Most analyses of Newton's method assume that $\displaystyle f'(a) \neq 0$,

where a is the solution of the nonlinear equation.

I'd like to investigate what happens in the special case where $\displaystyle f'(a) = 0$ but $\displaystyle f''(a) \neq 0$

I'd like to show that:

$\displaystyle $\lim_{x \to a}{f(x) \over [f'(x)]^2} = {1 \over 2f''(a)}$$

I'm pretty sure I need to use L'H^ospital's rule, but I haven't figured out how to make it work out...I keep getting the numerator to be zero, but I need to to be one.

After this is done, I have to show that in this case, the iteration function $\displaystyle g(x) = {x - {f(x) \over f'(x)}}$

generated by Newton's method is a contraction. Thanks in advance for any tips to get me going on this!