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 $0\leq \{x\} &lt; 1$

My approach : 

We know that $ \lim_{x\to 0}  \frac{\tan x}{x}=1$

and not getting any clue please suggest thanks.
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 $x&gt;0$

Now in $\displaystyle x\in \left(0,\frac{\pi}{2}\right)\;, \tan x&gt;x\Rightarrow \frac{ x}{\tan x }&lt;1$ and $\displaystyle \frac{ x}{\tan x }&gt;0$ 

So We get $$\lim_{x\rightarrow 0}\big\lfloor \frac{x}{\tan x}\big\rfloor =0$$
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 $0$.