I'm trying to proof that the function $\varphi\colon A\rightarrow A$ where $A$ is the set $(-\frac{\pi}{2},\frac{\pi}{2})$ and $\varphi(x)=\mathrm{arctan}\, x$ has an attracting fixed point.
Now in my book I've found this definition: If $\exists \epsilon\ge0$ | $\forall x\in \mathrm{(domain)} \cap(p-\epsilon,p+\epsilon)$ and $\lim_{n\rightarrow \infty} \varphi^n(x)$ than $p$ is an attracting fixed point.
I've just find that $0$ is a fixed point and I tried to apply that definition at my case but I'm not able to understand how $\varphi(\varphi(...\varphi(x)))=\varphi^n(x)=\mathrm{arctan}(\mathrm{arctan}(\mathrm{arctan}\dots \mathrm{arctan}(x)))$ can converge to $0$.
$\phi$ is striclty increasing. Also $\phi(x) <x$ for $0<x<\frac {\pi} 2$ [since $(\phi(x)-x)'=\frac 1{1+x^{2}} -1 <0$ and $(\phi(x)-x)(0)=0$]. It follow from these facts that $\phi^{n}(x)$ is a decreasing sequence. If its limit is $t$ then $\phi (t)=t$ and this implies that $t=0$. Thus $\phi^{n}(x) \to 0$ for al $x \in (0,\frac {\pi} 2)$. A similar argument works for $x <0$.