TITLE: Limit of infinite composition of sin(x)
QUESTION [24 upvotes]: I was playing around on desmos the other day, and noticed that $\sin\left(\sin\left(x\right)\right)$ is basically a version of sin with a lower amplitude (which makes intuitive sense). To me, it seems intuitve that this curve, when composed infinite times, becomes a straight line, as the values at $x=\frac{\pi}{2}\mathbb{Z}$ would move (slowly) towards the values at $x=\pi\mathbb{Z}$ by virtue of them moving away from the peaks, but is there a way to go about properly proving this?

REPLY [8 votes]: Evaluation of the Limit
Let $a_{n+1}=\sin(a_n)$ and suppose $a_0\gt0$.
As shown in this answer, for $x\gt0$, $0\lt\sin(x)\lt x$. Thus, $a_n$ is a decreasing sequence bounded below by zero. Therefore, $\lim\limits_{n\to\infty}a_n$ exists. Since $\sin(x)$ is continuous,
$$
\begin{align}
\lim_{n\to\infty}a_n
&=\lim_{n\to\infty}\sin(a_n)\tag{1a}\\
&=\sin\left(\lim_{n\to\infty}a_n\right)\tag{1b}
\end{align}
$$
Since $\sin(x)=x$ only at $x=0$, $(1)$ says that
$$
\bbox[5px,border:2px solid #C0A000]{\lim_{n\to\infty}a_n=0}\tag2
$$
Since $\sin(x)$ is an odd function, $(2)$ holds for $a_0\lt0$.
Of course, if $a_0=0$, then $a_n=0$ for all $n\ge0$, and $(2)$ holds for $a_0=0$.

Asymptotics
As shown in this answer
$$
\lim\limits_{x\to0}\frac{x-\sin(x)}{x^3}=\frac16\tag3
$$
If $a_0\ne0$, since $a_{n+1}=\sin(a_n)$, $(2)$ and $(3)$ say
$$
\lim_{n\to\infty}\frac{a_n-a_{n+1}}{a_n^3}=\frac16\tag4
$$
An immediate consequence of $(2)$ and $(4)$ is
$$
\begin{align}
\lim_{n\to\infty}\frac{a_{n+1}}{a_n}
&=\lim_{n\to\infty}\left(1-a_n^2\frac{a_n-a_{n+1}}{a_n^3}\right)\tag{5a}\\
&=1-0^2\cdot\frac16\tag{5b}\\[6pt]
&=1\tag{5c}
\end{align}
$$
Thus, $(4)$ and $(5)$ give
$$
\begin{align}
\lim_{n\to\infty}\left(\frac1{a_{n+1}^2}-\frac1{a_n^2}\right)
&=\lim_{n\to\infty}\frac{a_n^2-a_{n+1}^2}{a_n^2a_{n+1}^2}\tag{6a}\\
&=\lim_{n\to\infty}\frac{a_n-a_{n+1}}{a_n^3}\lim_{n\to\infty}\frac{a_n+a_{n+1}}{a_{n+1}}\lim_{n\to\infty}\frac{a_n}{a_{n+1}}\tag{6b}\\
&=\frac16\cdot2\cdot1\tag{6c}\\[3pt]
&=\frac13\tag{6d}
\end{align}
$$
Finally, Stolz-Cesàro and $(6)$ say that $\lim\limits_{n\to\infty}\frac1{na_n^2}=\frac13$; that is,
$$
\bbox[5px,border:2px solid #C0A000]{\lim_{n\to\infty}na_n^2=3}\tag7
$$
In fact, starting with $a_0=1$, we get
$$
\begin{array}{r|l}
n&na_n^2\\\hline
1&0.7080734183\\
10&2.1433001582\\
100&2.8511162950\\
1000&2.9803925383\\
10000&2.9976147637\\
100000&2.9997198782\\
1000000&2.9999678410
\end{array}
$$

Bounds
Substituting $x\mapsto x/\pi$ in $(21)$ from this answer, we get
$$
\cot(x)=\sum_{k\in\mathbb{Z}}\frac1{k\pi+x}\tag8
$$
Subtracting $(8)$ from $\frac1x$ and taking the derivative, we get
$$
\frac1{\sin^2(x)}-\frac1{x^2}=\sum_{\substack{k\in\mathbb{Z}\\k\ne0}}\frac1{(k\pi+x)^2}\tag9
$$
Evaluating $(9)$ at $x=0$ gives $\frac2{\pi^2}\zeta(2)=\frac13$, which agrees with $(6)$.
Taking two derivatives of $(9)$ gives
$$
\begin{align}
\frac{\mathrm{d}^2}{\mathrm{d}x^2}\left(\frac1{\sin^2(x)}-\frac1{x^2}\right)
&=\sum_{\substack{k\in\mathbb{Z}\\k\ne0}}\frac6{(k\pi+x)^4}\\
&\ge0\tag{10}
\end{align}
$$
Thus, $\frac1{\sin^2(x)}-\frac1{x^2}$ is an even convex function with a minimum of $\frac13$ at $x=0$.
Since $a_k^2$ is decreasing,
$$
\frac13\le\frac1{a_{k+1}^2}-\frac1{a_k^2}\le\frac1{\sin^2(a_0)}-\frac1{a_0^2}\tag{11}
$$
Summing $(11)$ gives
$$
\frac n3\le\frac1{a_n^2}-\frac1{a_0^2}\le n\left(\frac1{\sin^2(a_0)}-\frac1{a_0^2}\right)\tag{12}
$$
Since $\sin^2(x)\le1$, assume that $a_0^2\le1$, whence $\frac1{\sin^2(a_0)}-\frac1{a_0^2}\le\cot^2(1)$.
Solving $(12)$ for $a_n^2$ yields
$$
\frac{\tau a_0^2}{\tau+na_0^2}\le a_n^2\le\frac{3a_0^2}{3+na_0^2}\tag{13}
$$
where $\tau=\tan^2(1)\approx2.42551882$.

Graphical Verification
Limit $(7)$ holds for any $a_0\ne0$. Since $\sin(x)$ is an odd function, changing the sign of $a_0$, changes the sign of all $a_n$. Thus, we have that
$$\newcommand{\sgn}{\operatorname{sgn}}
\lim_{n\to\infty}\sqrt{n}\,a_n=\sqrt3\sgn(a_0)\tag{14}
$$
Therefore, if we define $\sin^{\ast n}(x)$ to be the $n$-fold composition of $\sin(x)$, we get
$$
\bbox[5px,border:2px solid #C0A000]{\lim_{n\to\infty}\sqrt{n}\sin^{\ast n}(x)=\sqrt3\sgn(\sin(x))}\tag{15}
$$
Here is graphical verification of $(15)$:

Asymptotically, $(15)$ says that
$$
\bbox[5px,border:2px solid #C0A000]{\sin^{\ast n}(x)\sim\sqrt{\frac3n}\sgn(\sin(x))}\tag{16}
$$
which would be a more precise way of saying what you were trying to say.