real analysis - How to evaluate $lim_{xto 0} frac{x^2sin {frac{1}{x}}}{sin x}$

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

Answer

Just observe that
$$\left|x^2\sin {\dfrac{1}{x}}\right|\le x^2$$ giving, as $x \to 0$,
$$\left|\dfrac{x^2\sin {\dfrac{1}{x}}}{\sin x}\right|\le \frac{x^2}{\left|\sin x\right|}=\left|\frac{x}{\sin x}\right|\times|x|\to 1\times0=0.$$