While practicing for my high school calculus exam, I went through the following limit problem: $$\lim_{x \to 0}x\cdot \sqrt{\cos{\frac{1}{x}}}$$
We haven't covered any similar example, and although we learned about the L'hopital's rule, our teacher says that we are not supposed to use for this exam..
Hint:
For real $x,$ $$\left|\cos\dfrac1x\right|\le1$$
and for $x\to0,\dfrac1x\to+\infty,\cos\dfrac1x$ may not be $\ge0$
$$\min(x,0)\le x\sqrt{\cos\frac1x}\le \max(x,0)$$ squeezes to $0$.
Notice that the fact that $\sqrt{\cos\dfrac1x}$ is not defined everywhere does not matter.
The product of a sequence with limit zero and a bounded sequence has limit zero. Here $\sqrt{cos(1/x)}$ is just a red, bounded herring.
observe that since $\sqrt {\cos(x)} \leq 1$
$0\leq x\sqrt {\cos(\frac{1}{x})}\leq x$.then by sandwich theorem,
$\lim \limits _{x\to 0}x\sqrt {\cos(\frac{1}{x})}=0$
