Is the fixpoint $x=0$ stable, attracting, repelling for $f(x)= x\sin(1/x)$? I'm working on exercise 1.5.8 from Goodson's Chaotic Dynamics.
We have the function $f(x) = x\sin(\frac{1}{x})$ for $x\neq 0$ and $f(0)=0$. I have already proven that $x=0$ is not an isolated point. Now I need to prove if $x=0$ is a stable, attracting or repelling point.
I know I have to use the definition
$\forall\epsilon>0, \exists\delta>0, |x-0|<\delta\implies |f^n(x)-0|<\epsilon\forall n\in\mathbb{N}$.
I think $x=0$ is an unstable point, so I would have to find a counterexample to the definition given above. However I've got no idea how to solve this problem. Can somebody help me?
 A: You have to be careful here with which type of stability you are considering.
From scholarpedia, we have Lyapunov stability

Lyapunov stability:
$x^e$ is a stable equilibrium if for every neighborhood U of $x^e$ there is a neighborhood V ⊆ U such that every solution x(t) starting in V remains in U for all t≥0. Notice that x(t) need not approach $x^e$.

and asymptotic stability (among others).

Asymptotic stability:
An equilibrium $x^e$ is asymptotically stable if it is Lyapunov stable and additionally V can be chosen so that |x(t)−$x^e$|→0 as t→∞ for all x(0) ∈ V.

The definition you have stated is Lyapunov stability. For this system, $x=0$ is Lyapunov stable, but it is not asymptotically stable.
Lyapunov stability can be established by noticing that $|f(x)| \leq |x|$ for all $x$, and taking $\delta < \epsilon$. Then, you can directly see that $|f^n(x)-0|  \leq |x| < \delta < \epsilon$ for all n.
Asymptotic stability can be disproven by noticing that $f(x)$ has a fixed point or 2-cycle whenever $\sin\left(\frac{1}{x}\right) = \pm 1$. As this occurs whenever $x = \frac{2}{k\pi}$ for $k$ odd, there are infinitely many fixed points arbitrarily close to $0$ (they are counterexamples that cannot converge to $0$).
