*Zach Pino*

# Applying L'Hopital's Rule in Complex Cases

## Quick Review of L'Hôpital's Rule

**L'Hôpital's rule** says that if you're trying to find a limit and you end up with something like 0/0 or infinity/infinity, you can try looking at the derivatives to calculate the limit. That is, the limit as *x* goes to *C* of *f(x)*/*g(x)*. If that gives you 0/0 or infinity/infinity, you can say that the limit that you were originally looking for is equal to the limit as *x* goes to *C* of *f'(x)*/*g'(x)*.

## Example of L'Hôpital's Rule in Complex Cases

Sometimes though, L'Hôpital's rule doesn't quite give you what you're looking for. Like in the case of the limit as *x* goes to zero of (1 - cos(*x*)) / *x*^2. Now at *x*=0, I have zero on the top and zero on the bottom, so this tells me that I should use L'Hôpital's rule. I do this, and I find that the limit as *x* goes to zero of (1 - cos(*x*)) / *x*^2 is the same as the limit as *x* goes to zero of sin(*x*) / 2*x*, because sin(*x*) is the derivative of 1 - cos(*x*), and 2*x* is the derivative of *x*^2. Here's the problem though; the limit as *x* goes to zero of sin(*x*) is 0, and the limit as *x* goes to zero of 2*x* is 0. Hmm, well, this is a problem. We use L'Hôpital's rule to avoid this in the first place. Now what?

Well, sin(*x*) / 2*x* is really just another function, right? So we can use L'Hôpital's rule on this. Now my new *f(x)* is sin(*x*), and my new *g(x)* is 2*x*. If I use that in L'Hôpital's rule, then I say the limit as *x* goes to zero of sin(*x*) / 2*x* is equal to the limit as *x* goes to zero of cos(*x*) - which is the derivative of sin(*x*) - divided by 2, which is the derivative of 2*x*. Finally, as *x* goes to zero, cos(*x*) goes to 1, and as *x* goes to zero, 2 stays 2. So I finally have a limit that's not 0/0. What I end up with is the limit as *x* goes to zero of (1 - cos(*x*)) / *x*^2 is equal to 1/2, and I found this just by repeatedly applying L'Hôpital's rule.

## Second Example

So we see this kind of thing a lot when we're looking at things like trigonometric functions and polynomials. Like the example limit as *x* goes to zero of (*x*^3 - *x*^2) / (2*x*^2). Now at *x*=0, both the top and the bottom are zero, so this limit is 0/0. I apply L'Hôpital's rule using the top - *f(x)*=*x*^3 - *x*^2 - and the bottom, *g(x)*, being 2*x*^2. I find the derivatives of those, and I plug them in. I end up with the limit as *x* goes to zero of (3*x*^2 - 2*x*) / 4*x*. Well, once again this limit is giving me 0/0, so I'm going to apply L'Hôpital's rule again. My new *f(x)* is 3*x*^2 - 2*x* from the top here, and my new *g(x)* is 4*x*. I find the derivatives, I plug them in, and I get the limit as *x* goes to zero of (6*x* - 2)/4. Well that limit exists, and that limit is -2/4, or -1/2. So from all this, these repeated applications of L'Hôpital's rule, I find that the limit as *x* goes to zero of (*x*^3 - *x*^2) / (2*x*^2) is -1/2.

## Third Example

Let's look at another case, like the limit as *x* goes to zero of *x*(1 - cos(*x*)) / (*x* - sin(*x*)). At *x* goes to zero, I have the limit of 0/0, so I use L'Hôpital's rule. Okay, so taking the derivative of the top, I get first, *x*, times the derivative of the second, which is sin(*x*), plus the derivative of the first (which is just 1) times the second, 1 - cos(*x*). I take the derivative of the bottom, and I end up with the derivative of *x*, which is 1, minus the derivative of sin(*x*), which is cos(*x*). All right, well as *x* goes to zero, the top is zero and the bottom is zero. Well let's use L'Hôpital's rule again. So the derivative at the top is going to be the derivative of *x*(sin(x)), so first times the derivative of the second plus the second times the derivative of the first, which is 1, plus the derivative of 1 - cos(*x*), which is just sin(*x*). I'm going to divide that by the derivative of 1 - cos(*x*), which is sin(*x*).

Because I have two sin(*x*)s on the top, I'm going to collect terms and write this as the limit as *x* goes to zero of (*x*(cos(*x*)) + 2sin(*x*)) / sin(*x*). Alright, here we go; we've used L'Hôpital's rule twice, the limit of the top is 0 + 0 - uh oh - over sin(*x*): 0. Huh. Well let's try using L'Hôpital's rule again! The derivative of *x*(cos(*x*) is -*x*(sin(*x*)) + cos(*x*). The derivative of 2sin(*x*) is 2cos(*x*), and I'm dividing that by the derivative of sin(*x*), or cos(*x*). Again, I'm going to simplify terms because I have a lot of cosines on the top, and I end up with (3cos(*x*) - *x*sin(*x*)) / cos(*x*). All right, what's the limit of this as *x* goes to zero? Well this first term goes to 3, this goes to 0 and cos(*x*) goes to 1. Ah, finally! 3/1, or just 3. So the limit as *x* goes to zero of (*x*(1 - cos(*x*)) / (*x* - sin(*x*)) is 3. We just kept applying L'Hôpital's rule every time we saw 0/0.

## Lesson Summary

Let's review. **L'Hôpital's rule** says that the limit as *x* goes to *C* of *f(x)*/*g(x)* is equal to the limit as *x* goes to *C* of *f'(x)*/*g'(x)* as long as your original limit gave to 0/0 or infinity/infinity. Now if your new limit gives you 0/0 or infinity/infinity, you can keep applying L'Hôpital's rule until you get something that makes a little more sense, like 0/3 or 1/2.

To unlock this lesson you must be a Study.com Member.

Create your account

### Register to view this lesson

### Unlock Your Education

#### See for yourself why 30 million people use Study.com

##### Become a Study.com member and start learning now.

Become a MemberAlready a member? Log In

Back