## L'Hôpital's Rule

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L'Hôpital's Rule ( aka L'Hospital's Rule or Bernoulli's Rule )

If the limit $$\displaystyle{ \lim_{x \to c}{\frac{n(x)}{d(x)}} }$$ is indeterminate of the type $$0/0$$ then

$$\displaystyle{ \lim_{x \to c}{\frac{n(x)}{d(x)}} = \lim_{x \to c}{\frac{n'(x)}{d'(x)}} }$$.

### Quick Notes

 This rule can be repeated until a determinate form is found. $$x$$ can approach a finite value $$c$$, $$\infty$$ or $$-\infty$$ For details about determinate and indeterminate forms, click here. This is not the quotient rule. For information about the difference, click here. Do not use this on determinate forms. It will often give you the wrong answer.

L'Hôpital's Rule is used to evaluate a limit when other techniques will not work, like factoring and rationalizing. It especially convenient to use with exponentials and, sometimes, limits involving trig functions. In fact, L'Hôpital's Rule can usually be used on limits where those other techniques work and it is often easier than factoring. So it is a nice tool to have when evaluating limits.

Some Things To Notice

1. Be careful to use L'Hôpital's Rule only on limits of indeterminate form. If you use it on a determinate form, you may ( and probably will ) get an incorrect answer.
2. L'Hôpital's Rule can be used multiple times. So, if it doesn't work the first time, check that you still have an indeterminate form, and then use it again.
3. One of the big mistakes that students make when they are first learning L'Hôpital's Rule is to confuse it with the derivative of the function itself. Read this next section to clarify it in your mind.

### Difference Between L'Hôpital's Rule and the Quotient Rule

It is very easy to confuse L'Hôpital's Rule with the quotient rule since they are so similar. Here is a rundown of the two for comparison.

1. Quotient Rule - The quotient rule is used on a quotient or ratio of terms to calculate the derivative of a function. The result of the quotient rule is the slope of the original function at all points along the curve. So you can use the result to determine the slope, calculate the equation of a tangent line, find extrema and continue on to determine the second derivative, to name a few uses.
To calculate the quotient rule, we use the following equations. Given that we have a function in the form $$\displaystyle{ f(x) = \frac{n(x)}{d(x)}}$$, the derivative $$f(x)$$ using the quotient rule is $$\displaystyle{ f'(x) = \frac{d(x)n'(x) - n(x)d'(x)}{[d(x)]^2}}$$

2. L'Hôpital's Rule - L'Hôpital's Rule is used only when we want to calculate a limit and can be used only under very specific circumstances (see the discussion below for a complete explanation). L'Hôpital's Rule works like this.
If we have a limit that goes to an indeterminate form, for example $$\displaystyle{ \lim_{x \to c}{\left[ \frac{n(x)}{d(x)} \right]}}$$ where $$\displaystyle{ \lim_{x \to c}{[n(x)]} = 0}$$ and $$\displaystyle{ \lim_{x \to c}{[d(x)]} = 0}$$ giving us $$0/0$$ in the original limit, then L'Hôpital's Rule tells us that $$\displaystyle{ \lim_{x \to c}{\frac{n(x)}{d(x)}} = \lim_{x \to c}{\frac{n'(x)}{d'(x)}}}$$.

Okay, now let's compare the two using $$\displaystyle{ f(x) = \frac{n(x)}{d(x)}}$$

Quotient Rule L'Hôpital's Rule Context: Derivatives Context: Limits One of several ways to calculate the derivative of a function. Can be used only on indeterminate limits. The resulting equation has many uses including calculating the slope of a curve. The resulting fraction $$[n'(x)/d'(x)]$$ has no meaning other than its limit is the same as the limit of the original function.

Study Note - When you are learning two similar concepts or you are learning a new concept that is similar to one you already know and you are getting confused, try the technique we used above (setting them side-by-side and comparing and contrasting them). This will help you separate them in your mind and know when to use which technique. This also works for learning similar words in foreign languages. Click here for more study techniques.

If you haven't already, now would be a good time to study what is meant by an indeterminate form, since the following discussion requires you to know it. You can find the complete discussion on the indeterminate forms page.

