How to evaluate $\lim_{x\to 0} \frac{x^2\sin {\frac{1}{x}}}{\sin x}$

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

Hint:

The required limit can be written as: $$\lim_{x \to 0} \frac{\sin \frac{1}{x}}{\frac{1}{x}} \times \frac{x}{\sin x} = 0 \times 1 =0$$