Can somebody help me with following: Let $f(x)=x^5-x+1$ and let $x$ be the limit of the Newton series $x_n=x_{n-1}-\frac{f(x_{n-1})}{f'(x_{n-1})}$. How do I prove that $x$ is a root of $f$, i.e. $f(x)=0$.
I have tried to proove that $|f(x_N)|<\frac{1}{n}$ for some $N$ big enough, by using the triangle inequality, but without succes.
hint
Assume that $$\lim_{n\to+\infty}x_n=x \in \Bbb R$$
$ f $ and its derivative $ f' $ are continuous at $ \Bbb R $.
So
$$\lim_{n\to+\infty}x_{n-1}=x$$ $$\lim_{n\to+\infty}f(x_{n-1})=f(x)$$ $$\lim_{n\to+\infty}f'(x_{n-1})=f'(x)$$
and $$\lim_{n\to+\infty}x_n=x=x-\frac{f(x)}{f'(x)}$$
By MVT, it is easy to check that the root of $ f(x)=0 $ is in $ [-2,-1]$, so $ f'(x)\ne 0$.
