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?
 A: $x\neq0$ 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, $$\lim_{x\rightarrow0}\left[\frac{2\cos{x}}{\cos2x}\right]=2$$
A: 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 $2\le f(x)<3$.
A: 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 $2$.
A: 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. 
