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

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\(\lim\limits_{n\rightarrow \infty}\sqrt{n}x_n=\sqrt{3}\).
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\(\lim\limits_{n\rightarrow \infty}x_n=0\).
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\(n\rightarrow \infty\)时，\(\{x_n\}\) 极限存在。
设 \(\lim\limits_{n\rightarrow \infty} x_n = a\), 则有\(\lim\limits_{n\rightarrow \infty} x_{n+1} = a\)，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 \(\lim\limits_{n\rightarrow \infty}[\dfrac{1}{x_{n+1}^2}-\dfrac{1}{x_{n}^2}] = \dfrac{1}{3}.\)
因为 \(x_{n+1}^2 = (\sin x_n)^2 = \left(x_n-\dfrac{x_n^3}{6}+o(x_n^3)\right)^2\),

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

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

由等价无穷小，\(\sin x\sim x,x\rightarrow 0\), 得\(\sin x_n\sim x_n,n\rightarrow \infty\), 忽略高阶无穷小，可得

\[\lim\limits_{n\rightarrow \infty}[\dfrac{1}{x_{n+1}^2}-\dfrac{1}{x_{n}^2}]=\lim\limits_{n\rightarrow \infty}\dfrac{x_n^4/3}{x_n^4}=\dfrac{1}{3}. \]

3、由Stolz 定理，若数列 \(\{b_n\}\) 极限存在, 且 \(\lim\limits_{n\rightarrow \infty} b_n = b\) ,则 \(\lim\limits_{n\rightarrow \infty}\dfrac{b_1+b_2+\cdots+b_n}{n}=b\).
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

\[\lim\limits_{n\rightarrow \infty}\dfrac{b_1+b_2+\cdots+b_n}{n}= \lim\limits_{n\rightarrow \infty}\dfrac{ \dfrac{1}{x_{n}^2}-\dfrac{1}{x_{0}^2}}{n}=b=\dfrac{1}{3}. \]

即$$\lim\limits_{n\rightarrow \infty}\dfrac{ \dfrac{1}{x_{n}^2}}{n}=b=\dfrac{1}{3},$$
亦即 $$\lim\limits_{n\rightarrow \infty}\sqrt{n}x_n=\sqrt{3}.$$