For $f(x)=-x^3$ the sequence $f^n(x)$ converges to $0$

Question

Let $f(x) = -x^3$ and $S_n(x) = \{x, f(x), f^2(x), ..., f^n(x)\}$.

I know that $S_n(x_o)$ converges to $0$ when $x_o \in (-1,1)$ , but how can I prove this?

The fact that with each interaction the function changes the signal left me confused.

My attempt was think that these facts could help to prove that $S_n(x_o)$ converges, but I don't know how:

$0 \leq |f^n(x_o)| \leq |f^{n-1}(x_o)|$

$0 \leq |f^{n}(x_o) + f^{n-1}(x_o)| \leq |f^{n-1}(x_o) + f^{n-2}(x_o)|$

Answer 1

Hints:

$$\begin{cases}(1)&f^n(x)=\pm x^{3^n}\\{}\\(2)& \forall\;|x|<1\;,\;\;x^m\xrightarrow[m'\to\infty]{}0\end{cases}$$

Answer 2

Using definition of convergent sequence, consider $\epsilon >0$:

$f^n(x_{0})=(-1)^nx_{0}^{3^n} \Rightarrow | f^n(x_{0})-0|=|x_{0}|^{3^n} < \epsilon $

Can you now find $N\in \mathbb{N}$ such that for $n \geq N$ it holds the inequality above?