# 判断 $y$ $=$ $\frac{x^{2} + 1}{x + 1} \cdot e^{\frac{1}{x-1}}$ 的渐近线的条数和类型

## 一、题目

$$y = \frac{x^{2} + 1}{x + 1} \cdot e^{\frac{1}{x-1}}$$

## 二、解析

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### 1. 水平渐近线

$$\lim_{x \rightarrow + \infty} \frac{x^{2} + 1}{x + 1} \cdot e^{\frac{1}{x-1}} =$$

$$\lim_{x \rightarrow + \infty} \frac{x^{2}}{x} \cdot e^{\frac{1}{x}} =$$

$$\lim_{x \rightarrow + \infty} x \cdot e^{0} = + \infty.$$

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$$\lim_{x \rightarrow – \infty} \frac{x^{2} + 1}{x + 1} \cdot e^{\frac{1}{x-1}} =$$

$$\lim_{x \rightarrow – \infty} \frac{x^{2}}{x} \cdot e^{\frac{1}{x}} =$$

$$\lim_{x \rightarrow – \infty} x \cdot e^{0} = – \infty.$$

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### 2. 倾斜渐近线

$$\lim_{x \rightarrow \infty} \frac{x^{2} + 1}{x(x + 1)} \cdot e^{\frac{1}{x-1}} =$$

$$\lim_{x \rightarrow \infty} \frac{x^{2} + 1}{x^{2} + x} \cdot e^{\frac{1}{x-1}} =$$

$$\lim_{x \rightarrow \infty} \frac{x^{2}}{x^{2}} \cdot e^{\frac{1}{x}} =$$

$$\lim_{x \rightarrow \infty} 1 \cdot e^{0} = 1.$$

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### 3. 垂直渐近线

$x$ $=$ $1$ 和 $x$ $=$ $-1$.

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$$\lim_{x \rightarrow 1^{+}} \frac{x^{2} + 1}{x + 1} \cdot e^{\frac{1}{x-1}} =$$

$$\lim_{x \rightarrow 1^{+}} 1 \cdot e^{\frac{1}{0^{+}}} =$$

$$\lim_{x \rightarrow 1^{+}} 1 \cdot e^{+ \infty} = + \infty$$

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$$\lim_{x \rightarrow 1^{-}} \frac{x^{2} + 1}{x + 1} \cdot e^{\frac{1}{x-1}} =$$

$$\lim_{x \rightarrow 1^{-}} 1 \cdot e^{\frac{1}{0^{-}}} =$$

$$\lim_{x \rightarrow 1^{-}} 1 \cdot e^{- \infty} = 1 \cdot 0 = 0.$$

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$$\lim_{x \rightarrow (-1)} \frac{x^{2} + 1}{x + 1} \cdot e^{\frac{1}{-2}} =$$

$$\lim_{x \rightarrow (-1)} \frac{2}{x + 1} \cdot e^{\frac{1}{-2}} = \infty.$$

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