一、题目
$$
I = \lim_{ x \rightarrow + \infty} \left[ \frac{x^{1+x}}{(1+x)^{x}}-\frac{x}{\mathrm{e}} \right] = ?
$$
二、解析 
$$
\begin{aligned}
I & = \lim_{\textcolor{yellow}{ x \rightarrow + \infty }} \left[ \frac{x^{1+x}}{(1+x)^{x}} – \frac{x}{\mathrm{e}} \right] \\ \\
& = \lim_{ \textcolor{yellow}{x \rightarrow + \infty }} \left[ \frac{x \cdot \textcolor{orange}{x^{x}}}{(1+x)^{x}} – \frac{x}{\mathrm{e}} \right] \\ \\
& = \lim_{\textcolor{yellow}{x \rightarrow + \infty}} \left[ \frac{x}{ \frac{(1+x)^{x}}{\textcolor{orange}{x^{x}}}} – \frac{x}{\mathrm{e}} \right] \\ \\
& = \lim_{\textcolor{yellow}{x \rightarrow + \infty}} \left[ \frac{x}{ \left( \frac{1+x}{\textcolor{orange}{x}} \right)^{\textcolor{orange}{x}}} – \frac{x}{\mathrm{e}} \right] \\ \\
& = \lim_{\textcolor{yellow}{x \rightarrow + \infty}} \left[ \frac{x}{ \left(1 + \frac{1}{x} \right)^{x}} – \frac{x}{\mathrm{e}} \right]
\end{aligned}
$$
我们知道，当 $x \rightarrow + \infty$, 式子 $\left(1 + \frac{1}{x} \right)^{x}$ 的极限为 $\mathrm{e}$. 但是，严格地说，即便 $x \rightarrow + \infty$, $\left(1 + \frac{1}{x} \right)^{x}$ 和 $\mathrm{e}$ 之间也不是真正的相等关系。
所以，$\lim_{x \rightarrow + \infty} \left[ \frac{x}{ \left(1 + \frac{1}{x} \right)^{x}} – \frac{x}{\mathrm{e}} \right]$ $\textcolor{red}{\boldsymbol{\neq}}$ $\lim_{x \rightarrow + \infty} \left[ \frac{x}{ \mathrm{e} } – \frac{x}{\mathrm{e}} \right]$ $\textcolor{red}{\boldsymbol{\neq}}$ $0$.
令 $\textcolor{violet}{ \frac{1}{x} }$ $\textcolor{violet}{ = }$ $\textcolor{violet}{t}$, $\textcolor{violet}{x}$ $\textcolor{violet}{=}$ $\textcolor{violet}{\frac{1}{t}}$, 则：
$$
\begin{aligned}
I & = \lim_{\textcolor{lightgreen}{ t \rightarrow 0^{+} }} \left[ \frac{\textcolor{violet}{ \frac{1}{t} }}{(1 + \textcolor{violet}{t})^{\textcolor{violet}{\frac{1}{t}}}} – \frac{\textcolor{violet}{\frac{1}{t}}}{\mathrm{e}} \right] \\ \\
& = \lim_{\textcolor{lightgreen}{ t \rightarrow 0^{+} }} \textcolor{violet}{\frac{1}{t}} \left[ \frac{1}{(1 + \textcolor{violet}{t})^{\textcolor{violet}{\frac{1}{t}}}} – \frac{1}{\mathrm{e}} \right] \\ \\
& = \lim_{\textcolor{lightgreen}{ t \rightarrow 0^{+} }} \textcolor{violet}{\frac{1}{t}} \left[ \frac{\mathrm{e}}{\mathrm{e}(1 + t)^{\frac{1}{t}}} – \frac{(1 + t)^{\frac{1}{t}}}{\mathrm{e} (1 + t)^{\frac{1}{t}}} \right] \\ \\
& = \lim_{\textcolor{lightgreen}{t \rightarrow 0^{+}}} \frac{\mathrm{e} – \textcolor{pink}{ (1+t)^{\frac{1}{t}}}}{\textcolor{violet}{t} \textcolor{tan}{ \mathrm{e} } (1+t)^{\frac{1}{t}}} \\ \\
& = \textcolor{tan}{ \frac{1}{\mathrm{e}} } \lim_{ \textcolor{lightgreen}{t \rightarrow 0^{+}}} \frac{\mathrm{e} – \textcolor{pink}{ \mathrm{e}^{\frac{1}{t} \ln (1+t)}}}{t \textcolor{orange}{ (1+t)^{\frac{1}{t}}}} \\ \\
& = \textcolor{tan}{ \frac{1}{\mathrm{e}} } \lim_{ \textcolor{lightgreen}{t \rightarrow 0^{+}}} \frac{ \mathrm{e} \left( 1 – \textcolor{pink}{ \mathrm{e}^{\frac{1}{t} \ln (1+t) – 1}} \right)}{t \textcolor{orange}{ \mathrm{e} }} \\ \\
& = \textcolor{tan}{ \frac{-1}{\mathrm{e}} } \lim_{ \textcolor{lightgreen}{t \rightarrow 0^{+}}} \frac{ \textcolor{pink}{ \mathrm{e}^{\frac{1}{t} \ln (1+t) – 1} – 1} }{t} \\ \\
& \textcolor{lightgreen}{ \leadsto } \textcolor{gray}{\lim_{t \rightarrow 0^{+} } \left[ \frac{1}{t} \ln (1+t) – 1 \right] = 0} \\ \\
& = \frac{-1}{\mathrm{e}} \lim_{ \textcolor{lightgreen}{t \rightarrow 0^{+}}} \frac{ \textcolor{pink}{ \mathrm{e}^{\frac{1}{t} \ln (1+t) – 1}} – 1 }{t} \\ \\
& = \frac{-1}{\mathrm{e}} \lim_{ \textcolor{lightgreen}{t \rightarrow 0^{+}}} \frac{ \mathrm{e}^{\frac{1}{t} \left[ \textcolor{red}{ \ln (1+t) – t } \right]} – 1 }{t} \\ \\
& = \frac{-1}{\mathrm{e}} \lim_{\textcolor{lightgreen}{t \rightarrow 0^{+}}} \frac{ \textcolor{red}{ \ln(1 + t ) – t } }{t^{2}} \\ \\
& = -\frac{1}{\mathrm{e}} \lim_{\textcolor{lightgreen}{t \rightarrow 0^{+}}} \frac{\textcolor{red}{ -\frac{1}{2} t^{2} } + o \left( t^{2} \right) } { t^{2}} \\ \\
& = \textcolor{springgreen}{\boldsymbol{ \frac{1}{2 \mathrm{e}} }}
\end{aligned}
$$