Limit of iteration of sine function Let $f(x) =\sin x $, and denote by $f^n = f\circ f\circ ...\circ f$ the n-th iteration of function $f.$ Find the limit (if exist) :$$\lim_{n\to\infty} n\cdot f^n (n^{-1} )$$
 A: Define $f_{n+1}(x)=\sin(f_n(x))$ where $f_0(x)=x$.
Since $\frac{\sin(x)}x\le1$ we have that $\frac{f_{n+1}(x)}{f_n(x)}\le1$, which implies, via induction, that
$$
\frac{f_n(x)}x\le1\tag{1}
$$
For all $x\in\mathbb{R}$, we have
$$
\frac{\sin(x)}x\ge1-\frac{x^2}6\tag{2}
$$
Combining $(1)$ and $(2)$, we get
$$
\begin{align}
\frac{f_{n+1}(x)}{f_n(x)}
&\ge1-\frac16f_n(x)^2\\
&\ge1-\frac{x^2}6\tag{3}
\end{align}
$$
which implies
$$
\frac{f_n(x)}x\ge\left(1-\frac{x^2}6\right)^n\tag{4}
$$
Using $(1)$ and $(4)$, we can deduce that
$$
\left(1-\frac1{6n^2}\right)^n\le nf_n\left(\frac1n\right)\le1\tag{5}
$$
Since
$$
\lim_{n\to\infty}\left(1-\frac1{6n^2}\right)^n=1\tag{6}
$$
the Squeeze Theorem and $(5)$ yield
$$
\lim_{n\to\infty}nf_n\left(\frac1n\right)=1\tag{7}
$$
A: As a heuristic, use
$$\sin(x -a x^3) = x + \left(a-\frac{1}{6}\right)x^3+ O(x^4),\quad x \rightarrow 0$$
to get 
$$f^n(x)= x -\frac{n}{6}x^3 + O(x^4),\quad x \rightarrow 0$$
As @robjohn pointed out, the $O()$ term depends on $n$, but it suggest the general form
$$f^n\left(\frac{1}{n}\right) = \frac{1}{n} +O(n^{-2}),\quad n \rightarrow \infty$$ and because 
$$\sin\left(\frac{1}{n} +O(n^{-2})\right)= \frac{1}{n} +O(n^{-2}),\quad n \rightarrow \infty$$ 
it follows 
$$nf^n\left(\frac{1}{n}\right) = n\left(\frac{1}{n} + O(n^{-2})\right)= 1 + O(n^{-1}),\quad n \rightarrow \infty$$ 
and therefore the limit is 1.
A: Incidentally, it looks as if $$\lim_{n\to\infty} nf^{n^2}\left(\frac1n\right)=\sqrt{\frac34}$$
We also have $$\frac1{\sin^2x}=\frac1{x^2}+\frac13+O(x^2)$$, so each sine adds 1/3 and a bit to $1/x^2$.  After $kn$ sines, it has only increased from $n^2$ to $n^2+kn/3+o(n)$, so your limit is still 1.
It takes $kn^2$ sines before the limit moves away from 1.  Then $1/x^2$ has reached $n^2+kn^2/3+o(n^2)$, so $$\lim_{n\to\infty}nf^{kn^2}\left(\frac1n\right)=\sqrt\frac3{3+k}$$
