<s>[INST] Calculating $\lim_{x\to0} \left\lfloor\frac{x^2}{\sin x \tan x}\right\rfloor$ 
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.
 [/INST] Let $[.]$ denote GIF.
If $x \to 0$, then $\sin x=x-x^3/6+O(x^5)$ and $\tan x =x+x^3/3+O(x^4)$
then $$f(x)=\left(\frac{x^2}{\sin x \tan x}\right)=\left(\frac{x^2}{x^2(1-x^2/6+...)(1+x^2/3+...)}\right)=\left(\frac{1}{1+x^2/6-x^4/18}\right)$$
Using $(1+z)^k =1+kz+O(z^2)$, we can write
$$1-x^2/6 \le f(x) \le 1-x^2/6+x^4/18 \implies [1-x^2/6] \le [f(x)] \le [1-x^2/6+x^4/18].$$
Hence $$\lim_{x\to 0} [f(x)]=0.$$
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