题目
求极限：
$$
\lim_{x \rightarrow + \infty} \frac{ \int_{1}^{x}[t^{2}(e^{\frac{1}{t}}-1)-t] dt }{x^{2} \ln (1+\frac{1}{x})}.
$$
解析
方法一
当 $x \rightarrow + \infty$ 时，$\frac{1}{x} \rightarrow 0$. 于是：
$$
\ln (1+\frac{1}{x}) \sim \frac{1}{x}.
$$
进而，有：
$$
\lim_{x \rightarrow + \infty} \frac{ \int_{1}^{x}[t^{2}(e^{\frac{1}{t}}-1)-t] dt }{x^{2} \ln (1+\frac{1}{x})} \Rightarrow
$$
$$
\lim_{x \rightarrow + \infty} \frac{ \int_{1}^{x}[t^{2}(e^{\frac{1}{t}}-1)-t] dt }{x^{2} \cdot \frac{1}{x}} \Rightarrow
$$
$$
\lim_{x \rightarrow + \infty} \frac{ \int_{1}^{x}[t^{2}(e^{\frac{1}{t}}-1)-t] dt }{x} \Rightarrow 洛必达 \Rightarrow
$$
$$
\lim_{x \rightarrow + \infty} \frac{x^{2}(e^{\frac{1}{x}}-1)-x}{1} \Rightarrow
$$
$$
\lim_{x \rightarrow + \infty} x^{2}(e^{\frac{1}{x}}-1)-x \Rightarrow 麦克劳林公式 \Rightarrow
$$
$$
\lim_{x \rightarrow + \infty} [x^{2} (\frac{1}{x} + \frac{1}{2x^{2}}) – x] \Rightarrow
$$
$$
\lim_{x \rightarrow + \infty} (x + \frac{1}{2} – x) \Rightarrow
$$
$$
\lim_{x \rightarrow + \infty} \frac{1}{2} = \frac{1}{2}.
$$
方法二
$$
\lim_{x \rightarrow + \infty} \frac{ \int_{1}^{x}[t^{2}(e^{\frac{1}{t}}-1)-t] dt }{x^{2} \ln (1+\frac{1}{x})} \Rightarrow
$$
$$
\lim_{x \rightarrow + \infty} \frac{ \int_{1}^{x}[t^{2}(e^{\frac{1}{t}}-1)-t] dt }{x} \cdot \frac{1}{x \ln (1+\frac{1}{x})}\Rightarrow
$$
$$
\lim_{x \rightarrow + \infty} \frac{ \int_{1}^{x}[t^{2}(e^{\frac{1}{t}}-1)-t] dt }{x} \cdot \frac{1}{x} \cdot \frac{1}{\ln (1+\frac{1}{x})}\Rightarrow
$$
$$
\lim_{x \rightarrow + \infty} \frac{ \int_{1}^{x}[t^{2}(e^{\frac{1}{t}}-1)-t] dt }{x} \cdot \frac{\frac{1}{x}}{\ln (1+\frac{1}{x})}.
$$
当 $x \rightarrow + \infty$ 时，$\frac{1}{x} \rightarrow 0$. 于是：
$$
\ln (1+\frac{1}{x}) \sim \frac{1}{x}.
$$
进而：
$$
\lim_{x \rightarrow + \infty} \frac{ \int_{1}^{x}[t^{2}(e^{\frac{1}{t}}-1)-t] dt }{x} \cdot \frac{\frac{1}{x}}{\ln (1+\frac{1}{x})}\Rightarrow
$$
$$
\lim_{x \rightarrow + \infty} \frac{ \int_{1}^{x}[t^{2}(e^{\frac{1}{t}}-1)-t] dt }{x} \cdot 1 \Rightarrow
$$
$$
\lim_{x \rightarrow + \infty} \frac{ \int_{1}^{x}[t^{2}(e^{\frac{1}{t}}-1)-t] dt }{x} \Rightarrow 洛必达 \Rightarrow
$$
$$
\lim_{x \rightarrow + \infty} \frac{x^{2}(e^{\frac{1}{x}}-1)-x}{1} \Rightarrow
$$
$$
\lim_{x \rightarrow + \infty} x^{2}(e^{\frac{1}{x}}-1)-x \Rightarrow
$$
$$
\lim_{x \rightarrow + \infty} x^{2}(e^{\frac{1}{x}}-1) – x^{2} \cdot \frac{1}{x} \Rightarrow
$$
$$
\lim_{x \rightarrow + \infty} x^{2}(e^{\frac{1}{x}} – 1 – \frac{1}{x}).
$$
接着，令 $A = \frac{1}{x}$, $x = \frac{1}{A}$, 则有：
$$
\lim_{A \rightarrow 0} \frac{1}{A^{2}} (e^{A} – 1 – A) \Rightarrow
$$
$$
\lim_{A \rightarrow 0} \frac{e^{A} – 1 – A}{A^{2}} \Rightarrow 洛必达 \Rightarrow
$$
$$
\lim_{A \rightarrow 0} \frac{e^{A}-1}{2A} \Rightarrow
$$
$$
\lim_{A \rightarrow 0} \frac{A}{2A} = \frac{1}{2}.
$$