I have been studying some numerical analysis and am currently looking at the order of convergence for Newton's method for finding roots with multiplicity more than 1; however, I have some misunderstandings regarding how this works.
For example, say $f(x)$ has a root of multiplicity $2$ at $x = 1$. We know that $x_n=x_{n-1}-\frac{f(x)}{f'(x)}$, so to apply the fixed point theorem, we have $g(x)=x-\frac{f(x)}{f'(x)}$. Similarly, the modified Newton's method will be $g(x)=x-2\frac{f(x)}{f'(x)}$.
Here's the part that I don't understand: $g(1)$ does not exist, but $\lim_{x \to 1} g(x) = 1$. Likewise, for the standard Newton's method, $g'(1)$ does not exist, but $\lim_{x \to 1} g'(x) = 1/2$, and for the modified Newton's method, $g'(1)$ and $g''(1)$ do not exist, but $\lim_{x \to 1} g'(x) = 0$ and $\lim_{x \to 1} g''(x) = 1/6$.
These are all of the numbers one would expect to show that the standard Newton's method has linear convergence, while the modified Newton's method has quadratic convergence, but how can we actually use the fixed point theorem if these values of $g$ and $g'$ do not exist? Is $g(1)$ really a fixed point, and is $\left | g'(x) \right | < 1$ around $x = 1$ true if it does not exist there? And is it appropriate to simply substitute in the limit values into the Taylor expansion of $g(x_{n+1})$ around $x=1$?
I'll also note that when I numerically applied these methods, I observed the expected linear and quadratic convergence, so I feel as though I am misunderstanding something about the fixed point theorem.
Any help would be greatly appreciated!