# 2018年考研数二第01题解析

## 题目

$$A. a = \frac{1}{2}, b = -1$$

$$B. a = – \frac{1}{2}, b = -1$$

$$C. a = \frac{1}{2}, b = 1$$

$$D. a = – \frac{1}{2}, b = 1$$

## 解析

$$\lim_{x \rightarrow 0} (1+x)^{\frac{1}{x}} = \lim_{x \rightarrow \infty} (1 + \frac{1}{x})^{x} = e.$$

$$\lim_{x \rightarrow 0} (e^{x} + ax^{2} + bx)^{\frac{1}{x^{2}}} = 1 \Rightarrow$$

$$\lim_{x \rightarrow 0} [ (1+ e^{x} + ax^{2} +bx -1) ^\frac{1}{e^{x} + ax^{2} +bx -1}]^{\frac{e^{x} + ax^{2} +bx -1}{x^{2}}} = 1 \Rightarrow$$

$$e^{\lim_{x \rightarrow 0} \frac{e^{x} + ax^{2} +bx -1}{x^{2}}} = 1 \Rightarrow$$

$$\lim_{x \rightarrow 0} \frac{e^{x} + ax^{2} +bx -1}{x^{2}} =0 \Rightarrow 洛必达 \Rightarrow$$

$$\lim_{x \rightarrow 0} \frac{e^{x} + 2ax + b}{2x} = 0 \Rightarrow 洛必达 \Rightarrow ①$$

$$\lim_{x \rightarrow 0} \frac{e^{x} + 2a}{2} = 0$$

$$e^{x} + 2a = 0$$

$$1 + 2a = 0.$$

$$a = – \frac{1}{2}$$

$$\lim_{x \rightarrow 0} \frac{e^{x} – x + b}{2x} = 0 \Rightarrow$$

$$\frac{1-0+b}{0}=0 \Rightarrow$$

$$\frac{1+b}{0}=0 ②$$

$$1+b=0$$

$$b=-1$$

EOF