<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] The function in question is even and hence it is sufficient to deal with $x\to 0^{+}$. And then we have the famous inequality $$\sin x<x<\tan x\tag{1}$$ so that $$\frac{\tan 2x}{\sin x} =\frac{\tan 2x}{2x}\cdot 2\cdot\frac{x}{\sin x} > 1\cdot 2\cdot 1=2$$ And clearly from the above equation it is evident that $(\tan 2x)/\sin x\to 2$ so that as $x\to 0^{+}$ we have $(\tan 2x)/\sin x<5/2$. It follows that the function $[(\tan 2x)/\sin x] $ is constant with value $2$ as $x\to 0^{+}$ and hence the desired limit is $2$.
Your approach is also fine and you may do without inequality $(1)$. You just have to note that as $x\to 0^{+}$ we have $x<2x$ and $\cos$ is decreasing so that $\cos x>\cos 2x$ and then $$\frac{2\cos x} {\cos 2x}>2$$ and the limit of the fraction is $2$ so that it is less than $5/2$ as $x\to 0^{+}$. And as before the desired limit is $2$. The question is much simpler than it appears. 
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