# 2022考研数二第01题解析：等价无穷小相减会产生更高阶的无穷小，反之也成立

## 二、解析

### 命题 ①

$$\frac{\alpha (x)}{\beta (x)} = 1$$

\begin{aligned} \frac{\alpha ^{2} (x)}{\beta ^{2} (x)} \\ \\ & = \frac{\alpha (x)}{\beta (x)} \cdot \frac{\alpha (x)}{\beta (x)} \\ \\ & = 1 \cdot 1 \\ \\ & = 1 \end{aligned}

$$\alpha ^{2} (x) \sim \beta ^{2} (x)$$

### 命题 ②

\begin{aligned} \frac{\alpha ^{2} (x)}{\beta ^{2} (x)} = 1 \\ \\ & \Rightarrow \frac{\alpha (x)}{\beta (x)} \cdot \frac{\alpha (x)}{\beta (x)} \\ \\ & \Rightarrow 1 \cdot 1 \\ \\ & = 1 \end{aligned}

\begin{aligned} \frac{\alpha ^{2} (x)}{\beta ^{2} (x)} = 1 \\ \\ & \Rightarrow \frac{\alpha (x)}{\beta (x)} \cdot \frac{\alpha (x)}{\beta (x)} \\ \\ & \Rightarrow \left( -1 \right) \cdot \left( -1 \right) \\ \\ & = 1 \end{aligned}

### 命题 ③

$$\lim_{x \to 0} \frac{\alpha (x)}{\beta (x)} = 1$$

$$\lim_{x \to 0} \frac{\beta (x)}{\beta (x)} = 1$$

\begin{aligned} \lim_{x \to 0} \frac{\alpha (x)}{\beta (x)} – \lim_{x \to 0} \frac{\beta (x)}{\beta (x)} \\ \\ & = \lim_{x \to 0} \frac{\alpha (x) – \beta (x)}{\beta (x)} \\ \\ & = 1 – 1 \\ \\ & = 0 \end{aligned}

$$\alpha (x) – \beta (x) = o \left( \alpha (x) \right)$$

### 命题 ④

$$\lim_{x \to 0} \frac{\alpha (x) – \beta (x)}{\alpha (x)} = 0$$

\begin{aligned} \lim_{x \to 0} \frac{\alpha (x)}{\alpha (x)} – \lim_{x \to 0} \frac{\beta (x)}{\alpha (x)} \\ \\ & = 1 – \lim_{x \to 0} \frac{\beta (x)}{\alpha (x)} \\ \\ & = 0 \end{aligned}

$$\lim_{x \to 0} \frac{\beta (x)}{\alpha (x)} = 1$$