Newton's Function Confusion "Suppose that $r$ is a double root of $f(x) = 0$, that is $f(x)=f'(x)=0$, $f''(x) \neq 0$, and suppose that $f$ and all derivatives up to and including the second are continuous in some neighborhood of $r$. Show that $\epsilon_{n+1} \approx \frac{1}{2}\epsilon_{n}$ for Newton's method and therby conclude that the rate of convergence is $\textit{linear}$ near a double root. (If the root has multiplicity $m$, then $\epsilon_{n+1} \approx \left [ \frac{(m-1)}{m}\right ]\epsilon_n $)".
I'm a good amount of confused on this problem. So I know that $\epsilon_n = -\frac{f(x_n)}{f'(x_n)}$  (our error) and that a function with a double root can be written as $f(x) = (x-r)^2g(x)$ where $r$ is our double root.
I just don't really know how to do this / start this. If I calculate $\epsilon_n$, I get $-\frac{(x-r)^2g(x)}{2(x-r)g(x) + (x-r)^2g'(x)}$, but what use is that? I think I need a decent push forward in the right direction. Help?
Maybe, the $x$'s in my $\epsilon_n$ calculation are supposed to be $x_n$'s? Since we know that as $x_n \to r$, $(x_n - r) \to 0$. Then we could do something with that? That would just make it $0$ though which doesn't help us.

This solution was shown to me by a friend. I understand now!
The solution is as follows:
Let us state the statement of Newton's Method: $x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$
We know that (our initial guess) is $x_{n} = r + \epsilon$ where $r$ is our double-root and $epsilon$ is our $\textit{very}$ small error. Then our error is $\epsilon = r - x_n$. 
Our statement for linear convergence is $|e_{n+1}| \leq |e_n|C$ where $C\in \left[ 0, 1\right )$
We can then write $\underbrace{r - x_{n+1}}_{e_{n+1}} = \underbrace{r - x_n}_{e_n}  + \frac{f(x_n)}{f'(x_n)}$ to get $e_{n+1} = e_n + \frac{f(x_n)}{f'(x_n)}$.
From the Taylor Series around $r$, we can write a function with the double root $r$ as $f(x) = (x-r)^2g(x)$ where $g(x)$ is defined in: $(x-r)^2\left [ \underbrace{\frac{f''(r)}{2!} + \frac{f'''(r)(x-r)}{3!} + \dots}_{g(x)} \right ]$ since $f(r) = 0$ and $f'(r) = 0$.
We then calculate $\frac{f(x_n)}{f'(x_n)} = \frac{(x_n - r)^2g(x_n)}{2(x_n - r)g(x_n) + (x_n -r)^2g'(x_n)} = \frac{(x_n - r)}{2 + \frac{g'(x)}{g(x)}}$.
We can then make the appropriate substitutions $\frac{-e_n}{2 - e_n\frac{g'(x)}{g(x)}}$
$\lim_{n \to \infty} \frac{g'(x)}{g(x)} = \frac{g'(r)}{g(r)} = K$ where $K$ is a constant.
Then $e_{n+1} = e_n - \frac{e_n}{2 - e_nK} = e_n\frac{(1-e_nK)}{2-e_nK}$.
When we look at $n$ as it approaches infinity, then $\frac{(1-e_nK)}{2-e_nK} \to \frac{1}{2}$. This leaves us with the conclusion that $e_{n+1} \approx \frac{1}{2}e_n$
