TITLE: Calculating $\lim_{x\to0} \left\lfloor\frac{x^2}{\sin x \tan x}\right\rfloor$
QUESTION [17 upvotes]: Find $$\lim_{x\to0}  \left\lfloor\frac{x^2}{\sin x \tan x}\right\rfloor$$ where $\lfloor\cdot\rfloor$ is greatest integer function

I am a high school teacher. One of my students came up to ask this limit. 
For $\lfloor\frac{\sin x}{x}\rfloor$, I have used $\sin x > x$ using increasing decreasing functions.
I tried to prove $x^2 > \sin x \tan x$ using increasing /decreasing
 function but I am not getting it.

REPLY [11 votes]: Apply GM - HM to $\sin x$ and $\tan x$ (both positive for $x\geq 0$), we get that
$$ \sqrt{ \sin x \tan x } \geq \frac{2} { \frac{1} {\sin x} + \frac{ 1}{ \tan x} } = \frac{2 \sin x} { 1 + \cos x } = 2 \tan \frac{x}{2} \geq x$$
The only 'calc' that you need is the last inequality, though it has an easy geometric solution.
For $x < 0$, both $\sin x, \tan x$ have the same sign, and you can do the above with absolute values instead, no difference.