# L'Hopitale's rule

1. Apr 19, 2012

### Hernaner28

This is just a question. We have:

$$\mathop {\lim }\limits_{x\to 0} \frac{{{x^2}\sin \frac{1}{x}}}{x}$$

Which is indeterminate of type 0/0 but we can apply L'Hopitale. Well, in fact we could (and must) cross out the x to solve the limit but let's assume that we forgot to do that and we derivate it so we get:

$$\mathop {\lim }\limits_{x\to 0} \frac{{2x\sin \frac{1}{x}-\cos \frac{1}{x}}}{1}$$

And the limit doesn't exist! Why is not L'Hopitale applicable in this case not crossing out the x's?
Thanks!

2. Apr 19, 2012

### PAllen

Because the derivative of the numerator doesn't have a limit as x->0. The function has a limit but not the derivative. The rule does not apply if the limits of the derivatives don't exist.

3. Apr 19, 2012

### Credulous

Hmm, well you know you can cancel the x to get: $$x \sin{\frac{1}{x}}.$$ From here it is just the sandwich theorem, we know $$-1 \leq \sin{\frac{1}{x}} \leq 1$$ so then $$\lim \limits_{x \to 0} -x \leq \lim \limits_{x \to 0} x \sin{\frac{1}{x}} \leq \lim \limits_{x \to 0} x$$

And you can take it from here.

4. Apr 19, 2012

### Dick

What you did wrong is doing a terrible job of differentiating the numerator. You have to use the chain rule.

5. Apr 20, 2012

### PAllen

No, I think the OP is correct, after simplification. (At a glance, I also thought they were wrong...but actually not).

6. Apr 20, 2012

### PAllen

It seemed to me the OP already knew this and referred to it. The question was why L'Hospital didn't work.

7. Apr 20, 2012

### Dick

Right you are.

8. Apr 20, 2012

### Citan Uzuki

L'hopital's rule says that whenever you have two functions f and g that each converge to zero, that if lim f'/g' exists, then lim f/g exists and the two limits are equal. It does not say that if lim f/g exists then lim f'/g' exists -- as you can see from your own example, it is entirely possible that the original limit exists and the limit of the derivatives does not.

9. Apr 20, 2012

### Hernaner28

This is what I wanted to know. So then, how can I realize that f'/g' doesn't exist but f/g does (let's keep assuming I forgot to cancel the x)?

THanks!

10. Apr 20, 2012

### Citan Uzuki

Unfortunately I can think of no simpler way to recognize this situation than to try to compute the limit of f'/g' directly, thus showing it doesn't exist, and then to use some alternate method (such as cancelling the x and applying the squeeze theorem) to show that the limit of f/g does.