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

求极限：

$$\underset{x\to +\mathrm{\infty }}{lim}\frac{{\int }_1^x[t^2(e^{\frac{1}{t}}-1)-t]dt}{x^2\ln (1+\frac{1}{x})}.$$

解析

方法一

当 $x\to +\mathrm{\infty }$ 时，$\frac{1}{x}\to 0$. 于是：

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

进而，有：

$$\underset{x\to +\mathrm{\infty }}{lim}\frac{{\int }_1^x[t^2(e^{\frac{1}{t}}-1)-t]dt}{x^2\ln (1+\frac{1}{x})}\Rightarrow$$

$$\underset{x\to +\mathrm{\infty }}{lim}\frac{{\int }_1^x[t^2(e^{\frac{1}{t}}-1)-t]dt}{x^2\cdot \frac{1}{x}}\Rightarrow$$

$$\underset{x\to +\mathrm{\infty }}{lim}\frac{{\int }_1^x[t^2(e^{\frac{1}{t}}-1)-t]dt}{x}\Rightarrow \mathrm{洛}\mathrm{必}\mathrm{达}\Rightarrow$$

$$\underset{x\to +\mathrm{\infty }}{lim}\frac{x^2(e^{\frac{1}{x}}-1)-x}{1}\Rightarrow$$

$$\underset{x\to +\mathrm{\infty }}{lim}{x}^2(e^{\frac{1}{x}}-1)-x\Rightarrow \mathrm{麦}\mathrm{克}\mathrm{劳}\mathrm{林}\mathrm{公}\mathrm{式}\Rightarrow$$

$$\underset{x\to +\mathrm{\infty }}{lim}[x^2(\frac{1}{x}+\frac{1}{2x^2})–x]\Rightarrow$$

$$\underset{x\to +\mathrm{\infty }}{lim}(x+\frac{1}{2}–x)\Rightarrow$$

$$\underset{x\to +\mathrm{\infty }}{lim}\frac{1}{2}=\frac{1}{2}.$$

方法二

$$\underset{x\to +\mathrm{\infty }}{lim}\frac{{\int }_1^x[t^2(e^{\frac{1}{t}}-1)-t]dt}{x^2\ln (1+\frac{1}{x})}\Rightarrow$$

$$\underset{x\to +\mathrm{\infty }}{lim}\frac{{\int }_1^x[t^2(e^{\frac{1}{t}}-1)-t]dt}{x}\cdot \frac{1}{x\ln (1+\frac{1}{x})}\Rightarrow$$

$$\underset{x\to +\mathrm{\infty }}{lim}\frac{{\int }_1^x[t^2(e^{\frac{1}{t}}-1)-t]dt}{x}\cdot \frac{1}{x}\cdot \frac{1}{\ln (1+\frac{1}{x})}\Rightarrow$$

$$\underset{x\to +\mathrm{\infty }}{lim}\frac{{\int }_1^x[t^2(e^{\frac{1}{t}}-1)-t]dt}{x}\cdot \frac{\frac{1}{x}}{\ln (1+\frac{1}{x})}.$$

当 $x\to +\mathrm{\infty }$ 时，$\frac{1}{x}\to 0$. 于是：

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

进而：

$$\underset{x\to +\mathrm{\infty }}{lim}\frac{{\int }_1^x[t^2(e^{\frac{1}{t}}-1)-t]dt}{x}\cdot \frac{\frac{1}{x}}{\ln (1+\frac{1}{x})}\Rightarrow$$

$$\underset{x\to +\mathrm{\infty }}{lim}\frac{{\int }_1^x[t^2(e^{\frac{1}{t}}-1)-t]dt}{x}\cdot 1\Rightarrow$$

$$\underset{x\to +\mathrm{\infty }}{lim}\frac{{\int }_1^x[t^2(e^{\frac{1}{t}}-1)-t]dt}{x}\Rightarrow \mathrm{洛}\mathrm{必}\mathrm{达}\Rightarrow$$

$$\underset{x\to +\mathrm{\infty }}{lim}\frac{x^2(e^{\frac{1}{x}}-1)-x}{1}\Rightarrow$$

$$\underset{x\to +\mathrm{\infty }}{lim}{x}^2(e^{\frac{1}{x}}-1)-x\Rightarrow$$

$$\underset{x\to +\mathrm{\infty }}{lim}{x}^2(e^{\frac{1}{x}}-1)–x^2\cdot \frac{1}{x}\Rightarrow$$

$$\underset{x\to +\mathrm{\infty }}{lim}{x}^2(e^{\frac{1}{x}}–1–\frac{1}{x}).$$

接着，令 $A=\frac{1}{x}$, $x=\frac{1}{A}$, 则有：

$$\underset{A\to 0}{lim}\frac{1}{A^2}(e^A–1–A)\Rightarrow$$

$$\underset{A\to 0}{lim}\frac{e^A–1–A}{A^2}\Rightarrow \mathrm{洛}\mathrm{必}\mathrm{达}\Rightarrow$$

$$\underset{A\to 0}{lim}\frac{e^A-1}{2A}\Rightarrow$$

$$\underset{A\to 0}{lim}\frac{A}{2A}=\frac{1}{2}.$$