<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] Let's consider
$$
f(x)=\frac{2\cos x}{\cos2x}
$$
over $(-\pi/2,\pi/2)$. Then
$$
f'(x)=2\frac{-\sin x\cos2x+2\cos x\sin2x}{\cos^22x}=
2\frac{\sin x(2\cos^2x+1)}{\cos^22x}
$$
Therefore $f$ has a local minimum at $0$ and $f(0)=2$.
In a suitable neighborhood of $0$ we have $2\le f(x)<3$.
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