### Indeterminate Forms Table

Determinate-Indeterminate Forms Table

Indeterminate Forms

Determinate Forms

$$0/0$$

$$\infty + \infty = \infty$$

$$\pm \infty / \pm \infty$$

$$- \infty - \infty = - \infty$$

$$\infty - \infty$$

$$0^{\infty} = 0$$

$$0 (\infty)$$

$$0^{-\infty} = \infty$$

$$0^0$$

$$(\infty) \cdot (\infty) = \infty$$

$$1^{\infty}$$

$$\infty ^ 0$$

Use L'Hôpital's Rule

Do Not Use L'Hôpital's Rule

Types of Indeterminate Forms

Notice in the L'Hôpital's Rule equation, we require the limit equation to be in a fraction. This is a strict requirement that cannot be broken. If the limit equation is not a fraction, we cannot use L'Hôpital's Rule. This section contains a discussion of the four main types of equations that you will run across where you need to determine if the limit is an indeterminate form. (This list is also found on the indeterminate forms page. )

 Case 1 - - Indeterminate Quotients In a perfect world, all your problems would be in this form because you can just apply L'Hôpital's Rule directly. These are problems where you already have a limit in the form $$\displaystyle{ \lim_{x \to a}{\frac{n(x)}{d(x)}} }$$ And, when you plug in $$x=a$$, you get either $$0/0$$ or $$\pm \infty / \pm \infty$$. In the last indeterminate form with $$\infty$$, the signs can be anything, so you have 4 possible cases. In any case, your problem is all set up to use L'Hôpital's Rule and you just directly apply the rule. Case 2 - - Indeterminate Products In this case, you have an determinate form that looks like $$0(\infty)$$ or $$0(-\infty)$$, in which case you just need to do a little bit of algebra to convert one of the pieces of the product into a denominator and you end up with case 1. This sometimes takes some thinking. For example, if you have $$\displaystyle{ \lim_{x \to 0^+}{x\ln(x)} }$$, you can rewrite $$x$$ as $$1/x^{-1}$$ and the limit now is $$\displaystyle{ \lim_{x \to 0^+}{\ln(x)/x^{-1}} }$$ where $$n(x) = \ln(x)$$ and $$d(x) = x^{-1}$$ and you have the indeterminate form $$\infty / \infty$$ and you can use L'Hôpital's Rule. Case 3 - - Indeterminate Differences In this case, you have the indeterminate form that looks like $$\infty - \infty$$ and we need a fraction to be able to apply L'Hôpital's Rule. To get the right form, just find a common a denominator and combine the terms to get an indeterminate form that looks like case 1. Case 4 - - Indeterminate Powers In the previous three cases, you either started with a fraction in the right form (case 1) or did some algebra to get your function into the right form (cases 2 and 3). In this case, you need another technique. The idea is that you have an indeterminate form that looks like $$0^0$$, $$1^{\infty}$$ or $$\infty ^ 0$$ and you need to get some kind of fraction in order to apply L'Hôpital's Rule. To do this we use the natural logarithm. Let's go through a general example. We have the limit $$\displaystyle{ \lim_{x \to a}{[f(x)]^{g(x)}} }$$ and $$[f(a)]^{g(a)}$$ is an indeterminate power. Set $$y = [f(x)]^{g(x)}$$ and take the natural log of both sides. So that $$\ln(y) = \ln( [f(x)]^{g(x)} )$$ and using the power rule for logarithms, we have $$\ln(y) = g(x) \ln[ f(x) ]$$. We take the limit of $$\ln(y)$$ using one cases 1-3 above to get a value $$L$$, which may be finite or infinite. To get the final answer, we 'undo' the natural log by putting $$L$$ in the exponent of $$e$$ giving $$e^L$$.

Okay, now that you know something about L'Hôpital's Rule, under what conditions you can use it and various situations you will encounter, let's watch some videos to help explain how it works and see some examples.

This first video is in-depth and a bit long but is well worth watching. The instructor is easy to listen to and he does a good job of explaining the details of L'Hôpital's Rule as well as giving examples and supporting arguments.

 MIT OCW - Lec 35 | MIT 18.01 Single Variable Calculus, Fall 2007

Here is a quick introductory explanation of L'Hôpital's Rule. This is good to watch if you are still a little fuzzy on how to use this rule. There is nothing new here but it gives you another perspective.

 Khan Academy - Introduction to L'Hopital's Rule

### Search 17Calculus

Practice Problems

Instructions - - Evaluate the following limits using L'Hôpital's Rule ( if valid ), giving your answers in exact form.

