Valid L'Hopitals method for this limit?

[Silversonic]

Homework Statement

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.

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?

 

[QuarkCharmer]

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

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

 

[vela]

As long as the limits exist, of course.

 

[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.