问题及解答

求极限 \[\lim\limits_{x\rightarrow+\infty}\frac{e^x}{(1+\frac{1}{x})^{x^2}}.\]

考虑 $\ln\dfrac{e^x}{\Bigl(1+\frac{1}{x}\Bigr)^{x^2}}$,

\[
\begin{split}
\ln\frac{e^x}{\Bigl(1+\frac{1}{x}\Bigr)^{x^2}}&=\ln e^x-\ln\Bigl(1+\frac{1}{x}\Bigr)^{x^2}=x-x\ln\Bigl(1+\frac{1}{x}\Bigr)^{x}\\
&=x\Bigl[1-\ln\Bigl(1+\frac{1}{x}\Bigr)^{x}\Bigr]=\frac{1-x\ln(1+\frac{1}{x})}{\frac{1}{x}},
\end{split}
\]

于是令 $t=\frac{1}{x}$,

\[
\lim_{x\rightarrow+\infty}\ln\frac{e^x}{\Bigl(1+\frac{1}{x}\Bigr)^{x^2}}=\lim_{t\rightarrow 0}\frac{1-\frac{1}{t}\ln(1+t)}{t}
\]

分子分母同趋于零, 可以用洛必达法则计算.

\[
\begin{split}
\text{上式}&=\lim_{t\rightarrow 0^+}\frac{\frac{1}{t^2}\ln(1+t)-\frac{1}{t}\frac{1}{1+t}}{1}\\
&=\lim_{t\rightarrow 0^+}\biggl[\frac{\ln(1+t)}{t^2}-\frac{1}{t(1+t)}\biggr]\\
&=\lim_{t\rightarrow 0^+}\frac{(1+t)\ln(1+t)-t}{t^2(1+t)}\\
&=\lim_{t\rightarrow 0^+}\frac{(1+t)\ln(1+t)-t}{t^2}\\
&=\lim_{t\rightarrow 0^+}\frac{\ln(1+t)+(1+t)\cdot\frac{1}{1+t}-1}{2t}\\
&=\lim_{t\rightarrow 0^+}\frac{\ln(1+t)}{2t}\\
&=\frac{1}{2},
\end{split}
\]

因此

\[
\lim_{x\rightarrow+\infty}\ln\frac{e^x}{\Bigl(1+\frac{1}{x}\Bigr)^{x^2}}=\frac{1}{2},
\]

这推出

\[
\ln\biggl(\lim_{x\rightarrow+\infty}\frac{e^x}{\Bigl(1+\frac{1}{x}\Bigr)^{x^2}}\biggr)=\frac{1}{2},
\]

故

\[
\lim_{x\rightarrow+\infty}\frac{e^x}{\Bigl(1+\frac{1}{x}\Bigr)^{x^2}}=e^{\frac{1}{2}}.
\]

(法二)

\[
\Bigl(1+\frac{1}{x}\Bigr)^{x^2}=e^{x^2\ln(1+\frac{1}{x})},
\]

回顾 $\ln(1+x)$ 的 Taylor 展开式, 有

\[
\begin{split}
x^2\ln(1+\frac{1}{x})&=x^2\cdot\Bigl[\frac{1}{x}-\frac{1}{2}(\frac{1}{x})^2+\frac{1}{3}(\frac{1}{x})^3+o((\frac{1}{x})^3)\Bigr]\\
&=x-\frac{1}{2}+\frac{1}{3x}+o(\frac{1}{x}),
\end{split}
\]

于是

\[
\begin{split}
\lim_{x\rightarrow+\infty}\frac{e^x}{\Bigl(1+\frac{1}{x}\Bigr)^{x^2}}&=\lim_{x\rightarrow+\infty}\frac{e^x}{e^{x^2\ln(1+\frac{1}{x})}}\\
&=\lim_{x\rightarrow+\infty}\frac{e^x}{e^{x-\frac{1}{2}+\frac{1}{3x}+o(\frac{1}{x})}}\\
&=\lim_{x\rightarrow+\infty}e^{\frac{1}{2}-\frac{1}{3x}+o(\frac{1}{x})}\\
&=e^{\frac{1}{2}}.
\end{split}
\]