Double faced

From Maclaurin series of the functions $f(x)=\sin x$ and $f(x)=\tan x$ , we know that,

$x\to 0^+\implies \sin x\lt x\lt \tan x~,~\sin x, \tan x \approx x\\ x\to 0^-\implies \tan x\lt x\lt \sin x~,~\sin x, \tan x \approx x$

We make a simple substitution: $\sin x= x-\delta_1$ and $\tan x=x+\delta_2$ where $\delta_1,\delta_2\to 0^+$ . Using this substitution, the right hand limit can be evaluated as,

$\lim_{x\to 0^+}\left\lfloor \frac{x^2}{\sin x \tan x}\right\rfloor=\lim_{x\to 0^+}\left\lfloor \frac{x^2}{(x-\delta_1)(x+\delta_2)}\right\rfloor$

We see that the limit is now defined if we simply put $x=0$ in the function inside the floor. Also, in this case, commuting the floor and the limit doesn't affect the limiting value since the entire function is inside the floor and the limit is defined just by simple value plugging. Thus, we can write the right hand limit as,

$RHL=\left\lfloor \lim_{x\to 0} \frac{x^2}{(x-\delta_1)(x+\delta_2)}\right\rfloor=\left\lfloor \frac{0^2}{-\delta_1\delta_2}\right\rfloor = \lfloor 0 \rfloor = \boxed{0}$

For the left hand limit, we simply manipulate the limit using the trigonometric identities $\sin (-x)=-\sin x$ and $\tan (-x)=-\tan x$ and proceed as we did for evaluation of right hand limit.

$LHL=\left\lfloor \lim_{x\to 0} \frac{x^2}{(\delta_1-x)(-x-\delta_2)}\right\rfloor=\left\lfloor \frac{0^2}{-\delta_1\delta_2}\right\rfloor = \lfloor 0 \rfloor = \boxed{0}$

Both LHL and RHL agree and are equal to $\boxed{0}$ .

Note: In general, you cannot commute the limit and floor operator with each other.