<s>[INST] Finding limit for given function 
If $f(x)=\lim\limits_{n \to \infty} \frac{x}{n} \left(\frac{1}{1+e^{-x/n}}+\frac{1}{1+e^{-2x/n}}+\dots+\frac{1}{1+e^{-x}} \right)$
  then what is  $\lim\limits_{x\to 0} \left(\frac{2f(x)-x}{x^2}\right)$?

My solution is : 
$f(x)=0$ since $\lim\limits_{n \to \infty} \frac{x}{n}$ part will be zero for $n=\infty$, making entire term $0$.
then $\lim\limits_{x\to 0} \left(\frac{2f(x)-x}{x^2}\right)$ will become $\lim\limits_{x\to 0} \left(\frac{-x}{x^2}\right)$ or $\lim\limits_{x\to 0} \left(\frac{-1}{x}\right)$ = $\infty$.  
This answer is not avaiable in the given options. What should be the correct answer for this questions?
 [/INST] Your answer is not correct because $f$ is not identically zero.
Note that the definition of $f$ is the limit of a Riemann sum:
$$\begin{align}f(x)&=\lim\limits_{n \to \infty} \frac{x}{n} \left(\frac{1}{1+e^{-x/n}}+\frac{1}{1+e^{-2x/n}}+\dots+\frac{1}{1+e^{-x}} \right)\\&=x\int_0^1\frac{dt}{1+e^{-tx}}=\ln\left(\frac{e^x+1}{2}\right).\end{align}$$
Therefore, as $x\to 0$,
$$\begin{align}
2f(x)-x&=2\ln\left(\frac{1+x+\frac{x^2}{2}+o(x^2)+1}{2}\right)-x\\
&=2\ln\left(1+\frac{x}{2}+\frac{x^2}{4}+o(x^2)\right)-x\\
&=2\left(\frac{x}{2}+\frac{x^2}{4}-\frac{x^2}{8}+o(x^2)\right)-x
=\frac{x^2}{4}+o(x^2)
\end{align}$$
and we may conclude that the given limit is $1/4$.
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