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

Let me continue where you stopped in your post.
Expand the denominator for a few terms and perform the long division. You should arrive to $$\frac{x^2}{\sin x\cdot \tan x} =1-\frac{x^2}{6}-\frac{7 x^4}{120}+O\left(x^5\right)$$  which is definitely smaller than $1$.
I am sure that you can take from here.