 Level A - Basic

Practice A01

$$\displaystyle{\lim_{x\to\pi/2}{\left[\frac{\tan(2x)}{x-\pi/2}\right]}}$$

solution

Practice A02

$$\displaystyle{\lim_{x\to2}{\left[\frac{3x^2-x-10}{x^2-4}\right]}}$$

solution

Practice A03

$$\displaystyle{\lim_{x\to3}{\left[\frac{x^4-81}{2x^2-5x-3}\right]}}$$

solution

Practice A04

$$\displaystyle{\lim_{x\to27}{\left[\frac{x-27}{x^{1/3}-3}\right]}}$$

solution

Practice A05

$$\displaystyle{\lim_{x\to1}{\left[\frac{x^{1/3}-1}{x^{1/4}-1}\right]}}$$

solution

Practice A06

$$\displaystyle{\lim_{x\to0}{\left[\frac{x+\tan(x)}{\sin(x)}\right]}}$$

solution

Practice A07

$$\displaystyle{\lim_{z\to\pi}{\left[\frac{z-\pi}{\sin(z)}\right]}}$$

solution

Practice A08

$$\displaystyle{\lim_{y\to0}{\left[\frac{\sin(3y)}{\sin(4y)}\right]}}$$

solution

Practice A09

$$\displaystyle{\lim_{x\to-\infty}{\left[\frac{x+5}{3x+7}\right]}}$$

solution

Practice A10

$$\displaystyle{\lim_{x\to\infty}{\left[\frac{7x^3+x+12}{2x^3-5x}\right]}}$$

solution

Practice A11

$$\displaystyle{\lim_{x\to\infty}{\left[\frac{7x^2-3x+12}{x^3+4x+127}\right]}}$$

solution

Practice A12

$$\displaystyle{\lim_{x\to-\infty}{\left[\frac{7x^2+x+21}{11-x}\right]}}$$

solution

Practice A13

$$\displaystyle{\lim_{x\to0}{\frac{\sin(x)}{x}}}$$

solution

Practice A14

$$\displaystyle{\lim_{x\to\infty}{\frac{\ln(x)}{x}}}$$

solution

Practice A15

$$\displaystyle{\lim_{x\to0^+}{\frac{\ln(x)}{x}}}$$

solution

Practice A16

$$\displaystyle{\lim_{x\to\infty}{\frac{x}{\ln(1+2e^x)}}}$$

solution

Practice A17

$$\displaystyle{\lim_{x\to\infty}{(x^2-x)}}$$

solution

 Level B - Intermediate

Practice B01

$$\displaystyle{\lim_{x\to0}{\left[\frac{\tan(x)-x}{\sin(x)-x}\right]}}$$

solution

Practice B02

$$\displaystyle{\lim_{n\to\infty}{n^{2/n}}}$$

solution

Practice B03

$$\displaystyle{\lim_{x\to\infty}{\left[\frac{x^5}{e^{5x}}\right]}}$$

solution

Practice B04

$$\displaystyle{\lim_{n\to\infty}{\sqrt[n]{k}}}$$ where k is a positive constant

solution

Practice B05

$$\displaystyle{\lim_{n\to\infty}{\sqrt[n]{n^k}}}$$ where k is a positive constant

solution

Practice B06

$$\displaystyle{\lim_{n\to\infty}{\sqrt[n]{\ln(n)}}}$$

solution

Practice B07

$$\displaystyle{\lim_{x\to0^+}{\sqrt{x}\ln(x)}}$$

solution

Practice B08

$$\displaystyle{\lim_{x\to0}{\cot(2x)\sin(6x)}}$$

solution

Practice B09

$$\displaystyle{\lim_{x\to\infty}{(xe^{1/x}-x)}}$$

solution

Practice B10

$$\displaystyle{\lim_{x\to1}{\left[\frac{1}{\ln(x)}-\frac{1}{x-1}\right]}}$$

solution

Practice B11

$$\displaystyle{\lim_{x\to\infty}{(e^x+x)^{1/x}}}$$

solution

Practice B12

$$\displaystyle{\lim_{x\to0}{(\cos(3x))^{5/x}}}$$

solution

$$\displaystyle{ \lim_{x \to 0^-}{ \frac{-e^{1/x}}{x} } }$$