<s>[INST] Evaluation of $ \lim_{x\to 0}\left\lfloor \frac{x^2}{\sin x\cdot \tan x}\right\rfloor$ Evaluation of $\displaystyle \lim_{x\to 0}\left\lfloor \frac{x^2}{\sin x\cdot \tan x}\right\rfloor$ where $\lfloor x \rfloor $ represent floor function of $x$.
My Try:: Here $\displaystyle f(x) = \frac{x^2}{\sin x\cdot \tan x}$ is an even function.
So we will calculate for $\displaystyle \lim_{x\to 0^{+}}\left\lfloor \frac{x^2}{\sin x\cdot \tan x}\right\rfloor$
Put $x=0+h$ where $h$ is a small positive quantity, and using series expansion
So limit convert into $\displaystyle \lim_{h\to 0}\left\lfloor \frac{h^2}{\sin h\cdot \tan h}\right\rfloor = \lim_{h\to 0}\left\lfloor \dfrac{h^2}{\left(h-\dfrac{h^3}{3!}+\dfrac{h^5}{5!}- \cdots\right)\cdot \left(h+\dfrac{h^3}{3}+\dfrac{2}{15}h^5+ \cdot\right)}\right\rfloor$
Now how can i solve after that, Help me 
Thanks
 [/INST] Hint?
$$
\lim_{x\to 0} \frac{x^2}{\sin x \cdot \tan x}
= \lim_{x\to 0} \cos x \left( \frac{x}{\sin x} \right)^2
$$
Now we can split $\lim ab=\lim a \cdot \lim b$, since both limits exists. Alternatively one can use L'ôpitals rule twice. $\lim_{x\to a} f/g = \lim_{x\to a} f'/g'$. Somewhat of a hazzle but it works
$$
\lim_{x\to 0} \frac{x^2}{\sin x \cdot \tan x}
= \lim_{x\to 0} \frac{2x}{\cos x \tan x + \sin x \sec^2x}
= \lim_{x\to 0} \frac{2}{\cos x + 1/\cos x + 2 \sin^2x \sec^2x}
$$
The rest now follows from $\sin 0 = 0$ and $\cos 0 = 1$.
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