# 2018年考研数二第09题解析

## 题目

$$\lim_{x \rightarrow + \infty} x^{2}[\arctan (x+1) – \arctan x] = ?$$

## 解析

$$\arctan x \sim x.$$

$$\lim_{x \rightarrow + \infty} x^{2}[\arctan (x+1) – \arctan x] =$$

$$\lim_{x \rightarrow + \infty} x^{2}[(x+1) – x]=$$

$$\lim_{x \rightarrow + \infty} x^{2} + 1 = +\infty .$$

$$\lim_{x \rightarrow + \infty} x^{2}[\arctan (x+1) – \arctan x] =$$

$$\lim_{x \rightarrow +\infty} \frac{\arctan (x+1) – \arctan x}{x^{-2}} \Rightarrow 洛必达 \Rightarrow$$

$$\lim_{x \rightarrow +\infty} \frac{(\arctan (x+1) – \arctan x)^{‘}}{(x^{-2})^{‘}} =$$

$$\lim_{x \rightarrow +\infty} \frac{\frac{1}{1+(x+1)^{2}} – \frac{1}{1+x^{2}}}{-2x^{-3}}=$$

$$\lim_{x \rightarrow +\infty} \frac{-1-2x}{(1+x^{2}+1+2x)(1+x^{2})} \times (-\frac{1}{2}) \cdot x^{3}. ①$$

$$① \Rightarrow \lim_{x \Rightarrow + \infty} \frac{x^{4}}{x^{4}} = \lim_{x \Rightarrow + \infty} \frac{1}{1} = 1.$$

$$\arctan (x+1) – \arctan x = f(x+1) – f(x).$$

$$\frac{f(x+1) – f(x)}{(x+1) – x} = f^{‘}(\xi) \Rightarrow$$

$$f(x+1) – f(x) = f^{‘}(\xi) \Rightarrow$$

$$\arctan (x+1) – \arctan x = f^{‘}(\xi) = \frac{1}{1+ \xi ^{2}}.$$

$$\lim_{x \rightarrow + \infty} x^{2}[\arctan (x+1) – \arctan x] =$$

$$\lim_{x \rightarrow + \infty} \frac{x^{2}}{1+\xi ^{2}}.$$

$$\frac{x^{2}}{1+(x+1)^{2}} < \frac{x^{2}}{1+\xi ^{2}} < \frac{x^{2}}{1+x^{2}}$$

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

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

$$\lim_{x \rightarrow + \infty} x^{2}[\arctan (x+1) – \arctan x] = 1.$$

EOF