Taking the limit to infinity of composition of functions I was reading this question Limit of $(\sin\circ\sin\circ\cdots\circ\sin)(x)$ and nowhere did it talk about how to prove the existence of the limit.
In general, if you're given a function $f(x)$, how do you check for the existence of $\lim\limits_{n\to \infty} \underbrace{(f\circ f \circ \cdots \circ f)}_{n\text{ times}}(x)$?
When $f(x) = \sin(x)$, if the limit $c$ exists , then $f(c)$ must equal $c$ so $c = 0$ right? But isn't it also possible that the limit doesn't exist?
Or is it the case that when you have a monotonically decreasing function that's bounded below by some value, then $\lim\limits_{n\to\infty} \underbrace{(f \circ f \circ \cdots \circ f)}_{n\text{ times}}(x)$ is defined and necessarily a value that satisfies the equation $f(x) = x$?
Oh, and as a side note, is there a better way of denoting $f(x)$ composed with itself $n$ times?
 A: As mentioned in the comments, Ross's answer in the linked question does prove that the limit exists. But with respect to your last comment, it might not be apparent. So I explain.
In general, when you have $\lim\limits_{n\to \infty} \underbrace{(f\circ f \circ \cdots \circ f)}_{n\text{ times}}(x)$, a standard way of showing this exists is to actually consider the sequence $f(x), f(f(x)), f^{(3)} (x), ...$ and so on. Then for any fixed x, this is just a sequence of real numbers. And then you do whatever you would do for that sequence of real numbers.
There is the possibility that there will be radically different behavior for different starting x. If that's the case, then you might have to be witty. Perhaps do something different, or perhaps break it up into cases and still consider the sequence above. It's a bit hard to say (giving general methods in mathematics is always very hard).
Implicitly, Ross did this. He also noted that when you iterate a few times, the sequence will monotonically decrease. And it's bounded. So it must converge.
A: For positive $x$, $0 > \sin(x) > x$, so the iterated sequence is decreasing and bounded. Such a sequence always has a limit. Similarly for negative $x$ (except that the sequence is increasing and bounded.
