# Valid L'Hopitals method for this limit?

1. Dec 11, 2011

### Silversonic

1. The problem statement, all variables and given/known data

Find the limit of

$\frac{x^{2}cos(1/x)}{sin(x)}$ as x tends to 0.

It doesn't actually say with L'Hopitals rule, but up until now that's what we've been dealing with.

3. The attempt at a solution

Wolfram alpha tells me the limit is zero. I know already how to compute the limit of $x^{2}cos(1/x)$ as x tends to 0, and this gives zero.

By L'Hopitals rule;

lim(x->0) $\frac{f(x)}{g(x)}$ = lim(x->0) $\frac{f'(x)}{g'(x)}$

Can I represent lim(x->0) $\frac{f(x)}{g(x)}$ as lim(x->0) f(x) * lim(x->0) $\frac{1}{g(x)}$?

And thus if I show f(x) tends to a value, and 1/g(x) tends to a value, then the whole thing tends to a value? Taking f(x) = $x^{2}cos(1/x)$ and 1/g(x) as $\frac{1}{sin(x)}$, then I can compute the limit of $\frac{1}{sin(x)}$ as the limit of $\frac{e^{x}}{e^{x}sin(x)}$. Using L'hopitals rule once on this gives me a limit of one. Thus the limit of the whole thing I've wanted is 0/1, so 0.

Is this correct?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Dec 11, 2011

### QuarkCharmer

Sure you can, it's a property of limits:

3. Dec 11, 2011

### vela

Staff Emeritus
As long as the limits exist, of course.

4. Dec 11, 2011

### Dick

I would think of it as (x/sin(x))*(x*cos(1/x)). I'd use l'Hopital for the first factor, if you don't already know it. The second factor isn't all that indeterminant, is it? Use another test on the second factor.