2023年考研数二第01题解析：渐近线、等价无穷小

一、题目

$y=x \ln \left(e+\frac{1}{x-1}\right)$ 的斜渐近线方程是 ( )

难度评级：

二、解析

首先，由选项可知，该斜渐近线的斜率一定为：

$$
\textcolor{orangered}{
k = 1
}
$$

当然，我们也可以按照传统方法，先求解 $k$:

$$
k=\lim \limits_{x \rightarrow \infty} \frac{y}{x}=
$$

$$
\lim \limits_{x \rightarrow \infty} \frac{x \ln \left(e+\frac{1}{x-1}\right)}{x}=
$$

$$
\lim \limits_{x \rightarrow \infty} \ln \left(e+\frac{1}{x-1}\right) = \ln e = 1
$$

接着，开始求解截距 $b$:

$$
\textcolor{orangered}{b} = \lim_{x \rightarrow \infty} (y = kx) =
$$

$$
\lim_{x \rightarrow \infty} \big[ x \ln (e+\frac{1}{x – 1}) – x \big] =
$$

$$
\lim_{x \rightarrow \infty} x \big[ \ln (e + \frac{1}{x – 1}) – \textcolor{orangered}{1} \big] =
$$

$$
\lim_{x \rightarrow \infty} x \big[ \ln (e + \frac{1}{x – 1}) – \textcolor{orangered}{\ln e} \big] =
$$

$$
\lim_{x \rightarrow \infty} x \big[ \ln (1 + \frac{1}{e(x – 1)}) \big] =
$$

等价无穷小替换：

$$
\lim_{x \rightarrow \infty} \frac{x}{e(x – 1)} =
$$

$$
\lim_{x \rightarrow \infty} \frac{x}{ex} = \textcolor{orangered}{\frac{1}{e}}
$$

于是，斜渐近线为：

$$
y = kx + b \Rightarrow
$$

$$
\textcolor{springgreen}{
y = x + \frac{1}{e}
}
$$

综上可知，正确选项为：B.

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