$\displaystyle\lim_{x \to 1}\frac{(1+x)^\frac1{x}(1+\frac1{x})^x-4}{(x-1)^2}$

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源自知乎提问：zhihu.com/question/625788205/answer/3247473030

题： $\displaystyle\lim_{x \to 1}\frac{(1+x)^\frac1{x}(1+\frac1{x})^x-4}{(x-1)^2}$ .

处理分子，先，化为 $\frac{(1+x)^{\frac1x+x}}{x^x}-4$ ，得到 \begin{align*}
&\quad\,\lim_{x \to 1}\frac{(1+x)^\frac1{x}(1+\frac1{x})^x-4}{(x-1)^2}\\
&={\color{blue}{4}}\lim_{x\to 1}\frac{\frac{(1+x)^{\frac1{x}+x}}{4x^x}-{\color{blue}{1}}}{(x-1)^2}\\[1ex]
&=4\lim_{x\to 1}\frac{\mathrm e^{\ln\frac{(1+x)^{\frac1{x}+x}}{4x^x}}-{\color{blue}{1}}}{(x-1)^2}\\[1ex]
&=4\lim_{x\to 1}\frac{\ln\frac{(1+x)^{\frac1{x}+x}}{4x^x}}{(x-1)^2}\qquad ({\color{blue}{\mathrm e^y-1\sim y\,(y\to 0)}})\\[1ex]
&=4\lim_{x\to 1}\frac{\frac{x^2+1}x\ln(1+x)-x\ln x-2\ln 2}{(x-1)^2}\\[1ex]
&=4\lim_{x\to 1}\frac{\Big(\frac{(x-1)^2}x+2\Big)\ln(1+x)-x\ln x-2\ln 2}{(x-1)^2}\\[1ex]
&=4\ln 2+4\lim_{x\to1}\frac{2\ln(1+x)-x\ln x-2\ln 2}{(x-1)^2},
\end{align*} 置 ${\color{red}{x:\mapsto x+1}}$，在 $x=0$ 处泰勒展开 \begin{align*}
\cdots&=4\ln 2+4\lim_{{\color{blue}{x\to0}}}\frac{2\ln(2+x)-(1+x)\ln(1+x)-2\ln 2}{x^2}\\[1ex]
&=4\ln 2+4\lim_{x\to0}\frac{2\ln(1+\frac x2)-(1+x)(x-\frac{x^2}2+o(x^2))}{x^2}\\[1ex]
&=4\ln 2+4\lim_{x\to0}\frac{2\big(\frac x2-\frac{x^2}8\big)-x-\frac{x^2}2+o(x^2)}{x^2}\\[1ex]
&=4\ln 2-3.
\end{align*}