Calculating $\lim_{x\to0} \left\lfloor\frac{x^2}{\sin x \tan x}\right\rfloor$

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.

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.

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.

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!