I'm trying to see if the iterative method $x_n=g(x_{n-1})$ where $g(x)=2\sqrt{x-1}$ will converge to $2$, if I take $x_0$ that is sufficiently close to $2$.
Indeed notice that $g(2)=2$. and we have a theorem that states that if $|g'(2)|<1$ then there is a neighborhood of $2$ such that if we take an $x_0$ from that neighborhood, the method will converge to $2$.
We also know that if $|g'(2)|>1$ then there is no such neighborhood.
But notice now that $g'(x)=\frac{1}{\sqrt{x-1}}$, and so $g'(2)=1$.
What can we say about this situation? will it converge?