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Find the limit as x goes to 0 of (log (1+x)) / (e2x – 1)

Evaluate the limit.

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


Using the expansions as x \to 0 (p. 287)

    \[ \log(1+x) = x + o(x), \qquad \text{and} \qquad e^x = 1 + x + o(x), \]

we compute,

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

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