Find $\lim\limits_{n\rightarrow \infty} \left| \ln \frac{1}{\left| x_n-A \right|} \right|^{1/n}$ The sequence $\{x_n\}$ satisfies $x_{n+1}=x_n+\sin x_n,n=1,2,\cdots $.

*

*Prove $\{x_n\}$ is convergent and find its limit $A$.

*Let $x_1\neq A$. Find the limit $$\lim_{n\rightarrow \infty} \left| \ln \dfrac{1}{\left| x_n-A \right|} \right|^{1/n}$$
My Attempt: For the first question, let $k\in \mathbb Z$ and $f(x)=x+\sin x$. When $x=k\pi$, it is trivial that $A=k\pi$. When $x\in (k\pi,(k+1)\pi)$, since $f(x)\in (k\pi,(k+1)\pi)$ and $f(x)$ monotonously increases on $\mathbb R$, obviously $\{x_n\}$ monotonously increases and is bounded from above. Take the limits of both sides of the recurrence, and we get $A=A+\sin A$, so $A=(k+1)\pi$.
My Question is, did I correctly do the first question? And how to solve the second one? I don't have any ideas yet.
 A: The sequence $\;\{x_n\}_{n\in\mathbb{N}}\;$ satisfies
$x_{n+1}=x_n+\sin x_n\quad$ for all $\;n\in\mathbb{N}\;.$
There are four possibilities:
1) $\;x_1\in\big]2k\pi,\pi+2k\pi\big[\quad$ where $\;k\in\mathbb{Z}\;,$
2) $\;x_1\in\big]\pi+2k\pi,2\pi+2k\pi\big[\quad$ where $\;k\in\mathbb{Z}\;,$
3) $\;x_1=2k\pi\quad$ where $\;k\in\mathbb{Z}\;,$
4) $\;x_1=\pi+2k\pi\quad$ where $\;k\in\mathbb{Z}\;.$
First case
If $\;x_1\in\big]2k\pi,\pi+2k\pi\big[\;$ where $\;k\in\mathbb{Z}\;,\;$ then $\;2k\pi<x_n<x_{n+1}<\pi+2k\pi\quad$ for any $\;n\in\mathbb{N}\;.$
I am going to prove it by induction.
Since $\;x_1\in\big]2k\pi,\pi+2k\pi\big[\;$ and the function $\;f(x)=x+\sin x\;$ is monotonically increasing in $\;\left]-\infty,+\infty\right[\;,\;$ it follows that
$\begin{align}
2k\pi&<x_1<x_1+\sin x_1=x_2=f(x_1)<f\left(\pi+2k\pi\right)=\\
&=\pi+2k\pi\;.
\end{align}$
Now I suppose that $\;2k\pi<x_{n-1}<x_n<\pi+2k\pi\;$ and prove that $\;2k\pi<x_n<x_{n+1}<\pi+2k\pi\;.$
Since $\;x_n\in\big]2k\pi,\pi+2k\pi\big[\;$ and the function $\;f(x)=x+\sin x\;$ is monotonically increasing in $\;\left]-\infty,+\infty\right[\;,\;$ it follows that
$\begin{align}
2k\pi&<x_n<x_n+\sin x_n=x_{n+1}=f(x_n)<f\left(\pi+2k\pi\right)=\\
&=\pi+2k\pi\;.
\end{align}$
So I have proved by induction that
