subscript, x, \to, c, end subscript, start fraction, u, left parenthesis, x, right parenthesis, divided by, v, left parenthesis, x, right parenthesis, end fraction, equals, limit, start subscript, x, \to, c, end subscript, start fraction, u, prime, left parenthesis, x, right parenthesis, divided by, v, prime, left parenthesis, x, right parenthesis, end fraction

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

Let's find, for example, limit, start subscript, x, \to, 0, end subscript, start fraction, 7, x, minus, sine, left parenthesis, x, right parenthesis, divided by, x, squared, plus, sine, left parenthesis, 3, x, right parenthesis, end fraction.
Substituting x, equals, 0 into start fraction, 7, x, minus, sine, left parenthesis, x, right parenthesis, divided by, x, squared, 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 limit, start subscript, x, \to, 0, end subscript, start fraction, start fraction, d, divided by, d, x, end fraction, open bracket, 7, x, minus, sine, left parenthesis, x, right parenthesis, close bracket, divided by, start fraction, d, divided by, d, x, end fraction, open bracket, x, squared, plus, sine, left parenthesis, 3, x, right parenthesis, close bracket, end fraction actually exists.
Problem 1.1
limit, start subscript, x, \to, 0, end subscript, start fraction, e, start superscript, x, end superscript, minus, 1, divided by, 2, x, end fraction, equals, question mark

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, limit, start subscript, x, \to, 0, end subscript, 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. 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 limit, start subscript, x, \to, 0, end subscript, natural log, left parenthesis, y, right parenthesis. Once we find it, we will be able to find limit, start subscript, x, \to, 0, end subscript, 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
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 limit, start subscript, x, \to, 0, end subscript, natural log, left parenthesis, y, right parenthesis, equals, 2, which means limit, start subscript, x, \to, 0, end subscript, y, equals, e, squared.
Problem 2.1
limit, start subscript, x, \to, 0, end subscript, open bracket, cosine, left parenthesis, 2, pi, x, right parenthesis, close bracket, start superscript, start superscript, start fraction, 1, divided by, x, end fraction, end superscript, end superscript, equals, question mark