# 2020 年研究生入学考试数学一第 9 题解析

## 题目

\begin{align}
\lim_{x \rightarrow 0}[\frac{1}{e^{x}-1}-\frac{1}{\ln(1+x)}] =
\end{align}

## 方法一

\begin{align}
\lim_{x \rightarrow 0}[\frac{1}{e^{x}-1}-\frac{1}{\ln(1+x)}] \\
& =\lim_{x \rightarrow 0}[\frac{\ln(1+x)-(e^{x}-1)}{(e^{x}-1)\ln(1+x)}] \\
& =\lim_{x \rightarrow 0}[\frac{\ln(1+x)-(e^{x}-1)}{x^{2}}] \\
& =\lim_{x \rightarrow0}\frac{[\ln (1+x)-x]+[x-(e^{x}-1)]}{x^{2}} \\
& =\lim_{x \rightarrow 0}\frac{\ln(1+x)-x}{x^{2}}+\lim_{x \rightarrow 0}\frac{x-(e^{x}-1)}{x^{2}} \\
& =-\frac{1}{2}+\lim_{x \rightarrow 0}\frac{1-e^{x}}{2x} \\
& =-\frac{1}{2}+\frac{-x}{2x} \\
& =-\frac{1}{2}+(-\frac{1}{2}) \\
& =-1.
\end{align}

## 方法二

\begin{align}
\lim_{x \rightarrow 0}[\frac{1}{e^{x}-1}-\frac{1}{\ln(1+x)}] \\
& =\lim_{x \rightarrow 0}[\frac{\ln(1+x)-(e^{x}-1)}{(e^{x}-1)\ln(1+x)}] \\
& =\lim_{x \rightarrow 0}[\frac{\ln(1+x)-e^{x}+1}{x^{2}}] \\
& =\lim_{x \rightarrow 0}[\frac{\frac{1}{1+x}-e^{x}}{2x}] \\
& =\lim_{x \rightarrow 0}[\frac{\frac{-1}{(1+x)^{2}}-e^{x}}{2}] \\
& \Rrightarrow x=0 \\
& \Rrightarrow \frac{\frac{-1}{1}-1}{2} \\
& =-1.
\end{align}

EOF