高等數學隨記 - 一道極限計算題的簡化求解

srcres258發表於2024-07-07

題目

例1. 求\(\lim_{x \to 0} \frac{e(x-2)+2e^{\frac{\ln(x+1)}{x}}}{2x^2}\).

解法一:常規解法,洛必達法則(繁瑣,暫略)

解法二:簡化解法,利用麥克勞林公式(注意適用條件)

解. 取以下麥克勞林公式:(\(\ln(x+1)\)展開到3次項,\(e^x\)展開到2次項)

\[\ln(x+1) = x-\frac{x^2}{2}+\frac{x^3}{3}+\mathrm{o}(x^3), \]

\[e^x = 1+x+\frac{x^2}{2}+\mathrm{o}(x^2), \]

對原式有:

\[\lim_{x \to 0} \frac{e(x-2)+2e^{\frac{\ln(x+1)}{x}}}{2x^2} = \lim_{x \to 0} \frac{e(x-2)+2e^{\frac{x-\frac{x^2}{2}+\frac{x^3}{3}+\mathrm{o}(x^3)}{x}}}{2x^2} \]

\[= \lim_{x \to 0} \frac{e(x-2)+2e^{1-\frac{x}{2}+\frac{x^2}{3}+\mathrm{o}(x^2)}}{2x^2} \]

\[= \lim_{x \to 0} \frac{e(x-2)+2e\cdot e^{-\frac{x}{2}+\frac{x^2}{3}+\mathrm{o}(x^2)}}{2x^2} \]

\[= \lim_{x \to 0} \frac{e(x-2)+2e(1-\frac{x}{2}+\frac{x^2}{3}+\mathrm{o}(x^2)+\frac{(-\frac{x}{2}+\frac{x^2}{3}+\mathrm{o}(x^2))^2}{2}+\mathrm{o}((-\frac{x}{2}+\frac{x^2}{3}+\mathrm{o}(x^2))^2))}{2x^2} \]

\[= \lim_{x \to 0} \frac{2e(\frac{x^2}{3}+\frac{x^2}{8})+\mathrm{o}(x^2)}{2x^2} \]

\[= \lim_{x \to 0} \frac{2e\cdot\frac{11}{24}x^2}{2x^2} \]

\[= \frac{11}{24}e. \]

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