# 2014年考研数二第15题解析：极限、等价无穷小、麦克劳林公式

## 题目

$$\lim_{x \rightarrow + \infty} \frac{ \int_{1}^{x}[t^{2}(e^{\frac{1}{t}}-1)-t] dt }{x^{2} \ln (1+\frac{1}{x})}.$$

## 解析

### 方法一

$$\ln (1+\frac{1}{x}) \sim \frac{1}{x}.$$

$$\lim_{x \rightarrow + \infty} \frac{ \int_{1}^{x}[t^{2}(e^{\frac{1}{t}}-1)-t] dt }{x^{2} \ln (1+\frac{1}{x})} \Rightarrow$$

$$\lim_{x \rightarrow + \infty} \frac{ \int_{1}^{x}[t^{2}(e^{\frac{1}{t}}-1)-t] dt }{x^{2} \cdot \frac{1}{x}} \Rightarrow$$

$$\lim_{x \rightarrow + \infty} \frac{ \int_{1}^{x}[t^{2}(e^{\frac{1}{t}}-1)-t] dt }{x} \Rightarrow 洛必达 \Rightarrow$$

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

$$\lim_{x \rightarrow + \infty} x^{2}(e^{\frac{1}{x}}-1)-x \Rightarrow 麦克劳林公式 \Rightarrow$$

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

$$\lim_{x \rightarrow + \infty} (x + \frac{1}{2} – x) \Rightarrow$$

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

### 方法二

$$\lim_{x \rightarrow + \infty} \frac{ \int_{1}^{x}[t^{2}(e^{\frac{1}{t}}-1)-t] dt }{x^{2} \ln (1+\frac{1}{x})} \Rightarrow$$

$$\lim_{x \rightarrow + \infty} \frac{ \int_{1}^{x}[t^{2}(e^{\frac{1}{t}}-1)-t] dt }{x} \cdot \frac{1}{x \ln (1+\frac{1}{x})}\Rightarrow$$

$$\lim_{x \rightarrow + \infty} \frac{ \int_{1}^{x}[t^{2}(e^{\frac{1}{t}}-1)-t] dt }{x} \cdot \frac{1}{x} \cdot \frac{1}{\ln (1+\frac{1}{x})}\Rightarrow$$

$$\lim_{x \rightarrow + \infty} \frac{ \int_{1}^{x}[t^{2}(e^{\frac{1}{t}}-1)-t] dt }{x} \cdot \frac{\frac{1}{x}}{\ln (1+\frac{1}{x})}.$$

$$\ln (1+\frac{1}{x}) \sim \frac{1}{x}.$$

$$\lim_{x \rightarrow + \infty} \frac{ \int_{1}^{x}[t^{2}(e^{\frac{1}{t}}-1)-t] dt }{x} \cdot \frac{\frac{1}{x}}{\ln (1+\frac{1}{x})}\Rightarrow$$

$$\lim_{x \rightarrow + \infty} \frac{ \int_{1}^{x}[t^{2}(e^{\frac{1}{t}}-1)-t] dt }{x} \cdot 1 \Rightarrow$$

$$\lim_{x \rightarrow + \infty} \frac{ \int_{1}^{x}[t^{2}(e^{\frac{1}{t}}-1)-t] dt }{x} \Rightarrow 洛必达 \Rightarrow$$

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

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

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

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

$$\lim_{A \rightarrow 0} \frac{1}{A^{2}} (e^{A} – 1 – A) \Rightarrow$$

$$\lim_{A \rightarrow 0} \frac{e^{A} – 1 – A}{A^{2}} \Rightarrow 洛必达 \Rightarrow$$

$$\lim_{A \rightarrow 0} \frac{e^{A}-1}{2A} \Rightarrow$$

$$\lim_{A \rightarrow 0} \frac{A}{2A} = \frac{1}{2}.$$