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

Evalue the limit.

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


First, we have

    \[ (1+x)^{\frac{1}{x}} = e^{\frac{1}{x} \log (1+x)}. \]

As x \to 0 we use the expansion for \log (1+x) (page 287 of Apostol) to write,

    \[ \frac{1}{x} \log (1+x) = \frac{1}{x} \left(x - \frac{x^2}{2} + o(x^2)\right) = 1 - \frac{x}{2} + o(x) \quad \text{as} \quad x \to 0. \]

From this we see that \frac{1}{x} \log(1+x) \to 1 as x \to 0; and so,

    \[ \frac{1}{x} \log(1+x) - 1 \to 0 \quad \text{as} \quad x \to 0. \]

Hence, using the expansion for e^x as x \to 0 (page 287 of Apostol) we have

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

Therefore, we have

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

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