# 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

• For small values of $x$, $\sin x \approx x$ and $\tan x \approx x$. Jan 18, 2017 at 17:01
• means $\displaystyle \frac{\sin x}{x}<1$ and $\displaystyle \frac{\tan x}{x}>1$, but i did not understand what is approx value of theses
– DXT
Jan 18, 2017 at 17:06
• I wonder why wolfram alpha gives 4033 not 4034. Eager to know the answer of this question.
– user312097
Jan 18, 2017 at 17:06
• means one value is $2016$ and other is $2017,$ i did not understand how can i calculate it
– DXT
Jan 18, 2017 at 17:07
• The sin x part will be 2016, I think.
– user312097
Jan 18, 2017 at 17:08

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: $$1<\frac{\tan x}{x}<1+\frac{1}{2017} \implies 2017<\frac{2017\tan x}{x}<2018 \implies \left\lfloor\frac{2017\tan x}{x}\right\rfloor=2017.$$
• Thanks zipirovich, but i did not understand why $\displaystyle 1-\frac{1}{2017}<\frac{\sin x}{x}$ and $\displaystyle \frac{\tan x}{x}>1+\frac{1}{2017},$ explain me
• @DURGESHTIWARI: Because of the limit. Think of the $\varepsilon$-$\delta$ definition of limits. Pick $\varepsilon=\frac{1}{2017}$. Then there exists a $\delta>0$ such as as long as $0<|x|<\delta$ (that's the "sufficiently close" part), $|\frac{\sin x}{x}-1|<\varepsilon=\frac{1}{2017}$, or equivalently $1-\frac{1}{2017}<\frac{\sin x}{x}<1+\frac{1}{2017}$. The right-hand side of this double inequality is irrelevant, though, because we know that $\sin(x)/x<1$. So we get that $1-\frac{1}{2017}<\frac{\sin x}{x}<1$. Jan 18, 2017 at 18:48