Why can we not use L'Hopitals' rule to prove $\lim_{x \to 0} \dfrac{x^2 \sin \frac{1}{x}}{\sin x}=0$? This is a problem in Schaum's outline -- Advanced Calculus, page 89, problem 4.84.
Why we cannot use L'Hopitals' rule to prove $$\lim_{x \to 0} \dfrac{x^2 \sin \dfrac{1}{x}}{\sin x}=0\;?$$
I thought the numerator $x^2 \sin \dfrac{1}{x}$ goes to $0$ as $x$ goes to $0$,
since $x^2$ goes to 0 and $\sin \dfrac{1}{x}$is bounded. And the denominator goes to 0 too. But why we can not use L'Hopitals' rule here?
 A: L'Hopitals says that if $\lim\limits_{x\to a}\frac{f(x)}{g(x)} $ is of the form
$\frac{\infty}{\infty}$ or $\frac{0}{0}$ and if 
 $$\lim\limits_{x\to a}\frac{f^{\prime}(x)}{g^{\prime}(x)} $$ exists then the first limit exits and the two limits are equal.
In this case we have the form $\frac{0}{0}$. OK. But now we find that 
$$\frac{f^{\prime}(x)}{g^{\prime}(x)} =\frac{2x\sin\frac{1}{x}-\cos \frac{1}{x}}{\cos x}$$ and this limit does not exist, thus nothing can be concluded.
A: Take the derivative of the numerator:
$$
2x\sin\frac{1}{x}+x^2\left[\cos\frac{1}{x}\right]\cdot\left(-\frac{1}{x^2}\right).
$$
This would not get you anywhere if you want to apply L'Hopital directly, because the assumptions are not satisfied. 
For the sake of calculation, note that
$$
\lim_{x \to 0} \dfrac{x^2 \sin \dfrac{1}{x}}{\sin x}=
\lim_{x \to 0} \left(x\cdot\frac{x}{\sin x}\cdot\sin\frac{1}{x}\right).
$$
Also, you could try "simpler problems" first:


*

*Do you know what is the limit $$\lim_{x\to 0}\frac{x}{\sin x}?$$

*Can you show that $$\lim_{x \to 0} \left(x \cdot\sin\frac{1}{x}\right)=0?$$



For the part of calculating $\lim_{x\to 0}\frac{x}{\sin x}$ above, you can use L'Hopital rule. 
