Find the value of $\lim_{x\to 0} \sum^{2013}_{k=1} \frac{\{\frac{x}{tanx}+2k\}}{2013}$... Problem : 
$$\lim_{x\to 0} \sum^{2013}_{k=1} \frac{\{\frac{x}{\tan x}+2k\}}{2013}$$
where $\{.\}$ represent the fraction part function $0\leq \{x\} < 1$
My approach : 
We know that $ \lim_{x\to 0}  \frac{\tan x}{x}=1$
and not getting any clue please suggest thanks. 
 A: Let $$l=\frac{1}{2013}\lim_{x\rightarrow 0}\sum^{2013}_{k=1}\left\{2k+\frac{x}{\tan x}\right\}$$
Now Using $\displaystyle \{x+n\} = \{x\}\;,$ Where $n\in \mathbb{N}$
So $$l=\frac{1}{2013}\sum^{2013}_{k=1}2k+\frac{1}{2013}\lim_{x\rightarrow 0}\sum^{2013}_{k=1}\left\{\frac{x}{\tan x}\right\}$$
So $$l=\frac{2}{2013}\cdot \frac{2013\cdot 2014}{2}+\lim_{x\rightarrow 0}\left\{\frac{x}{\tan x}\right\}$$
So $$l=2014+\lim_{x\rightarrow 0}\frac{x}{\tan x}-\lim_{x\rightarrow 0}\big\lfloor \frac{x}{\tan x}\big\rfloor$$
So $$l=2014+1-0 = 2015$$
Bcz  Here $\displaystyle \frac{\tan x}{x}$ is an even Function .
So we will calculate limit for $x>0$
Now in $\displaystyle x\in \left(0,\frac{\pi}{2}\right)\;, \tan x>x\Rightarrow \frac{ x}{\tan x }<1$ and $\displaystyle \frac{ x}{\tan x }>0$ 
So We get $$\lim_{x\rightarrow 0}\big\lfloor \frac{x}{\tan x}\big\rfloor =0$$
A: Let's try again. First off, we can immediately eliminate $2k$ from the fractional part function, so we're looking at the limit of the sum
$$\sum_{k=1}^{2013}\frac{\left\{\frac{x}{\tan x}\right\}}{2013}=\frac{1}{2013}\sum_{k=1}^{2013}\left\{\frac{x}{\tan x}\right\}=\frac{1}{2013}\cdot 2013\cdot\left\{\frac{x}{\tan x}\right\}=\left\{\frac{x}{\tan x}\right\},$$
since after we factor out $\frac{1}{2013}$ the terms of summation don't depend on $k$. Now we know that $\displaystyle \lim_{x\to0}\frac{x}{\tan x}=1$, but that's not enough to determine the fractional part. We also need to use the fact that $x<\tan x$ for $\displaystyle x\in\left(0,\frac{\pi}{2}\right)$, and thus $\displaystyle 0<\frac{x}{\tan x}<1$ there, so $\displaystyle \left\{\frac{x}{\tan x}\right\}=\frac{x}{\tan x}$, and the limit is 1 as $x\to0^{+}$. Similarly we can show that it's true from the left (or just refer to the fact that $\frac{x}{\tan x}$ is even).
Note that the answer would be different if we flip the fraction in the questions, i.e. with $\displaystyle \left\{\frac{\tan x}{x}\right\}$.
A: See the $2k$ is a natural number so you can directly remove it out from the problem . So now we have $\sum \frac{x/tan(x)}{2013}$ now the sum is indirectly $\sum 1=2013$ so $2013$ cancels out. Now we are left with ${\frac{x}{tan(x)}}$ now we know that $lim {x \to 0} tan(x)/x=1$ also we know ${y}=y-[y]$ so we can write fractional part as $\frac{x}{tan(x)}-[\frac{x}{tan(x)}]$ so the limit becomes $1-1=  0$ hence the limit is $0$.
