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.
 A: So first of all, note that
$$
\lim_{x\to 0}\frac{x^2}{\sin x \tan x}=\dots=
\left(\lim_{x\to 0}\frac{x}{\sin x}\right)^2
\left(\lim_{x\to 0}\cos(x)\right)=1
$$
Now, this doesn't necessarily tell us anything about the original function, since the greatest integer function is not continuous. So, we actually have three possibilities:


*

*if $x^2\geq\sin x \tan x$ in a sufficiently small neighborhood of $x=0$, then the limit is $1$.

*if $x^2<\sin x \tan x$ in a sufficiently small neighborhood of $x=0$, then the limit is $0$.

*if neither of the above is true, the limit does not exist.


So, we must find out: which is it?

Here's an attempt to show that the limit becomes $0$:
$$
\begin{align}
x^2 &< \sin x \tan x\\
x^2\cos x &< \sin^2 x \\
x^2\cos x &<\frac12 (1-\cos(2x))\\
2x^2\cos x &< (1-\cos(2x))\\
2x^2\cos x + \cos(2x) &< 1\\
\end{align}
$$
From there, it might be possible to make an argument using Taylor series.
A: Since you already know how get the limit without the floor function, I will try to prove that inequality without Taylor series.
$$x^2<\sin x \tan x \quad as \; x \to 0$$
I made the substitution $x \to \arctan x$ . 
$\arctan^2 x<x\sin (\arctan x)$
$\arctan x < \large \frac{x}{(x^2+1)^{\frac 14}}$
There are two functions $f(x)$ and $g(x)$ . $f(0)=g(0)$ . If $f'(x)>g'(x)$ on the interval $(0, a)$ , then that implies that $f(x)>g(x)$ on the interval $(0, a)$ . Therefore if $RHS'>LHS'$ , then $RHS>LHS$ . 
$\large \large \large \frac {1}{x^2+1} <\frac {x^2+2}{2(x^2+1)^{\frac 54}}$
$1<\large \frac {x^2+2}{2(x^2+1)^{\frac 14}}$ 
Using standard techniques (such as first derivative test) we can show that the $RHS$ has a minimum at $(0, 1)$ so we have proved the inequality. Hope this helps!
A: 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.
A: Consider the function
$$f(x) = \frac{x^2\cos(x)}{\sin^2(x)}$$
Then we have that
$$f(x+h)\approx \cos(x) < 1$$
for every $h\neq 0$ with $|h|$ small enough. Thus the limit converges to $0$.

We want to show that $f(x)<1$ for $|x|\neq0$ small enough. We will do so by showing that $g(x) = \sin^2(x)>x^2\cos(x)=h(x)$. The two functions agree in $0$, thus it is enough to show that $g'(x)>h'(x)$ for $x>0$ (we don't have to consider $x<0$, since both $g$ and $h$ are even functions).
$$g'(x) = 2\sin(x)\cos(x)$$
$$h'(x) = x\cos(x)-x^2\sin(x)$$
Since both agree in $0$ (again) we can consider the next derivative:
$$g''(x) = 2(\cos^2(x) - \sin^2(x))$$
$$h''(x) = \cos(x) - x(3\sin(x) -x\cos(x))$$
Notice that $g''(0)=2>1=h''(0)$, thus by continuity of both functions there is a neighborhood of $0$ where $g''>h''$, and we're done.
A: First let us investigate the limit without the floor function:
We already know that
$$\lim_{x\to 0} \frac{\sin(x)}{x}= \lim_{x\to 0} \frac{x}{\sin(x)}=1$$
So we can easily rewrite
$$\lim_{x\to 0} \frac{x^2}{\sin(x)\tan(x)}$$
as
$$\lim_{x\to 0} \left(\frac{x}{\sin(x)}\right)^2\left(\frac{\cos(x)}{1}\right)$$
$$= \lim_{x\to 0} \left(\frac{\cos(x)}{1}\right)=1$$
Case 1: If $0<x<\frac{\pi}{6}$ then $x>\sin(x)$
now let
$$f(x)=x^2-\sin(x)\tan(x)$$
Bear with me as this is slightly long, I used symbolab to calculate the $n^{th}$ derivatives of $f(x)$
$$f'(x)=2x-\sin(x)(1+\sec^2(x))$$
$$f''(x)=2-\cos(x)-\sec(x)-2\sec^2(x)\tan(x)\sin(x)$$
$$f'''(x)=\sin(x)-sec(x)\tan(x)-2[\alpha]$$
where $\alpha$ is a positive expression when $x\in [0,\frac{\pi}{6}]$.
Clearly, and since  $\sec(x)>1$ over this interval,
$$\sin(x)-\sec(x)\tan(x)<\tan(x)-\sec(x)\tan(x)<0$$
which means that $f'''(x)<0$ and that $f''(x)$ is decreasing.
Now since $f''(0)=0$ and $f''(x)$ is decreasing over this interval, then $f''(x)<0$ which means that $f'(x)$ is decreasing
$f'(0)=0$ and $f'(x)$ is decreasing means that $f'(x)<0$ and, hence, $f(x)$ is decreasing
last but not least, knowing that $f(x)=0$ and $f(x)$ is decreasing lead us to conclude that $f(x)<0$
$$i.e.\;\; x^2<\sin(x)\tan(x)$$
$$\Rightarrow \frac{x^2}{\sin(x)\tan(x)}<1$$
And, hence, for a sufficiently small $x>0$
$$\left\lfloor \left(\frac{x}{\sin(x)}\right)^2\left(\frac{\cos(x)}{1}\right)\right\rfloor=0$$
Finally, and given that the function in question is an even function
i.e.
$$\left(\frac{-x}{\sin(-x)}\right)^2\left(\frac{\cos(-x)}{1}\right)= \left(\frac{x}{\sin(x)}\right)^2\left(\frac{\cos(x)}{1}\right)$$
We conclude that
$$\lim_{x\to 0} \left\lfloor \left(\frac{x}{\sin(x)}\right)^2\left(\frac{\cos(x)}{1}\right)\right\rfloor=0$$
A: 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 $$\lim_{x\to 0} [f(x)]=0.$$
