$\lim_{x \to 0} \frac{x^2 \sin \left(\frac{1}{x}\right)} {\sin x} \\=\frac{ \lim_{x \to 0} \frac{\sin \left(\frac{1}{x}\right)}{\left(\frac{1}{x}\right)}} {\lim_{x \to 0} \frac{ \sin x}{x}} \\=\frac{\lim_{ y \to \infty } \frac{\sin y} {y}}{1} \\= 0 (\because \sin y \leq 1, y \to \infty) .$