How can I show that the convergence of of Newton's method can't be of order $k$ where $k > 2$?
I have that $$\lim_{n \to \infty}\frac{x^{n+1}-\alpha}{(x^n-\alpha)^2}=\frac{f''(\alpha)}{2f'(\alpha)}.$$
And that $$f''(\alpha) \neq 0$$
I know that I need to show that there doesn't exist a constant $C > 0$ such that $|x^{n+1}-\alpha| \leq C|x^n-\alpha|^k$ for $k>2$.
Could anyone give me some pointers on how I could show this?
If $|x^{n+1}-\alpha| \le C|x^n - \alpha|^k$ where $k > 2$, then $$\frac{x^{n+1} - \alpha}{(x^n - \alpha)^2} \to 0,$$ which contradicts $f''(\alpha) \ne 0$.