原文地址 zhuanlan.zhihu.com
$$
\lim_{x \to + \infty} x^2[(1 + \frac{1}{x + 1})^{x + 1} - (1 + \frac{1}{x})^x]
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解：
$$
= \lim_{x \to +\infty} x^2 [e^{(x + 1)\ln{(1 + \frac{1}{x + 1})}} - e^{x\ln{(1 + \frac{1}{x})}}]
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$$
泰勒公式 ： ln(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \dots
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其中 \lim_{x \to +\infty} [e^{(x + 1)\ln{(1 + \frac{1}{x + 1})}} - e^{x\ln{(1 + \frac{1}{x})}}]
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$$
= \lim_{x \to \infty} {e^{(x + 1)[\frac{1}{x + 1} - \frac{1}{2(x + 1)^2} + \frac{1}{3(x + 1)^3}]} - e^{x[\frac{1}{x} - \frac{1}{2x^2} + \frac{1}{3x^3}]} }
$$
$$
= \lim_{x \to +\infty} {e^{[1 - \frac{1}{2(x + 1)} + \frac{1}{3(x + 1)^2}]} - e^{[1 - \frac{1}{2x} + \frac{1}{3x^2}]} }
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无穷小量代换：e^{\alpha} - e^{\beta} = e^{\beta}(e^{\alpha - \beta} - 1) \sim e^{\beta}(\alpha - \beta) ,当 (\alpha - \beta) \to 0 时
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$$
=\lim_{x \to +\infty} e^{1 - \frac{1}{2x} + \frac{1}{3x^2}} [e^{\frac{1}{2x} - \frac{1}{2(x + 1)} - \frac{1}{3x^2} + \frac{1}{3(x + 1)^2}} - 1]
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$$
= \lim_{x \to +\infty} e [\frac{1}{2x(x + 1)} - \frac{2x + 1}{3x^2(x + 1)^2}]
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$$
原式 = \lim_{x \to +\infty}x^2 e [\frac{1}{2x(x + 1)} - \frac{2x + 1}{3x^2(x + 1)^2}]
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$$
= \lim_{x \to +\infty} e [\frac{x}{2(x + 1)} - \frac{2x + 1}{3(x + 1)^2}]
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$$
= \frac{e}{2}
$$