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.

L Hopital rule says that this is $$ \frac{(x^2\sin(\frac{1}{x}))'}{\cos x}$$ in cero, but the numerator has a discontinuity in the derivate, and the value must be calculated by definition:
$$\lim _{x\to 0} \frac{x^2\sin(\frac{1}{x})}{x}=\lim _{x\to 0} x\sin(\frac{1}{x})=0$$
So the limit is cero.
Another way:
$$\lim _{x\to 0} \frac{x^2\sin(\frac{1}{x})}{\sin x}=\lim _{x\to 0}\frac{x}{sin(x)} x\sin(\frac{1}{x})=1*\lim _{x\to 0} x\sin(\frac{1}{x})=0$$
