Finding the limit of iterated function composition with itself $f_1 = \sin x$
$f_2 = \sin(\sin x)$
$f_3 = \sin(\sin(\sin x))$
$f_n = \sin(f_{n-1})$
Just from playing around on a graphing calculator, I was able to guess the limit of this would be the zero function as $n \rightarrow \infty$, but I'm not certain and I don't know how I could prove this.
 A: We proceed via the following steps:


*

*Show that the sequence $f_n$ has a limit, for any choice of $x$.

*Show that the limit is zero, again for any choice of $x$.
1:

Assume $x\in[0,1]$; the case $x\in[-1,0]$ can be dealt with separately, and we can reduce to one of these cases by starting after one iteration because $\sin x\in[-1,1]$. We may also assume without loss of generality that $x\neq 0$, because for this choice of $x$ the iterates $f_n = \sin^n(x)$ are all $0$.
Because $x\in(0,1]$, it follows that $\sin x\in(0,1]$ as well, and by induction we see that $f_n\in(0,1]$ for all $n\geq 0$. Moreover, from the mean-value theorem we know that for $0<y<1$, $$\sin y = \sin y - \sin 0 = y\cos c$$ for some $0<c<x$. Since $0 < \cos c \leq 1$ for $0<c<1$, we see that $\sin y \leq y$ when $y\in(0,1]$. In particular, we see that $$f_{n+1} = \sin(f_n) \leq f_n,$$ so $(f_n)$ is a monotone decreasing sequence. Since we also know that $f_n\geq 0$, and a monotone decreasing sequence that is bounded below has a limit, we conclude that $(f_n)$ tends to some nonnegative limit at $n\to\infty$. A similar proof works when $x\in[-1,0)$.

2:

Fix $x\in[-1,1]$, construct the corresponding sequence $f_n$, and let $L = \lim_{n\to\infty} f_n$ be its limit. Then by continuity of sine, $$L = \lim_{n\to\infty} f_n = \lim_{n\to\infty} \sin(f_{n-1}) = \sin\left(\lim_{n\to\infty} f_{n-1}\right) = \sin L.$$ But the only value of $L$ satisfying $\sin L = L$ is $L=0$. Therefore $f_n\to 0$ for any choice of starting point $x$.

Addendum:

Another way to derive the monotonicity of the sequence $f_n$ is via the fundamental theorem of calculus: for $x>0$ and corresponding sequence $f_n$,
  $$
f_{n+1} = \sin f_n = \sin f_n - 0 = \int_0^{f_n} \cos t~dt \leq \int_0^{f_n} 1~dt = f_n.
$$
  For $x<0$ the same idea works, but you need to take care of minus signs.