$2k\pi<x_n<x_{n+1}<\pi+2k\pi\quad$ for any $\;n\in\mathbb{N}\;.$
Consequently,
$\exists\lim\limits_{n\to\infty}x_n=A_1=\sup\limits_{n\in\mathbb{N}}\{x_n\}\in\big]2k\pi,\pi+2k\pi\big]\;.$
Moreover,
$\lim\limits_{n\to\infty}x_{n+1}=\lim\limits_{n\to\infty}x_n+\lim\limits_{n\to\infty}\sin x_n\;,$
$A_1=A_1+\sin A_1\;,$
$A_1=\pi+2k\pi\;.$
So far I have proved that
if $\;x_1\in\big]2k\pi,\pi+2k\pi\big[\;,$ then $\;\exists\lim\limits_{n\to\infty}x_n=A_1=\pi+2k\pi\;.$
Second case
If $\;x_1\in\big]\pi+2k\pi,2\pi+2k\pi\big[\;$ where $\;k\in\mathbb{Z}\;,\;$ then $\;\pi+2k\pi<x_{n+1}<x_n<2\pi+2k\pi\quad$ for any $\;n\in\mathbb{N}\;.$
I am going to prove it by induction.
Since $\;x_1\in\big]\pi+2k\pi,2\pi+2k\pi\big[\;$ and the function $\;f(x)=x+\sin x\;$ is monotonically increasing in $\;\left]-\infty,+\infty\right[\;,\;$ it follows that
$\begin{align}
\pi+2k\pi&=f\left(\pi+2k\pi\right)<f(x_1)=x_1+\sin x_1=\\
&=x_2<x_1<2\pi+2k\pi\;.
\end{align}$
Now I suppose that $\;\pi+2k\pi<x_n<x_{n-1}<2\pi+2k\pi\;$ and prove that $\;\pi+2k\pi<x_{n+1}<x_n<2\pi+2k\pi\;.$
Since $\;x_n\in\big]\pi+2k\pi,2\pi+2k\pi\big[\;$ and the function $\;f(x)=x+\sin x\;$ is monotonically increasing in $\;\left]-\infty,+\infty\right[\;,\;$ it follows that
$\begin{align}
\pi+2k\pi&=f\left(\pi+2k\pi\right)<f(x_n)=x_n+\sin x_n=\\
&=x_{n+1}<x_n<2\pi+2k\pi\;.
\end{align}$
So I have proved by induction that
$\pi+2k\pi<x_{n+1}<x_n<2\pi+2k\pi\quad$ for any $\;n\in\mathbb{N}\;.$
Consequently,
$\exists\lim\limits_{n\to\infty}x_n=A_2=\inf\limits_{n\in\mathbb{N}}\{x_n\}\in\big[\pi+2k\pi,2\pi+2k\pi\big[\;.$
Moreover,
$\lim\limits_{n\to\infty}x_{n+1}=\lim\limits_{n\to\infty}x_n+\lim\limits_{n\to\infty}\sin x_n\;,$
$A_2=A_2+\sin A_2\;,$
$A_2=\pi+2k\pi\;.$
So far I have proved that
if $\;x_1\in\big]2k\pi,\pi+2k\pi\big[\;,$ then $\;\exists\lim\limits_{n\to\infty}x_n=A_2=\pi+2k\pi\;.$
Third case
If $\;x_1=2k\pi\;,$ then $\;x_n=2k\pi\;$ for any $\;n\in\mathbb{N}\;.$
Consequently,
$\exists\lim\limits_{n\to\infty}x_n=A_3=2k\pi\;.$
Fourth case
If $\;x_1=\pi+2k\pi\;,$ then $\;x_n=\pi+2k\pi\;$ for any $\;n\in\mathbb{N}\;.$
Consequently,
$\exists\lim\limits_{n\to\infty}x_n=A_4=\pi+2k\pi\;.$

