[再寄小读者之数学篇](2014-06-22 求极限 [中国科学技术大学2011年高等数学B考研试题])

设数列 $\sed{x_n}$ 满足 $0<x_1<\pi$, $x_{n+1}=\sin x_n\ (n=1,2,\cdots)$. (1) 证明 $\dps{\vlm{n}x_n}$ 存在, 并求其极限; (2) 计算 $\dps{\vlm{n}\sex{\cfrac{x_{n+1}}{x_n}}^{\frac{1}{x_n^2}}}$; (3) 证明 $\dps{\vlm{n}\sqrt{\cfrac{n}{3}}x_n=1}$.

证明: (1) 由 $0<x_{n+1}=\sin x_n<x_n$ 知 $\sed{x_n}$ 递减有下界, 而 $\dps{\vlm{n}x_n=x_\infty}$ 存在. 于 $x_{n+1}=\sin x_n$ 中令 $n\to\infty$ 有 $$\bex x_\infty=\sin x_\infty\ra x_\infty =0 \eex$$ ($0<x_\infty<\pi\ra \sin x_\infty <x_\infty$). (2) $$\beex \bea \vlm{n}\sex{\cfrac{x_{n+1}}{x_n}}^\frac{1}{x_n^2} &=\exp\sex{\vlm{n}\cfrac{\ln x_{n+1}-\ln x_n}{x_n^2}}\\ &=\exp\sex{\vlm{n}\cfrac{\ln \sin x_n-\ln x_n}{x_n^2}}\\ &=\exp\sex{\lim_{x\to 0^+}\cfrac{\ln \sin x-\ln x}{x^2}}\\ &=\exp\sez{\lim_{x\to 0^+}\cfrac{\cfrac{1}{\xi_x}(\sin x-x)}{x^2}}\quad\sex{\sin x<\xi_x<x}\\ &=\exp\sez{ \lim_{x\to 0^+} \cfrac{-\cfrac{1}{6}x^3+o(x^3)}{x^2\xi_x} }\\ &=e^{-\frac{1}{6}}. \eea \eeex$$ (3) $$\beex \bea \vlm{n}nx_n^2&=\vlm{n}\cfrac{n}{\cfrac{1}{x_n^2}}\\ &=\vlm{n}\cfrac{1}{\cfrac{1}{x_{n+1}^2}-\cfrac{1}{x_n^2}}\\ &=\vlm{n}\cfrac{x_n^2x_{n+1}^2}{x_n^2-x_{n+1}^2}\\ &=\vlm{n}\cfrac{x_n^2\sin^2x_n}{x_n^2-\sin^2x_n}\\ &=\lim_{x\to 0}\cfrac{x^2\sin^2x}{x^2-\sin^2x}\\ &=\lim_{x\to 0}\cfrac{x^4}{(x-\sin x)(x+\sin x)}\\ &=\cfrac{1}{\cfrac{1}{3!}\cdot 2}\\ &=3. \eea \eeex$$