Suppose $f$ is some real analytic function. How can I show that every root of $f$ with multiplicity $ \geq2$ is an attracting fixed point of Newtons method? What I've tried is to use the criterion for an attracting fixed point $\varepsilon$ : $′(\varepsilon)<1$ . Plugging that in the Newton operator you get $_(\varepsilon)′=\frac{(\varepsilon)(\varepsilon)″}{((\varepsilon)′)^2}<1 $. Since this has a removable discontinuity at $\varepsilon$ I came to the idea of showing that the limit as $→\varepsilon$ the Newton operator is less than one. Here I didn't come any further.
I'm trying to show the same thing for complex functions too. Thanks for help in advance!