Proving $\sqrt{n}(x_n)$ converges when $x_n = \sin(x_{n-1}), x_1=1$ This is a problem that showed up on a qual exam that I have been stuck on for a while.
Let \begin{equation} x_n = \sin(x_{n-1}), x_1 = 1 \end{equation}
Prove $\lim_{n \rightarrow \infty} \sqrt{n} x_n$ exists and compute its value. The problem gave the following hint: Show that \begin{equation} \frac{1}{x^2_{n+1}} - \frac{1}{x^2_{n}}    \end{equation} converges to a constant.
I have shown that $x_n$ converges to $0$, but I am unsure on how to begin bounding $\sqrt{n}x_n$ or how the hint is helpful in this problem. I have tried using the MVT to show $\sin(x_n) \rightarrow 0$ faster than $\sqrt{n} \rightarrow \infty$, but I didn't get far. Any help will be appreciated. 
 A: Hints: Using elementary calculus, you can show that $\displaystyle\lim_{x \to 0}\dfrac{1}{\sin^2 x} - \dfrac{1}{x^2} = L$ where $L$ is some constant (which you can explicitly work out). 
Then, you can use the fact that $\displaystyle\lim_{n \to \infty}x_n = 0$ to show that $\displaystyle\lim_{n \to \infty}\dfrac{1}{x_{n+1}^2} - \dfrac{1}{x_n^2} = \lim_{n \to \infty}\dfrac{1}{\sin^2(x_n)} - \dfrac{1}{x_n^2} = \lim_{x \to 0}\dfrac{1}{\sin^2 x} - \dfrac{1}{x^2} = L$. 
Finally, you can use the result $\displaystyle\lim_{n \to \infty}\dfrac{1}{x_{n+1}^2} - \dfrac{1}{x_n^2} = L$, to show that $\dfrac{1}{x_n^2} \approx Ln + \text{const}$ for large $n$. 
A: Let
$$u_n=\frac1{x_n^2}.$$
Then
$$u_{n+1}=\frac1{\sin^2 (1/\sqrt{u_n})}=\frac{u_n}{1-1/(3u_n)+O(u_n^{-2})}
=u_n+\frac13+O(u_n^{-1}).$$
Then
$$u_{n+1}-u_n\to\frac13$$
as $n\to\infty$.
From this you can deduce that $u_n=n/3+o(n)$ etc.
A: Why the given hint is useful. It suffices to find the limit $L$ of $\frac{1/x_n^2}{n}$, then $\sqrt{n}x_n$ is convergent to $1/\sqrt{L}$. Now, by Stolz-Cesaro Theorem,
$$L=\lim_{n\to\infty}\frac{1/x_n^2}{n}=\lim_{n\to\infty}\left( \frac{1}{x^2_{n+1}} - \frac{1}{x^2_{n}}\right)=\lim_{n\to\infty}\left( \frac{1}{\sin^2(x_n)} - \frac{1}{x^2_{n}}\right).$$
Finally, in order to find $L$, show that $x_n\to 0$ and use the Taylor expansion of $\sin(x)$ at $x=0$.
