2021 上海交通大学硕士研究生入学考试数学分析第7题

求极限

$$ \lim_{x\to\infty}\dfrac{(1+\frac{1}{x})^x-e}{\sin\dfrac{1}{x}}
$$

Solution: $$\begin{aligned} &\lim_{x\to\infty}\dfrac{(1+\frac{1}{x})^x-e}{\sin\dfrac{1}{x}}\\
=&e\lim_{x\to\infty}\dfrac{e^{x\ln(1+\frac{1}{x})-1}-1}{\dfrac{1}{x}}\\
=&e\lim_{x\to\infty}x[x\ln(1+\dfrac{1}{x})-1]\\
=&e\lim_{x\to\infty}x[x(\dfrac{1}{x}-\dfrac{1}{2x^2}+o(\dfrac{1}{x^3}))-1]\\
=&-\dfrac{e}{2} \end{aligned}$$