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.

Main content

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 00 or .
In other words, it helps us find limxcu(x)v(x), where limxcu(x)=limxcv(x)=0 (or, alternatively, where both limits are ±).
The rule essentially says that if the limit limxcu(x)v(x) exists, then the two limits are equal:
limxcu(x)v(x)=limxcu(x)v(x)
Want to learn more about L'Hôpital's rule? Check out this video.

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

Let's find, for example, limx07xsin(x)x2+sin(3x).
Substituting x=0 into 7xsin(x)x2+sin(3x) results in the indeterminate form 00. So let's use L’Hôpital’s rule.
=limx07xsin(x)x2+sin(3x)=limx0ddx[7xsin(x)]ddx[x2+sin(3x)]L’Hopital’s rule=limx07cos(x)2x+3cos(3x)=7cos(0)2(0)+3cos(30)Substitution=2
Note that we were only able to use L’Hôpital’s rule because the limit limx0ddx[7xsin(x)]ddx[x2+sin(3x)] actually exists.
Problem 1.1
limx0ex12x=?
Choose 1 answer:

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, limx0(1+2x)1sin(x). Substituting x=0 into the expression results in the indeterminate form 1.
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=(1+2x)1sin(x), we will find limx0ln(y). Once we find it, we will be able to find limx0y.
ln(y)=ln(1+2x)sin(x)
Substituting x=0 into ln(1+2x)sin(x) results in the indeterminate form 00, so now it's L’Hôpital’s rule's turn to help us with our quest!
=limx0ln(y)=limx0ln(1+2x)sin(x)=limx0ddx[ln(1+2x)]ddx[sin(x)]L’Hopital’s rule=limx0(21+2x)cos(x)=(21)1Substitution=2
We found that limx0ln(y)=2, which means limx0y=e2.
Problem 2.1
limx0[cos(2πx)]1x=?
Choose 1 answer:

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

Want to join the conversation?