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\begin{array}{ll}

(1) &\lim_{x \to 0} \frac{e^x - 1}{x} = 1　　\because ネイピア数の定義式\\
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(2) &\lim_{x \to 0} \frac{log(1 + x)}{x} = 1\\
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&u = e^x - 1 とおくと、e^x = 1 + u　\therefore x = \log (1 + u)\\
&{x \to 0} のとき、{u \to 0} となるので、　\because \lim_{x \to 0} e^x - 1 = 1 - 1 = 0\\
&\lim_{x \to 0} \frac{e^x - 1}{x} = 1　からxを消して\\
&\lim_{u \to 0} \frac{u}{\log (1 + u)} = 1　逆数も成り立つので\\
&\lim_{u \to 0} \frac{\log (1 + u)}{u} = 1\\
& \therefore \lim_{x \to 0} \frac{\log (1 + x)}{x} = 1\\
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(3) &\lim_{x \to 0} (1 + x)^{\frac{1}{x}} = e\\
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& \lim_{x \to 0} \frac{\log (1 + x)}{x} = 1　より\\
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&\lim_{x \to 0} \frac{\log (1 + x)}{x} = \lim_{x \to 0} \frac{1}{x} \cdot \log (1 + x) = \lim_{x \to 0} \log (1 + x)^{\frac{1}{x}}\\
&1 = \log e\\
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&\therefore \lim_{x \to 0} \log (1 + x)^{\frac{1}{x}} = \log e\\
&\lim_{x \to 0} (1 + x)^{\frac{1}{x}} = e　　\because \log (1 + x)^{\frac{1}{x}} が e に近づく\\
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(4) &\lim_{x \to \pm \infty} (1 + \frac{1}{x})^{x} = e\\
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&\lim_{x \to 0} (1 + x)^{\frac{1}{x}} = e より、x = \frac{1}{t} (t = \frac{1}{x}) とすると\\
&{x \to 0} のとき、{t \to \pm \infty}\\
&\lim_{x \to 0} (1 + x)^{\frac{1}{x}} = \lim_{t \to \pm \infty} (1 + \frac{1}{t})^t\\
&\therefore \lim_{x \to \pm \infty} (1 + \frac{1}{x})^x\\
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\end{array}