$\displaystyle \lim \limits_{x\rightarrow0} \sum \limits_{k=1}^{2013}\frac{\{x/\tan x+2k\}}{2013}$
RHL: Let $x=a+h, h>0$ and $a=0$
$\displaystyle \lim \limits_{h\rightarrow0} \sum \limits_{k=1}^{2013}\frac{\{h/\tan h+2k\}}{2013}=\lim \limits_{h\rightarrow0} \sum \limits_{k=1}^{2013}\frac{\{1+2k\}}{2013}$ which will always be zero since $k$ is an integer.
LHL: Let $x=a-h, h>0$ and $a=0$
$\displaystyle \lim \limits_{h\rightarrow0} \sum \limits_{k=1}^{2013}\frac{\{-h/\tan (-h)+2k\}}{2013}=\lim \limits_{h\rightarrow0} \sum \limits_{k=1}^{2013}\frac{\{-h/-\tan (h)+2k\}}{2013}=\lim \limits_{h\rightarrow0} \sum \limits_{k=1}^{2013}\frac{\{1+2k\}}{2013}$ which will always be zero since $k$ is an integer.
So the limit should be zero, but my answer is wrong. Where am I going wrong?
$$\displaystyle \lim \limits_{x\rightarrow0} \sum \limits_{k=1}^{2013}\frac{\{x/\tan x+2k\}}{2013}=\displaystyle \lim \limits_{x\rightarrow0} \sum \limits_{k=1}^{2013}\frac{x/\tan x+2k}{2013}-\displaystyle \lim \limits_{x\rightarrow0} \sum \limits_{k=1}^{2013}\frac{[x/\tan x+2k]}{2013}$$
$$S_1=\lim \limits_{x\rightarrow0}\frac{x}{\tan x}+2014$$
$$S_2=\displaystyle\sum \limits_{k=1}^{2013}\frac{2k}{2013}=2014\;\;\;(*)$$
The difference yields the result.
where
1)$[.]$ denotes the GIF function
2) $(*)$ follows since $2k$ is a, positive integer and $\lim_{x\rightarrow0}\frac{x}{\tan x}=1^-$