Is the limit of the function is $0$? According to wolfram alpha,$$\lim _{x \to \ 0} \frac{\cos x}{\sin x} \cdot \sin\left(\frac 1 x\right)=0$$
In my question, the function is defined only in $(0,\infty)$
I want to prove it through the squeeze theorem.
I think that:
$$-x\le\frac{\cos x}{\sin x} \cdot \sin\left(\frac 1 x \right)\le x$$
And because $\lim_{x\to 0}-x=0,\lim_{x\to 0}-x=0$, according to the squeeze theorem, the limit equals $0$.
Is it correct?
P.S - the original question was about $\arctan x \cdot\sin\left(\frac 1 x \right) $ however through some calculations I wrote it as above.
Thanks,
Alan 

The easiest way to find $\displaystyle\lim_{x\to0} \left(\arctan x\cdot\sin \left( \frac 1 x\right)\right)$ is first to notice that $|\sin(\text{anything})|\le 1$ so that
$$
-|\arctan x|\le \arctan x\cdot\sin\left(\frac 1 x \right) \le |\arctan x|
$$
and then find $\lim\limits_{x\to0}\arctan x$.
