Let 

$$ f(x) = \lim_{n \rightarrow \infty}\frac{x}{n}\left(\frac{1}{1 + e^{-\frac{x}{n}}} + \frac{1}{1 + e^{-\frac{2x}{n}}} + ... + \frac{1}{1 + e^{-x}}\right)$$, x > 0. Then $$\lim_{x \rightarrow 0}\left(\frac{2f(x) - x}{x^2}\right)$$ is

A

0

B

$$\frac{1}{8}$$

C

$$\frac{1}{4}$$

D

$$\frac{1}{2}$$