Newton's method for $f(x)=\arctan (x)$ I am trying to apply Newton's method for $f(x)=\arctan (x)$. There is already a post showing how to solve this:

As $g(x)$ is an odd function, if $x_0$ is a root, then $x_1=-x_0$ is a
root as well.
The periodic sequence is
$(x_n)=x_0,-x_0,x_0,\ldots$
Apply Newton method to find the root of $f(x)=\arctan x$
$x_1=x_{0}-\frac{f(x_{0})}{f'(x_{0})}= -x_0$
$x_1=-x_0$ so we get
$-x_0=x_0-\frac{\arctan x_0}{\frac{1}{x_0^2+1}}$
expand
$2 x_0=\left(x_0^2+1\right) \arctan x_0$
Which proves that $\pm x_0$ are roots of $g(x)=\frac{2x}{x^2+1}-\arctan x$
$x_0\approx 1.39175$

My question is, what happens if we apply Newton’s method for the same problem with initial guesses $x_{0}+\epsilon$ and $x_{0}-\epsilon$?
 A: For $x_0-\epsilon$ you will converge to the actual root $x=0$, and for $x_0+\epsilon$ you get a sequence that diverges. In particular, the values are alternatingly positive and negative, and their size/modulus is increasing to infinity.
This can be seen as follows:
Let $y_0 = x_0+\epsilon>0$. Then the first iteration gives $y_1=y_0-(1+y_0^2)\arctan(y_0)$ and the size/modulus of $y_1$ is $(1+y_0^2)\arctan(y_0)-y_0$ because $y_1$ is negative.
One can show that the function $h(y)= \frac{2y}{1+y^2}-\arctan(y)$ has $h(y)<0$ for $y>x_0$, and one can show that $h(y) < 0$ is equivalent to $y < (1+y^2)\arctan(y)-y$. So we have that $|y_1|>y_0$, and similarly $|y_n|>|y_{n-1}|$ for all $n$.
To actually show that for $y_0>x_0$ we have $|y_n|\to\infty$, try proving that $|y_{n+1}|-|y_n| > |y_n|-|y_{n-1}|$, which implies that $|y_n|=\sum_{k<n}|y_{k+1}|-|y_k|\to\infty$.
Similarly, if $0<y_0=x_0-\epsilon<x_0$ one can show that $h(y_0) > 0$ and thus $|y_1| < y_0$. and similarly $|y_{n+1}|<|y_n|$.
To actually show that $y_k\to0$ suppose that $\lim_{k\to\infty}|y_k|\neq0$. Then this limit must be fixed in some sense by the newton iterations, which is not possible.
