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