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 

$$ - \tan^{-1}(x) \le  \tan^{-1}(x) \sin (\frac 1x ) \le \tan^{-1}(x) $$
and $$\lim _{x \to 0} \tan^{-1}(x) = 0 $$
