Evaluation of limit with confusing moment Evaluate the following limit:
$\lim \limits_{x\to 0}\dfrac{2\cos ^{-3}x\cdot\sin x}{2\sin x+x\cdot \cos x}\color {\blue}{=}\lim \limits_{x\to 0}\dfrac{2\cos ^{-3}x}{2++x\cdot \frac{\cos x}{\sin x}}$ but $$\lim \limits_{x\to 0}x\cdot \frac{\cos x}{\sin x}=1$$ then $\lim \limits_{x\to 0}\dfrac{2\cos ^{-3}x}{2++x\cdot \frac{\cos x}{\sin x}}=\frac{2}{3}.$ Right?
But I have one question. In the second equality I divide enumerator and denominator to $\sin x$. Does it legal? What about if $\sin x=0$? Can anyone explain it to me please?

Since, $x\ne 0$ hence, one can also cancel $x$ as factor in numerator & denominator   $$\lim_{x\to 0}\frac{\cos^{-3}x\cdot \sin x}{2\sin x+x\cdot \cos x}$$
$$=\lim_{x\to 0}\frac{\cos^{-3}x\cdot \frac{\sin x}{x}}{2\frac{\sin x}{x}+\cos x}$$
$$=\lim_{x\to 0}\frac{\frac{\sin x}{x}}{2\frac{\sin x}{x}+\cos x}\left(\frac{1}{\cos x}\right)^3$$
$$=\frac{1}{2\cdot 1+1}\left(1\right)^3=\color{red}{\frac{2}{3}}$$
