Convergence of the recursive sequence $x_n=\frac{x_{n-1}}2+\frac3{x_{n-1}},n\ge 1$ Consider the iterative scheme : 
$x_n=\frac{x_{n-1}}2+\frac3{x_{n-1}},n\ge 1$
with the initial point $x_0 \gt 0$. Then the sequence $\{x_n\}$
$(a)$ converges only if $x_o \lt 3$
$(b)$converges for any $x_0$
$(c)$does not converge for any  $x_0$
$(d)$converges only if $x_0\gt 1$
My attempt :
Since $x_0\gt 0$ ,we have $x_n\gt 0, \forall n\in \mathbb{N}$
Now , $x_n-x_{n-1}=\frac{x_{n-1}^2+6-2x_{n-1}^2}{x_{n-1}}=\frac{6-x_{n-1}^2}{2x_{n-1}}$
$\Rightarrow x_n\lt ,\gt x_{n-1}$ according as $x_{n-1}\gt ,\lt \sqrt{6}$.
Again,by the given recurrence relation ,
$x_{n-1}^2-2x_nx_{n-1}+6=0$ 
This is quadratic in $x_{n-1}$ and to have a real solution 
$4x_{n}^2-24\ge 0$ i.e $x_n\ge \sqrt6$ or $x_{n}\le -\sqrt6$ 
The latter is not possible, so $x_n\gt \sqrt6$
What does this mean?? Does this mean that even if $x_0$ be any value greater than zero and smaller than $\sqrt6$,  the sequence is ultimately monotone decreasing converging to $\sqrt6$ (limit obtained from the relation)? So, should my answer be $(b)$?
Please help me with your ideas. Thanks for your time.
 A: If $x_0>0$ we have $x_n>0$ and from here:
$$x_n-\sqrt6=\frac{(x_{n-1}-\sqrt6)^2}{2x_{n-1}}\geq0,$$ which by your work:
$$x_n-x_{n-1}=\frac{6-x_{n-1}^2}{2x_{n-1}}\leq0$$ gives that $x$ decreases.
Id est, our sequence is closed to $\sqrt6$ for any $x_0>0$.
A: Since finite terms of the sequence don't matter, so even if $x_0>\sqrt{6},$ the sequence still remains monotonically decreasing and bounded below and hence it converges for any $x_0.$
A: There is a nice substitution and technique applied here. It enables us to find a closed form of $x_n$ from which the correct alternative can be easily derived. Given
\begin{align*}
x_n=\frac{x_{n-1}}{2}+\frac{3}{x_{n-1}}\qquad\qquad n\geq 1
\end{align*}
we substitute
$\color{blue}{x_n=\sqrt{6}\,y_n}$
and we obtain
\begin{align*}
y_n&=\frac{y_{n-1}}{2}+\frac{1}{2y_{n-1}}\\
&=\frac{y_{n-1}^2+1}{2y_{n-1}}\tag{1}
\end{align*}
From (1) we get
\begin{align*}
y_n-1&=\frac{\left(y_{n-1}-1\right)^2}{2y_{n-1}}\\
y_n+1&=\frac{\left(y_{n-1}+1\right)^2}{2y_{n-1}}\\
\frac{y_n-1}{y_n+1}&=\left(\frac{y_{n-1}-1}{y_{n-1}+1}\right)^2=\cdots=\underbrace{\left(\frac{y_0-1}{y_0+1}\right)^{2^n}}_{C_n}\tag{2}
\end{align*}
We can now calculate $y_n$ from (2) and obtain a closed form of $x_n$ this way:
\begin{align*}
y_{n}-1&=C_n\left(y_n+1\right)\\
y_n\left(1-C_n\right)&=1+C_n\\
y_n&=\frac{1+C_n}{1-C_n}\qquad\Rightarrow\qquad
\color{blue}{x_n=\sqrt{6}\,\frac{1+\left(\frac{x_0-\sqrt{6}}{x_0+\sqrt{6}}\right)^{2^n}}
{1-\left(\frac{x_0-\sqrt{6}}{x_0+\sqrt{6}}\right)^{2^n}}\qquad n\geq 0}\tag{3}
\end{align*}

We conclude from (3)
\begin{align*}
\left\{x_n\right\}_{n\geq 0}\ \mathrm{convergent}\quad&\Leftrightarrow\quad\left|\frac{x_0-\sqrt{6}}{x_0+\sqrt{6}}\right|<1\\
&\Leftrightarrow\quad\left|x_0-\sqrt{6}\right|<\left|x_0-\left(-\sqrt{6}\right)\right|\\
&\,\,\color{blue}{\Leftrightarrow\quad x_0>0}
\end{align*}
and (b) is the correct answer.

