For what values of $x_0$, $x_n:=\frac{x_{n-1}}{2}+\frac{3}{x_{n-1}}$ where $n \geq 1$ with $x_0>0.$ converges? $x_n:=\frac{x_{n-1}}{2}+\frac{3}{x_{n-1}}$ where $n \geq 1$ with $x_0>0.$
For what values of $x_0$ does this sequence converge?

If $(x_n) \to L$, then assuming $L>0$, I have
$L=\lim_\limits{n \to \infty} x_n=
\lim_\limits{n \to \infty}\left[\frac{x_{n-1}}{2}+\frac{3}{x_{n-1}}\right]=
\lim_\limits{n \to \infty}\left[\frac{x_{n-1}^2+6}{2x_{n-1}}\right]=
\frac{\lim_\limits{n \to \infty}x_{n-1}^2+6}{\lim_\limits{n \to \infty}2x_{n-1}}=
\frac{L^2+6}{2L} \implies 2L^2=L^2+6 \implies L=\sqrt{6}.$
How to determine what values of $x_0$ will give the convergence of $(x_n)$?

$x_n > 0 $ for all $n$
$x_{n} \ge \sqrt{6}$ for all $n \ge 1$ by the AM-GM inequality
$x_n=\frac{x_{n-1}}{2}+\frac{3}{x_{n-1}}=x_{n-1}$ when $x_{n-1} = \sqrt{6}$
$x_n=\frac{x_{n-1}}{2}+\frac{3}{x_{n-1}}<x_{n-1}$ when $x_{n-1} > \sqrt{6}$
Basically its going decrease till $x_n = \sqrt{6}$
Therefore, it converges to $\sqrt{6}$ irrespective of the value of $x_0$
