一道群内高数题


求极限$\lim\limits_{n\rightarrow \infty}n\sin\left(2\pi n!\mathrm{e}\right)$的值.

解答:利用$\mathrm{e}$的估计式,有
$$
\mathrm{e}=\frac{1}{0!}+\frac{1}{1!}+\cdots+\frac{1}{n!}+\frac{1}{\left(n+1\right)!}+\frac{\theta_n}{\left(n+1\right)\left(n+1\right)!},0<\theta_n<1
$$
从而得到
$$
\lim\limits_{n\rightarrow \infty}n\sin\left(2\pi n!\mathrm{e}\right)=2\pi
$$


Author: 噻吩
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