极限的等价变换需要建立在函数连续性的基础之上

一、题目

$$
\lim\limits_{x \to +\infty} \left( 1+\frac{1}{x} \right)^{x+\sqrt{x}} = ?
$$

二、解析

首先，当 $x \to +\infty$ 的时候，原式可等价变换为：

$$
\begin{aligned}
& \ \textcolor{lightgreen}{ \left( 1+\frac{1}{x} \right)^{x+\sqrt{x}} } \\ \
= & \ \mathrm{e}^{\left( x+\sqrt{x} \right)\ln\left( 1+\frac{1}{x} \right)} \\ \
= & \ \mathrm{e}^{x\left( 1 + \textcolor{orangered}{ \frac{1}{\sqrt{x}} } \right)\ln\left( 1+\frac{1}{x} \right)} \\ \
= & \ \mathrm{e}^{x\left( 1 + \textcolor{orangered}{0} \right)\ln\left( 1+\frac{1}{x} \right)} \\ \
= & \ \mathrm{e}^{ \textcolor{lightblue}{ x \ln\left( 1+\frac{1}{x} \right) }}
\end{aligned}
$$

由于对数函数 $\ln$ 具有连续性，所以：

$$
\begin{aligned}
& \ \lim \limits_{x \to +\infty} \textcolor{lightblue}{ x \ln \left( 1+\frac{1}{x} \right) } \\ \
= & \ \lim\limits_{x \to +\infty} \ln\left( 1+\frac{1}{x} \right)^{x} \\ \
= & \ \ln\left[ \lim\limits_{x \to +\infty} \left( 1+\frac{1}{x} \right)^{x} \right] \\ \
= & \ \ln \mathrm{e} \\ \
= & \ 1
\end{aligned}
$$

又由于幂函数 $\mathrm{e}^{x}$ 都具有连续性，所以：

$$
\begin{aligned}
& \ \lim\limits_{x \to +\infty} \textcolor{lightgreen}{ \left( 1+\frac{1}{x} \right)^{x+\sqrt{x}} } \\ \
= & \ \mathrm{e}^{\lim\limits_{x \to +\infty} \textcolor{lightblue}{ x \ln\left( 1+\frac{1}{x} \right) }} \\ \
= & \ \textcolor{lightgreen}{ \mathrm{e} }
\end{aligned}
$$

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