# 2023年考研数二第03题解析：数列比较大小

## 一、题目

(A) $x_{n}$ 是 $y_{n}$ 的高阶无穷小

(B) $y_{n}$ 是 $x_{n}$ 的高阶无穷小

(C) $x_{n}$ 是 $y_{n}$ 的等价无穷小

(D) $x_{n}$ 是 $y_{n}$ 的同阶但非等价无穷小

## 二、解析

$$\lim_{x \rightarrow \infty} x_{n} = 0$$

$$\lim_{x \rightarrow \infty} y_{n} = 0$$

$$\frac{x_{n+1}}{x_{n}} = \frac{\sin x_{n}}{x_{n}} = \frac{x_{n}}{x_{n}} = \textcolor{orangered}{1}$$

$$\frac{y_{n+1}}{y_{n}} = \frac{y_{n}^{2}}{y_{n}} = y_{n} = \textcolor{orangered}{0}$$

$$y = \frac{2}{\pi} x$$

$$\frac{2}{\pi} x < \sin x$$

$$\sin x_{n} > \frac{2}{\pi} x_{n} \Rightarrow$$

$$\textcolor{springgreen}{ x_{n+1} > \frac{2}{\pi} x_{n} }$$

$$y_{1} = \frac{1}{2}$$

$$y_{2} = \frac{1}{4}$$

$$y^{3} = \frac{1}{16}$$

$$\textcolor{springgreen}{ y_{n + 1} < \frac{1}{2} y_{n} }$$

$$\frac{y_{n+1}}{x_{n+1}} < \frac{\frac{1}{2} y_{n}}{\frac{2}{\pi} x_{n}} \Rightarrow$$

$$\textcolor{springgreen}{ \frac{y_{n+1}}{x_{n+1}} < \frac{\pi}{4} \frac{y_{n}}{x_{n}} = \Big(\frac{\pi}{4} \Big)^{2} \frac{y_{n-1}}{x_{n-1}} = \cdots = \Big(\frac{\pi}{4} \Big)^{n} \frac{y_{1}}{x_{1}} = \Big(\frac{\pi}{4} \Big)^{n} = 0 }$$

$$\lim_{x \rightarrow \infty} \frac{y_{n+1}}{x_{n+1}} = \frac{y_{n}}{x_{n}} = 0$$