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 $1-$

• To dheeraj is your question is $\displaystyle \lim_{x\rightarrow 0}\lfloor \frac{\tan x\sin x}{x^2} \rfloor$, Where $\lfloor x \rfloor$ represent floor function of $x$ – juantheron Feb 18 '16 at 17:05
• Yah There is a greatest integer function. – Dhiraj Barnwal Feb 18 '16 at 17:06
• Then answer is $=1$ – juantheron Feb 18 '16 at 17:06
• How do we determine that ? As one quantity tries to reduce the value and the other tries to increase it. – Dhiraj Barnwal Feb 18 '16 at 17:10
• Using $f(x) = \tan x\sin x-x^2$ – juantheron Feb 18 '16 at 17:11

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