<s>[INST] 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?
 [/INST] Note that 
\begin{align}
\frac{2\cos x}{\cos 2x}-2&=\frac{2\cos x-2\cos 2x}{\cos 2x}\\
&=\frac{4\sin\frac{3x}{2}\sin\frac{x}{2}}{\cos 2x}\\
&>0
\end{align}
for $x\in(\frac{-\pi}{4},\frac{\pi}{4})$.
We also have
\begin{align}
3-\frac{2\cos x}{\cos 2x}&=\frac{3\cos 2x-2\cos x}{\cos 2x}\\
&=\frac{6\cos^2x-2\cos x-3}{\cos 2x}\\
&=\frac{36\cos^2x-12\cos x-18}{6\cos 2x}\\
&=\frac{(6\cos^2x-1)^2-19}{6\cos 2x}\\
&>0
\end{align}
for $x\in(-\alpha,\alpha)$, where $\alpha=\arccos(\frac{1+\sqrt{19}}{12})$
Note that $0<\alpha<\frac{\pi}{4}$.
For $x\in(-\alpha,\alpha)$, $\displaystyle\lfloor \frac{2\cos x}{\cos 2x}\rfloor=2$.
The limit is $2$.
</s>