Finding $\lim_{x\rightarrow 0}\lfloor \frac{2017 \sin x}{x} \rfloor +\lfloor \frac{2017 \tan x}{x} \rfloor $ Finding value of $\displaystyle \lim_{x\rightarrow 0}\bigg\lfloor \frac{2017 \sin x}{x}\bigg \rfloor +\bigg\lfloor \frac{2017 \tan x}{x}\bigg \rfloor,$ where $\lfloor x \rfloor $ is floor function of $x$
Attempt as we know $\sin x< x < \tan x$  
wan,t  be able to go further, could some help me, thanks
 A: Yes, the key is that $\sin x<x<\tan x$ for $x\to0^{+}$, which implies $\displaystyle \frac{\sin x}{x}<1<\frac{\tan x}{x}$. Since after division both expressions are even functions, this double inequality is true both as $x\to0^{+}$ and as $x\to0^{-}$. We also know that both $\displaystyle \lim_{x\to0}\frac{\sin x}{x}=1$ and $\displaystyle \lim_{x\to0}\frac{\tan x}{x}=1$, but due to the inequalities, $\sin(x)/x$ approaches $1$ from below, while $\tan(x)/x$ approaches $1$ from above.
Since $\displaystyle \frac{\sin x}{x}<1$ and $\displaystyle \lim_{x\to0}\frac{\sin x}{x}=1$, for $x$ sufficiently close to $0$ we have:
$$1-\frac{1}{2017}<\frac{\sin x}{x}<1 \implies 2016<\frac{2017\sin x}{x}<2017 \implies \left\lfloor\frac{2017\sin x}{x}\right\rfloor=2016.$$
Similarly, since $\displaystyle \frac{\tan x}{x}>1$ and $\displaystyle \lim_{x\to0}\frac{\tan x}{x}=1$, for $x$ sufficiently close to $0$ we have:
$$1<\frac{\tan x}{x}<1+\frac{1}{2017} \implies 2017<\frac{2017\tan x}{x}<2018 \implies \left\lfloor\frac{2017\tan x}{x}\right\rfloor=2017.$$
