How to evaluate $\lim_{x\to 0} \frac{x^2\sin {\frac{1}{x}}}{\sin x}$ 
Find the value of $\lim_{x\to 0} \dfrac{x^2\sin {\dfrac{1}{x}}}{\sin x}$.

Below I am showing my attempt at the question:
$x^2\sin {\dfrac{1}{x}}\to 0$ as $x\to 0$ since $\sin {\dfrac{1}{x}}$ is bounded in a neighbourhood of $0$ and $x^2\to 0$ as $x\to 0$.
Hence , we have a $\dfrac{0}{0}$ form.
By L'Hospital's Rule ,$\lim_{x\to 0} \dfrac{x^2\sin {\dfrac{1}{x}}}{\sin x}$ reduces to  $\dfrac{2x\sin {\dfrac{1}{x}}-\cos{\dfrac{1}{x}}}{\cos x}$ whose limit can't be computed at $x=0$ since $\lim _{x\to 0} \cos{\dfrac{1}{x}}$ does not exist.
How can I evaluate this correctly?
Will the answer be limit does not exist?
Do excuse me if I am unable to post a good question as this is my first question on MSE
 A: Hint:
The required limit can be written as: $$\lim_{x \to 0} \frac{\sin \frac{1}{x}}{\frac{1}{x}} \times \frac{x}{\sin x} = 0 \times 1 =0$$
A: Just observe that
$$
\left|x^2\sin {\dfrac{1}{x}}\right|\le x^2
$$ giving, as $x \to 0$,
$$
\left|\dfrac{x^2\sin {\dfrac{1}{x}}}{\sin x}\right|\le \frac{x^2}{\left|\sin x\right|}=\left|\frac{x}{\sin x}\right|\times|x|\to 1\times0=0.
$$
A: Note that:
$$-\frac{x^2}{\sin(x)} \leq \frac{x^2}{\sin x} \sin(\frac{1}{x}) \leq \frac{x^2}{\sin x}$$
So by the squeezing theorem, the limit is $0$.
A: Bounding the numerator in the way pointed out already
$$
|x^2\sin(1/x)|\leq x^2
$$
and Taylor expanding the denominator yields
$$
-\frac{x^2}{x+O(x^3)}<\frac{x^2\sin(1/x)}{\sin x}<\frac{x^2}{x+O(x^3)}
$$
and you may conclude that the limit is zero by the squeeze theorem.
A: $$\begin{align}L&=\lim_\limits{x\to0}\dfrac{x^2\sin\dfrac1x}{\sin x}\\&=\lim_\limits{x\to0}\dfrac x{\sin x}\cdot\lim_\limits{x\to0}x\sin\left(\dfrac1x\right)\\&=1\cdot\lim_\limits{x\to0}x\sin\left(\dfrac1x\right)\\\hline\text{We know}\\-1&\le\sin\left(\dfrac1x\right)\le1\\-x&\le x\sin\left(\dfrac1x\right)\le x\\\text{Since}\\\lim_\limits{x\to0}(-x)&=0=\lim_\limits{x\to0}x\\\text{Hence}\\&\lim_\limits{x\to0} x\sin\left(\dfrac1x\right)=0\end{align}$$
Using this result in $L$, we have $L=0$
