Calculating the limit of $\frac{x^2\sin(\frac{1}{x})}{\sin x}$ $$ \lim _{x\to 0} \frac{x^2\sin(\frac{1}{x})}{\sin x}$$
Okay since sine is bounded ${x^2\sin(\frac{1}{x})} \ \ \to 0$
$\sin x\to  0$ Thus we can apply l'hospitals to it .
Applying l'Hospital's we get :
$$\lim_{x\to 0}\frac{\sin(\frac{1}{x})(1-2x)}{\cos x}$$ Here We can't find a way out? What would you advice me to do? Is there anyway to compute the limit?
Does this imply that the limit doesn't exist btw? No I don't think so, I think the conditions of the l'Hospital's are just not met.

You might know that $\lim_{x\to 0} \frac{x}{\sin(x)} = 1$ and that $\lim_{x\to 0} x \sin(1/x) = 0.$ 
