Newton's method

We assume that f(x), f'(x) and f''(x) are continuous in [a,b], and that for some [tex]\alpha[/tex] [tex]\in [/tex] (a,b) , we have [tex] f( \alpha )= 0 [/tex] and [tex]f'( \alpha ) \neq 0 [/tex]. We show that if [tex]x_{0}[/tex] is chosen close enough to [tex]\alpha[/tex], the iterates
[tex]x_{n+1} = x_{n}- \frac{f(x_{n})}{f'(x_{n})}[/tex]
converge to [tex]\alpha[/tex].

I tried to use Taylor's expansion for [tex] f( \alpha )[/tex] (centered at [tex]x_{n}[/tex]), and I got to this expression

[tex]lim_{n \rightarrow \infty} (\alpha -x_{n+1})= lim_{n \rightarrow \infty} - \frac{1}{2} f''(c) \frac{( \alpha - x_{n} )^{2}}{f'(x_{n})}[/tex]

where [tex] c \in ( \alpha , x_{n} )[/tex]
and I guess I want the right hand side to be 0 to get to the answer. But I am not sure how to prove this.

 

I am sorry, I fixed the latex part. (hopefully it's readable now)

By using Taylor's expansion, I got an expression, and by letting n go to infinity, if I could make the RHS become 0, I guess I would have the answer to the problem.
The thing is ,to do that, I am not sure how to use the fact that [tex] x_{0}[/tex] is chosen close enough to [tex]\alpha[/tex].

 

I am just trying to prove that in this case,

[tex] lim_{n \rightarrow \infty} (\alpha -x_{n+1})=0 [/tex]

so then the iterate [tex] x_{n+1} [/tex] converges to [tex]\alpha[/tex].

 

In the hypothesis, we have that f ''(x) is continuous in [a,b], so it exists.
I guess there must be a reason why they wanted that hypothesis in the problem...

What would be another way to approach this problem?