Find $\lim_{n\to\infty}x_n\sqrt{\log n}$. 
Let $0<x_0<\pi$ and consider the sequence defined by $x_{n+1}=x_n-\frac{x_n-\sin(x_n)}{n+1}$. Calculate $\lim_{n\to\infty}x_n\sqrt{\log n}$.

I did a little code in C to guess an answer and it seems like this limit is $\sqrt{3}$.
I proved that $0<x_n<\pi$ for all $n$. However I couldn't do much more.
Can anyone help?
EDIT: I had, unfortunatelly wrote the limit wrong. It was written $\lim_{n\to\infty}x_n\sqrt{\log x_n}$.
 A: By writing $x_{n+1} = \frac{n}{n+1} x_n + \frac{1}{n}\sin x_n$, we can prove that $(x_n)$ is monotone decreasing. So $(x_n)$ converges to some value $\ell \in [0, x_0)$.
Now introduce the function $f(x) = \frac{x - \sin x}{x^3}$ and notice that (1) this can be continuated at $x = 0$ with value $f(0) = \frac{1}{6}$, and that (2) $f(x) \geq f(\pi) > 0$ on $[0, \pi$]. With this function, we may write
\begin{align*}
\frac{1}{x_{n+1}^2}
&= \frac{1}{x_n^2} \left( 1 - \frac{x_n^2 f(x_n)}{n+1} \right)^{-2} \\
&= \frac{1}{x_n^2} \left( 1 + \frac{2x_n^2 f(x_n)}{n+1} + \mathcal{O}\left( \frac{x_n^4}{n^2} \right) \right) \\
&= \frac{1}{x_n^2} + \frac{2f(x_n)}{n+1} + \mathcal{O}\left( \frac{1}{n^2} \right).
\end{align*}
So it follows from the Stolz-Cesàro theorem that
$$ \lim_{n\to\infty} \frac{x_n^{-2}}{\log n}
= \lim_{n\to\infty} \frac{x_{n+1}^{-2} - x_{n}^{-2}}{\log(n+1) - \log n}
= \lim_{n\to\infty} \frac{\frac{2f(x_n)}{n+1} + \mathcal{O}\left( \frac{1}{n^2} \right)}{\log\left(1+\frac{1}{n}\right)}
= 2f(\ell). $$
But this in particular tells that $x_n$ converges to $0$, i.e. $\ell = 0$. So we have
$$ \lim_{n\to\infty} x_n \sqrt{\log n}
= \sqrt{\frac{1}{2f(0)}}
= \sqrt{3}. $$
A: Look up "iterated sine".
It turns out that,
for any initial value $x_0$,
$\lim_{n \to \infty} \sqrt{n} \sin^{(n)} (x_0)
=\sqrt{3}
$.
