Calculus (3rd Edition)

Since $-1\leq \sin \frac{1}{x^2}\leq 1$, then we have $$-x\leq x\sin\frac{1}{x^2}\leq x.$$ Moreover, $\lim\limits_{x \to 0}x=\lim\limits_{x \to 0}-x=0$, then by the Squeeze Theorem, we have $$\lim\limits_{x \to 0}x\sin \frac{1}{x^2}=0.$$