Let $x \in \mathbb{R} \cap [0, 1]$.
Let's consider the following function:
$f : [0, 1] \to [0, 1]$ so that $x \mapsto 2x$ if $x \leq \dfrac{1}{2}$ and $x \mapsto 2(1 - x)$ otherwise.
It is easy to show that for all $x \in [0, 1]$, $f(x) \in [0, 1]$. But, for some $x$, if you compose $f$ as much time as you want, you will not be able to reach $0$. Whereas, $f(0) = 0$, $f(1) = 0$, and we can find as much numbers that will reach $0$ by solving the equations $f(x) = 1$, and so on.
Is there a quick way to show that, for a $x$ given, this number will reach $0$ after a certain number $n$ of compositions, that is $\exists n \in \mathbb{N}, f^n(x) = 0$.
As a non-trivial example, $\dfrac{2015}{2^{2015}}$ will reach $0$ eventually, in a general manner, $\dfrac{1}{2^n}$ for all $n \in \mathbb{N}$ will reach $0$.
Note that if $f(x)=0$, then $x=0,1$. Let's assume $x=1$, since $x=0$ is trivial.
Now note the following:
$f(x)=1\implies x=\frac12$
$f(x)=\frac12\implies x=\frac14,\frac34$
$f(x)=\frac14,\frac34\implies x=\frac18,\frac38,\frac58,\frac78$
$f(x)=\frac18,\frac38,\frac58,\frac78\implies x=\frac1{16},\frac3{16},\dots$
Now, this is only conjecture, but it appears to be the case that if $x=\frac{2n+1}{2^k}$, where $k,n\in\mathbb N$ and $0<x<1$, then after exactly $k+1$ compositions, you will have $0$. If $x=\frac{2n}{2^k}$, notice the fraction simplifies, and so simplify until you've found the form $x=\frac{2n+1}{2^k}$.
Since we've created these numbers directly from the assumption that $f^n(x)=0$, there exist no other numbers such that this is the case.