Everywhere I read a proof of Newton's method (that one with $x_{n+1} = x_n-f(x)/f'(x)$), it is used that the derivative of the function does not vanish at the root we are looking for.
There is a post here on the website where people say that, in this case, the convergence is usually slow, but I couldn't find anywhere if the hypothesis is necessary or not for the melhod to work, and I coudn't find an example by myself, if the root is isolated.
Is it possible to provide an example of a function, say $f: (-1,1)\to \mathbb{R}$ such that $f$ is of class $\mathcal{C^2}$ on $(-1,1)$ (i.e., $f''$ exixts and is continous in the interval), $f(0)=f'(0)=0$, with $0$ being an isolated root of $f$, and for all $\delta>0$, starting with $x_0\in (-\delta,\delta) \backslash{0}$, the method does not converge to the root $0$?
If it is not necessary that $f'(0)\neq 0$, where can I find a proof of the convergence of the method that does not use this?
I will appreciate any help!