Hence,
if $\color{blue}{\;x_1\in\big]2k\pi,2\pi+2k\pi\big[\;}$ where $\color{blue}{\;k\in\mathbb{Z}\;}$, then $\color{blue}{\;\exists\lim\limits_{n\to\infty}x_n=A=\pi+2k\pi\;}$.
if $\color{blue}{\;x_1=2k\pi\;}$ where $\color{blue}{\;k\in\mathbb{Z}\;}$, then $\color{blue}{\;\exists\lim\limits_{n\to\infty}x_n=A^*=2k\pi\;}$.

A: There is a more elementary way to solve the second part of your question.
Since $\;x_1\ne\lim\limits_{n\to\infty}x_n\;,$ by looking at the first and second cases of my previous answer, it follows that $\;\lim\limits_{n\to\infty}x_n=A=\pi+2k\pi\;$ where $\;k\in\mathbb{Z}\;$ and $\;x_n\ne A=\pi+2k\pi\;$ for all $\;n\in\mathbb{N}\;.$
$\begin{align}
\lim\limits_{n\rightarrow\infty}&\left|\ln\dfrac{1}{\left|x_n-A\right|}\right|^{1/n}=\lim\limits_{n\rightarrow\infty}\sqrt[n]{\big|-\ln\left|x_n-A\right|\big|}=\\
&=\lim\limits_{n\rightarrow\infty}\sqrt[n]{\big|\ln\left|x_n-A\right|\big|}
\end{align}$
and, by using an application of Stolz-Cesàro Theorem (geometric mean application), it is sufficient to calculate the following limit
$\lim\limits_{n\to\infty}\dfrac{\big|\ln\left|x_{n+1}-A\right|\big|}{\big|\ln\left|x_n-A\right|\big|}=\lim\limits_{n\to\infty}\left|\dfrac{\ln\left|x_n-A+\sin x_n\right|}{\ln\left|x_n-A\right|}\right|.$
It is possible to use that application because there exists $\;n_0\in\mathbb{N}\;$ such that $\;\big|\ln|x_n-A|\big|>0\;$ for all $\;n\ge n_0\;$, indeed $\;\lim\limits_{n\to\infty}\big|\ln|x_n-A|\big|=+\infty\;.$
First of all, I am going to calculate the limit
$\begin{align}
\lim\limits_{x\to A}&\dfrac{\ln\left|x-A+\sin x\right|}{\ln\left|x-A\right|}=\lim\limits_{x\to A}\dfrac{\ln\left|x-A-\sin(x-A)\right|}{\ln\left|x-A\right|}=\\
&=\lim\limits_{t\to0}\dfrac{\ln\left|t-\sin t\right|}{\ln\left|t\right|}=\lim\limits_{t\to0}\dfrac{\ln\left|t-\sin t\right|-3\ln|t|+3\ln|t|}{\ln\left|t\right|}=\\
&=\lim\limits_{t\to0}\left[\dfrac{\ln\left|\dfrac{t-\sin t}{t^3}\right|}{\ln\left|t\right|}+3\right]=\dfrac{\ln\left|\dfrac16\right|}{-\infty}+3=3\;.
\end{align}$
Hence,
$\lim\limits_{x\to A}\left|\dfrac{\ln\left|x-A+\sin x\right|}{\ln\left|x-A\right|}\right|=3$
and, by applying the Sequential Criterion for a limit of a function, it follows that
$\lim\limits_{n\to\infty}\dfrac{\big|\ln\left|x_{n+1}-A\right|\big|}{\big|\ln\left|x_n-A\right|\big|}=\lim\limits_{n\to\infty}\left|\dfrac{\ln\left|x_n-A+\sin x_n\right|}{\ln\left|x_n-A\right|}\right|=3\;.$
Finally, by using an application of Stolz-Cesàro Theorem (geometric mean application), it follows that
$\lim\limits_{n\rightarrow\infty}\left|\ln\dfrac{1}{\left|x_n-A\right|}\right|^{1/n}=\lim\limits_{n\rightarrow\infty}\sqrt[n]{\big|\ln\left|x_n-A\right|\big|}=3\;.$
A: To solve the second question, use the fact that for small $\delta$, $\sin(\pi-\delta)=\sin(\delta)\approx \delta-\frac{\delta^3}{6}$. Thus, starting at $\pi-x_1=\delta$ for small positive $\delta$, we obtain $\pi-x_2\approx\frac{\delta^3}{6}$, $\pi-x_3\approx\frac{\left(\frac{\delta^3}{6}\right)^3}{6}=\frac{\delta^9}{6^4}$, $\pi-x_4\approx\frac{\left(\frac{\delta^9}{6^4}\right)^3}{6}=\frac{\delta^{3^3}}{6^{3^2+3+1}}$ and so on. So $\pi-x_n\approx\frac{\delta^{3^n}}{6^{\frac{3^n-1}{2}}}$. You can make the approximations rigorous using Taylor series expansions, and show that the errors don't accumulate. Then you can solve for $\log(\pi-x_n)$ and $|\log(\pi-x_n)|^{1/n}$. By taking $n \to \infty$, you'll notice that it doesn't matter where $x_1$ is in the range $(0,\pi)$.
