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 $n oinfty$ 有 $$ex x_infty=sin x_infty a x_infty =0 eex$$ ($0<x_infty<pi a sin x_infty <x_infty$). (2) $$eex ea vlm{n}sex{cfrac{x_{n+1}}{x_n}}^frac{1}{x_n^2} &=expsex{vlm{n}cfrac{ln x_{n+1}-ln x_n}{x_n^2}}\ &=expsex{vlm{n}cfrac{ln sin x_n-ln x_n}{x_n^2}}\ &=expsex{lim_{x o 0^+}cfrac{ln sin x-ln x}{x^2}}\ &=expsez{lim_{x o 0^+}cfrac{cfrac{1}{xi_x}(sin x-x)}{x^2}}quadsex{sin x<xi_x<x}\ &=expsez{ lim_{x o 0^+} cfrac{-cfrac{1}{6}x^3+o(x^3)}{x^2xi_x} }\ &=e^{-frac{1}{6}}. eea eeex$$ (3) $$eex ea vlm{n}nx_n^2&=vlm{n}cfrac{n}{cfrac{1}{x_n^2}}\ &=vlm{n}cfrac{1}{cfrac{1}{x_{n+1}^2}-cfrac{1}{x_n^2}}\ &=vlm{n}cfrac{x_n^2x_{n+1}^2}{x_n^2-x_{n+1}^2}\ &=vlm{n}cfrac{x_n^2sin^2x_n}{x_n^2-sin^2x_n}\ &=lim_{x o 0}cfrac{x^2sin^2x}{x^2-sin^2x}\ &=lim_{x o 0}cfrac{x^4}{(x-sin x)(x+sin x)}\ &=cfrac{1}{cfrac{1}{3!}cdot 2}\ &=3. eea eeex$$