Find $\lim_{x\to 0} \left\lfloor \frac{\tan 2x}{\sin x} \right\rfloor $ 
Find the limit
$$\lim_{x\to 0} \left\lfloor \dfrac{\tan 2x}{\sin x} \right\rfloor $$

My try:
$$ \tan 2x =\dfrac{\sin 2x}{\cos 2x}$$
$$\sin 2x =2\cos x\sin x$$
So:
$$\dfrac{\tan 2x}{\sin x}=\dfrac{2\cos x}{\cos 2x}$$
So:
$$\lim_{x\to 0} \left\lfloor \dfrac{\tan 2x}{\sin x} \right\rfloor=
\lim_{x\to 0}\left\lfloor\dfrac{2\cos x}{\cos 2x}\right\rfloor $$
Now what?

$x\neq0$ and around $0$ we have
$$\frac{2\cos{x}}{\cos2x}>2$$ because it's
$$\frac{2\cos{x}}{2\cos^2x-1}>2$$ or
$$(1-\cos{x})(1+2\cos{x})>0.$$
Thus, $$\lim_{x\rightarrow0}\left[\frac{2\cos{x}}{\cos2x}\right]=2$$
