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