How can I prove that $x_{n+1} = x_n \sin(x_n)$ converges for any $x_0$? I have already prove that it is true if exist $\sin(x_m) = 1$ or $\sin(x_m) = -1$. So I need make proof only for absolute-decreasing progression. But I am afraid that $\sin(x_n)$ can has limit $1$ and can go to $1$ very fast. Is it a good way to count limit $\sin(x_n)$? And what solution is correct?
 A: Let $y_n = |x_n|$ 
Then $y_{n+1} = y_n |\sin y_n| \leq y_n \forall n$ (*)
So $y_n$ is decreasing and bounded below by $0$. Therefore it converges to some limit $l$.
By passing to the limit in (*) we get that either $l=0$ or $sin l= 1$ giving the family of solutions discussed in the comments, namely $l=2k\pi+\frac{\pi}{2}$
Now note that if $y_n$ converges to $0$ then $x_n$ also converges to $0$. If $y_n$ converges to a nonzero value $l$ then a little more work will also prove that $x_n$ also converges to either $l$ or $-l$ because for n large enough $y_n$ will be between $l$ and $l+\frac{\pi}{2}$ hence $x_n$ will have constant sign
Either way, $x_n$ will also be convergent
A: Note that $0\le|x_{n+1}|=|x_n\sin x_n|\le|x_n|$, so the sequence of absolute values, $|x_0|,|x_1|,|x_2|,\ldots$ definitely converges to a nonnegative limit $L(x_0)$ satisfying the equation $L(x_0)=L(x_0)|\sin L(x_0)|$. If $L(x_0)=0$, we're done: $|x_n|\to0$ implies $x_n\to0$.  If $L(x_0)\gt0$, we need only worry about the possibility that $x_n$ approaches both $L(x_0)$ and $-L(x_0)$.  But this can't happen: If $x_n\approx\sigma L(x_0)$ where $\sigma=\sin L(x_0)\in\{1,-1\}$, then, using the fact that $\sin(\sigma u)=\sigma\sin u$ for all $u$ and $\sigma^2=1$, we have 
$$x_{n+1}=x_n\sin x_n\approx\sigma L(x_0)\sin(\sigma L(x_0))=\sigma L(x_0)(\sigma\sin L(x_0))=\sigma L(x_0)(\sigma^2)=\sigma L(x_0)\approx x_n$$
Remarks: As ImNotTheGuy observes in a comment beneath the OP, $\pi/2+2k\pi$ (with $k\in\mathbb{Z}$) is a fixed point for the mapping $f(x)=x\sin x$, which means there are sequences that tend to each such limit, as well as sequences that tend to $0$.  There are, in fact, uncountably many sequences tending to each of the possible limits:  If $k\ge0$, for example, and $x_n=\pi/2+2k\pi+\epsilon$ with $\epsilon\gt0$ but very small, then $x_{n+1}=x_n\cos\epsilon =\pi/2+2k\pi+\epsilon'$ with 
$$\epsilon'=\epsilon\cos\epsilon-(1-\cos\epsilon)(\pi/2+2k\pi)\approx\epsilon-(\epsilon^2/2)(\pi/2+2k\pi)\lt\epsilon$$
(but still positive), so each positive limit has a "basin of attraction" that includes an open inteval of values for $x_0$ slightly greater than the limit (and likewise, for each negative limit, an open interval of values slightly less than the limit -- since $|x_{n+1}|=|x_n\sin x_n|\le|x_n|$, successive terms in any sequence always get closer to the origin).
A: In general If $x_0$ is an initial approximation of a ﬁxed point , the ﬁxed point iteration generates a sequence of approximants by $$x_{n+1}=\varphi (x_n),$$ where $\varphi$ is a continuous function. Suppose that at the ﬁxed point that we indicate with $\alpha$ we have $$\varphi '(\alpha)=\varphi ''(\alpha)=\cdots\varphi ^{p-1}(\alpha)=0, \quad \varphi ^p(\alpha)\neq 0 \qquad (*);$$ then we have the following theorem: 
Let $\alpha$ be  a fixed point of $\varphi$ and $I_{\epsilon}:=\{x\in\Bbb R: |x-\alpha|<\epsilon\}$. Asuume that $\varphi \in \Bbb C^p[I_{\epsilon}]$ satisfes $(*)$. If $$M(\epsilon):=|\varphi '(t)|<1$$ then the fixed point iteration converges to $\alpha$ for every $x_0\in I_{\epsilon}$ and the order the convergence is $p$. 
Note that to use this theorem you have to know tha fxed point $\alpha$ (you can plot the function for example and obtain an approximation of point). You can find this argument in this book.
