What is the value of $\lim _{x\to 0}\left\lfloor\frac{\tan x \sin x}{x^2}\right\rfloor$ How do I evaluate the limit
$$\lim _{x\to 0}\left\lfloor\frac{\tan x \sin x}{x^2}\right\rfloor$$ 
where $\lfloor\cdot\rfloor$ denotes greatest integer function.
I know that $x>\sin x$ and $x < \tan x$ but how do I use these results here?
$\tan x \over x$ tends to $1+$ whereas $\sin x \over x$ tends to $1-$ 

Using Taylor–Young expansions: we know that
$$\tan(x)=x+\frac{x^3}3+o(x^4),\qquad\sin(x)=x-\frac{x^3}6+o(x^4),$$
hence
$$\tan(x)\sin(x)=x^2+\frac{x^4}6+o(x^5),$$
and hence
$$\frac{\tan(x)\sin(x)}{x^2}=1+\frac{x^2}6+o(x^3).$$
From here, we conclude that there exists a punctured neighborhood $V$ of $0$ such that
$$\forall x\in V,\ \frac{\tan(x)\sin(x)}{x^2}>1.$$
Since
$$\lim_{x\to0}\frac{\tan(x)\sin(x)}{x^2}=1$$
we can assume that $V$ has been chosen such that
$$\forall x\in V,\ 1<\frac{\tan(x)\sin(x)}{x^2}<2.$$
Hence
$$\forall x\in V,\ \left\lfloor\frac{\tan(x)\sin(x)}{x^2}\right\rfloor=1,$$
from which we conclude that
$$\lim_{x\to0}\left\lfloor\frac{\tan(x)\sin(x)}{x^2}\right\rfloor=1.$$
