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