where are recursive limits of functions well defined? suppose we have a function $f(x)=x^2-1$. We want to know where $f^n(x)$ as $n$ tends to infinity has a well defined limit. This can only be the case for $x_0$ where $f^n(x_0) = f(x_1) = x_1$, for any $x_1$ in the domain. Otherwise, the function keeps changing and never settles. For $f(x)=x^2-1$, this leaves a set of discontinuous points where $y=\phi$ (golden ratio). However, if we use $f^2(x)$ as the recursive function instead we also get stable flat lines at $y=0,-1$. Another example is $f(x)= x^2$. When we take this to infinity the 'limit function' will be $y = 0$ in the domain $[-1,1]$ and $y=1$ at $x=1,-1$ and undefined elsewhere.
Here is a projection of the set of graphs for this function https://www.desmos.com/calculator/zpdweq69u1
My question is how do we find where the limit function is defined for $x^2-1$? It seems to me that the only point where $y=x$ is at $\phi$, so can we essentially solve $f^n(x) = \phi$? Is there a more general approach? Is this too chaotic to be interesting?
The following parts is not necessary to answer the question:
The limit function points can be pictured by drawing the $y=x$ graph and extending a horizontal line to from the function to the graph. Then proceeding to draw out a 'staircase' between these two functions, we can see if there's a limit point or if it diverges. But these are not always easy to see. A good illustration of this is the example of tetration. For $f(x) = a^x$, for a>1 we have the limit function to be a flat line on the smaller solution of $a^x=x$ and a point at the larger solution with undefined beyond the larger solution. You can see that there is no limit function beyond $e^{(\frac{x}{e})}$. for $a<1$, it seems that the limit function is a flat line across all real $x$ for any $a$. However this is only true for $a>e^{-1}$. For smaller $a$ the staircase becomes a square and does not truly approach the limit from either side. The staircase picture method is not enough to prove this fact. How can we prove this radius of convergence rigorously?
 A: Here's a brief analysis of the limiting behavior for the function $f(x)=x^2-1$, which more-or-less confirms the guesses in your question.
To start with, as you have noticed the values of $x$ such $f(x)=x$ can all be found with simple algebra: the roots of $x^2-1=x$ are the same as the roots of $x^2-x-1$ which by the quadratic formula are 
$$x = \frac{1}{2} \pm \frac{\sqrt{5}}{2} \approx -0.618 \quad\text{and}\quad 1.618
$$ 
There are further values of $x$ which have the property that for some $N$ we have $f^N(x) = \frac{1}{2} \pm \frac{\sqrt{5}}{2}$, from which it follows that for all $n \ge N$ the values of $f^n(x)$ are constant, equal to one of $\frac{1}{2} \pm \frac{\sqrt{5}}{2}$. For the values of $x$ such that $N=2$, one must solve the equations $f^2(x) = \frac{1}{2} - \frac{\sqrt{5}}{2}$ and $f(x) = \frac{1}{2} + \frac{\sqrt{5}}{2}$. You will get two values of $x$ for the solution of the first equation, and for each of those two values the sequence $f^n(x)$ for $n=0,1,2,3,4,...$ will be 
$$x,\frac{1}{2} - \frac{\sqrt{5}}{2},\frac{1}{2} - \frac{\sqrt{5}}{2},\frac{1}{2} - \frac{\sqrt{5}}{2},\frac{1}{2} - \frac{\sqrt{5}}{2},...
$$
You can continue this similarly for $N=3$ with $f^2(x)$ replaced by $f^3(x)$, then for $N=4$ and so on.
This gives a complete description of all points $x$ such that $f^n(x)$ is eventually constant, and as you say in your question, if $x$ is not one of those points then $f^n(x)$ is not eventually constant and in fact does not approach any limit; proving this rigorously requires some calculus.
There is another interesting possiblity for $f^n(x)$, namely it can be "periodic" which means that it can oscillate between some finite set of values; for example, if you start with $x=0$ then $f^n(x)$ for $n=0,1,2,3,4,5,...$ oscillates as $0,-1,0,-1,0,-1,...$. Also $f^n(x)$ can be "eventually periodic" such as for $x=1$ and then $f^n(x)$ for $n=0,1,2,3,4,5,...$ is $1,0,-1,0,-1,0,-1,...$
However, this is somewhat unsatisfactory because the answer is specifically tailored towards the function $f(x)=x^2-1$. Your question seems to be groping for a more general and more satisfactory answer. For that, I recommend learning about kneading theory as I wrote in my comment.
