设 $0<x_0<1,x_{n+1}=sin x_n(n=1,2,cdots)$, 求证$limlimits_{n ightarrow infty}sqrt{n}x_n=sqrt{3}$.

一学生问我一道题：设 (0<x_0<1,x_{n+1}=sin x_n(n=1,2,cdots)), 求证(limlimits_{n ightarrow infty}sqrt{n}x_n=sqrt{3}).
下面证明主要有我教研室王艳秋副教授提供。
证明：1、先证明(limlimits_{n ightarrow infty}x_n=0).
因为当 (0<x_0<1)时,(x_{n+1}=sin x_n<x_n(n=1,2,cdots)), 故(x_n) 单调递减，且有下界0，故当(n ightarrow infty)时，({x_n}) 极限存在。
设 (limlimits_{n ightarrow infty} x_n = a), 则有(limlimits_{n ightarrow infty} x_{n+1} = a)，由(sin x) 的连续性，对等式(x_{n+1}=sin x_n)
两边取极限 (a =sin a), 所以 (a = 0).

2、再证明 (limlimits_{n ightarrow infty}[dfrac{1}{x_{n+1}^2}-dfrac{1}{x_{n}^2}] = dfrac{1}{3}.)
因为 (x_{n+1}^2 = (sin x_n)^2 = left(x_n-dfrac{x_n^3}{6}+o(x_n^3) ight)^2),

[dfrac{1}{x_{n+1}^2}-dfrac{1}{x_{n}^2} = dfrac{1}{(x_n-dfrac{x_n^3}{6}+o(x_n^3))^2}-dfrac{1}{x_{n}^2} ]

[=dfrac{x_n^2-(x_n-dfrac{x_n^3}{6}+o(x_n^3))^2}{x_n^2(x_n-dfrac{x_n^3}{6}+o(x_n^3))^2}, ]

由等价无穷小，(sin xsim x,x ightarrow 0), 得(sin x_nsim x_n,n ightarrow infty), 忽略高阶无穷小，可得

[limlimits_{n ightarrow infty}[dfrac{1}{x_{n+1}^2}-dfrac{1}{x_{n}^2}]=limlimits_{n ightarrow infty}dfrac{x_n^4/3}{x_n^4}=dfrac{1}{3}. ]

3、由Stolz 定理，若数列 ({b_n}) 极限存在, 且 (limlimits_{n ightarrow infty} b_n = b) ,则 (limlimits_{n ightarrow infty}dfrac{b_1+b_2+cdots+b_n}{n}=b).
令 (b_n = dfrac{1}{x_{n}^2}-dfrac{1}{x_{n-1}^2},n=1,2,cdots.), 则 (b_1+b_2+cdots+b_n = dfrac{1}{x_{n}^2}-dfrac{1}{x_{0}^2}), 故

[limlimits_{n ightarrow infty}dfrac{b_1+b_2+cdots+b_n}{n}= limlimits_{n ightarrow infty}dfrac{ dfrac{1}{x_{n}^2}-dfrac{1}{x_{0}^2}}{n}=b=dfrac{1}{3}. ]

即$$limlimits_{n ightarrow infty}dfrac{ dfrac{1}{x_{n}^2}}{n}=b=dfrac{1}{3},$$
亦即 $$limlimits_{n ightarrow infty}sqrt{n}x_n=sqrt{3}.$$