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 $$\lim_{x\to0}\left\lfloor\frac{\tan(x)\sin(x)}{x^2}\right\rfloor=1.$$