Neighborhoods necessary for convergence of a sequence. This is a follow-on to a question I posed yesterday. It pertains to
Edwards's Advance Calculus of Several Variables, Chapter III.
Given a real-valued function $f(x)$, continuously differentiable
and monotonic on the interval $[a,b]\subset\mathbb{R}$ such that
it changes sign between $a$ and $b$, the sequence${\{x_{n}\}_{n=0}^{\infty}}$
defined by $x_{n+1}=x_{n}-\frac{f(x_{n})}{f^{\prime}(x_{n})}$ and
$x_{0}\in[a,b]$ converges to a root of $f$. Edwards doesn't offer
a proof, and the fact is only used as a motivating example. 
Using a diagram, he demonstrates that the modified sequence $x_{n+1}=x_{n}-\frac{f(x_{n})}{f^{\prime}(x_{0})}$
may fail to converge. 
I concocted the example that $f(x)=\sin x$ will fail to converge
if $x_{0}=\frac{\tan x_{0}}{2}$ because $x_{1}=x_{0}-\frac{\sin x_{0}}{\cos x_{0}}=-x_{0}$
so $x_{2}=-x_{0}+\frac{\sin x_{0}}{\cos x_{0}}=x_{0}$. 
Edwards uses the contraction mapping theorem to prove that for $f^{\prime}>0$
and $M=\max f^{\prime}([a,b])$, the modification $x_{n+1}=x_{n}-\frac{f(x_{n})}{M}$
will produce a convergant sequence under the above stated conditions.
He then uses the contraction mapping theorem to show that it is possible
to solve $f[x]=y$ for $x$ when $y$ is given ``near'' some $f[x_{0}]=y_{0}$
with known values for $x_{0}$, $f^{\prime}(x_{0})$ and $y_{0}$.
In this case he uses $x_{n+1}=x_{n}-\frac{f(x_{n})-y}{f^{\prime}(x_{0})}$
as the generic term in the sequence. But I observe that $g(x)=f(x)-y$
leads to $x_{n+1}=x_{n}-\frac{g(x_{n})}{g^{\prime}(x_{0})}$. The
sequence of these terms will converge to the same value of $x$, where
$f(x)=y$ and $g(x)=0$. So $x_{n+1}=x_{n}-\frac{f(x_{n})-y}{f^{\prime}(x_{0})}$
is suseptible to the same vulnerability as is $x_{n+1}=x_{n}-\frac{f(x_{n})}{f^{\prime}(x_{0})}$.

Theorem 1.3 Let $f:\mathbb{R\rightarrow\mathbb{R}}$ be a $\mathscr{C^{1}}$
  function such that $f(a)=b$ and $f^{\prime}(a)\ne0$. Then there
  exist neighborhoods $\mathit{U}=[a-\delta,a+\delta]$ of $a$ and
  $\mathit{V}=[b-\epsilon,b+\epsilon]$ of $b$ such that, given $y_{*}\in\mathit{V}$,
  the sequence ${\{x_{n}\}_{0}^{\infty}}$ defined inductively
  by 
  $$x_{0}=a, x_{n+1}=x_{n}-\frac{f(x_{n})-y}{f^{\prime}(a)}$$
   converges to a (unique) point $x_{*}\in\mathit{U}$ such that $f(x_{*})=y_{*}$.
PROOF Chose $\delta>0$ so small that $|f^{\prime}(a)-f^{\prime}(x)|\le\frac{1}{2}|f^{\prime}(a)|$
  if $\mathit{x\in U}=[a-\delta,a+\delta]$. Then let $\epsilon=\frac{1}{2}\delta|f^{\prime}(a)|$. ...

The proof uses the contraction mapping theorem with these stated assumptions.
My question is: how does restricting the domain and range in this
way eliminate the previously demonstrated pitfall?  Put another way: what are the least restrictive neighborhoods necessary for convergence?
The generic problematic condition appears to be $-f(x)=f(-x)$ with
 $2x_{n}f'(x_{n})=f(x_{n})$.
 A: The problem arises when the mapping is symmetric about the origin.
Taking the case where $f^{\prime}[x]>0$, the ``loop'' condition
occurs when $2xf^{\prime}[x]=f[x]$. This is because 
$x_{n+1}=x_{n}-\frac{f[x_{n}]}{f^{\prime}[x_{n}]}$, $x_{n+1}=x_{n}-2x_{n}=-x_{n}$.
By the symmetry of the function, that leads to $x_{n+2}=x_{n}$, etc.
Let $x_{p}$ be that value of $x$ where the loop occurs.
Note that the chord drawn from the origin to $\{x_{p},f[x_{p}]\}$
has a slope $f[x_{p}]/x_{p}=2f^{\prime}[x_{p}]<f^{\prime}[0]$. So
$\frac{1}{2}f^{\prime}[0]>f^{\prime}[x_{p}]$. By inspection it is
apparent that the graph is concave-down to the right of the origin.
So for $0<x_{n}<x_{p}$, $f^{\prime}[x_{n}]>f^{\prime}[x_{p}]$ and
$x_{n+1}$will safely land somewhere between $-x_{p}$ and the origin.
From this, it can be seen that restricting $|x|<\delta\implies|f^{\prime}[x]-f^{\prime}[0]|\le\frac{1}{2}|f^{\prime}[0]|$
avoids the loop condition.
I know this is not a rigorous argument.  I will post a follow-on question, time permitting.

An example showing that a sequence with $x_{0}\in[a,b]$ might not converge. 
