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Compute the limit of the given function

by RoRi

Evaluate the limit.

    \[ \lim_{x \to 0^+} \left( \frac{\log x}{(1+x)^2} - \log \left( \frac{x}{1+x} \right) \right). \]

First, we simplify a bit

    \begin{align*} \lim_{x \to 0^+} \left( \frac{\log x}{(1+x)^2} - \log \left( \frac{x}{1+x} \right) \right) &= \lim_{x \to 0^+} \left( \frac{\log x}{(1+x)^2} - \log x + \log(1+x) \right) \\[9pt] &= \lim_{x \to 0^+} \left( \frac{\log x - (1+x)^2 \log x}{(1+x)^2} \right) + \lim_{x \to 0^+} \left( \log (1+x) \right) \\[9pt] &= \lim_{x \to 0^+} \left( \frac{(-2x-x^2) \log x}{(1+x)^2} \right) + 0 \\[9pt] &= \lim_{x \to 0^+} \frac{ (-2-x)\log x^x }{(1+x)^2}. \end{align*}

But, by Example #3 (page 302 of Apostol), we know that \lim_{x \to 0^+} x^x = 1. Since the logarithm is continuous we then have

    \[ \lim_{x \to 0^+} \log x^x = \log 1 = 0. \]

Therefore,

    \[ \lim_{x \to 0^+} \frac{(-2-x) \log x^x}{(1+x)^2} = 0. \]

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