(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Find the limit of

[itex]\frac{x^{2}cos(1/x)}{sin(x)}[/itex] 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 [itex]x^{2}cos(1/x)[/itex] as x tends to 0, and this gives zero.

By L'Hopitals rule;

lim(x->0) [itex]\frac{f(x)}{g(x)}[/itex] = lim(x->0) [itex]\frac{f'(x)}{g'(x)}[/itex]

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

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) = [itex]x^{2}cos(1/x)[/itex] and 1/g(x) as [itex]\frac{1}{sin(x)}[/itex], then I can compute the limit of [itex]\frac{1}{sin(x)}[/itex] as the limit of [itex]\frac{e^{x}}{e^{x}sin(x)}[/itex]. 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

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# Valid L'Hopitals method for this limit?

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