Finding the limit of a recursive sequence $x_{n+1}=\frac{1}{2}(x_{n}+\frac{M}{x_{n}})$ 
Define the sequence $\{x_n\}_{n\in\mathbb{N}}$ by
  $$x_0=\frac{M}{2}, \qquad x_{n+1}=\frac{1}{2}\left(x_{n}+\frac{M}{x_{n}}\right)$$
  where $M\in\mathbb{R}$, $M\geq 0$. Find $\lim\limits_{x_n\to\infty}x_{n}$ and prove it exists.

My idea was, if I can prove that $x_{n}\geq x_{n+1}$ then I can say
that
$$x_{n}\geq x_{n+1}\iff x_{n}\geq\dfrac{1}{2}\left(x_{n}+\dfrac{M}{x_{n}}\right)\iff 2x_{n}\geq x_{n}+\dfrac{M}{x_{n}}\iff$$
$$ x_{n}\geq\frac{M}{x_{n}}\Longleftrightarrow x_{n}^{2}\geq M\Longleftrightarrow x_{n}\geq\sqrt{M}$$
but I couldn't find a way to prove that $x_{n}\geq x_{n+1}$.
 A: Assuming that $M>0$:
First we show that $(x_n)$ is bounded below:
Note $x_n$ satisfies  $x_n^2-2x_{n+1}x_n+M=0$. This quadratic has a real root, so $4x_{n+1}^2-4M\ge 0$. 
So $x_{n+1}^2\ge M$.
Next, we show that $(x_n)$ is eventually  decreasing:
For $n\ge 2$
$$x_n-x_{n+1}=x_n-{1\over2}\Bigl(x_n+{M\over x_n}\Bigr)={1\over2}{x_n^2-M\over x_n}\ge 0$$
Thus, since $(x_n)$ is bounded below by $\sqrt M$ and is eventually decreasing, $(x_n)$ converges to some $L$ with $L\ge \sqrt M>0$. But then
$$
x_{n+1}\rightarrow{1\over2}\Bigl( L+{M\over L}\Bigr).
$$
We then must have
$$
L={1\over2}\Bigl( L+{M\over L}\Bigr);
$$
which implies $L=\sqrt M$.
If $M=0$, then $(x_n)$ obviously converges to 0.
A: For any positive $M$, this method seems to converge and is decreasing towards a solution after the first iteration (for $x_0 > 2$, it is always decreasing, but for $x_0 < 2$, it takes one iteration to increase and then decreases towards the solution. I'll explain that below).  To have $x_n \ge x_{n+1}$, you need
$$
x_n \ge x_{n+1} = \frac 12 \left( x_n + \frac M{x_n} \right) \qquad \Longleftrightarrow \qquad x_n^2 - M \ge 0 \qquad \Longleftrightarrow \qquad x_n \ge \sqrt M
$$
if we assume that $x_0$ is positive to begin with. (Otherwise, one can prove that a convergent sequence following this pattern will be a root of $x^2 - M$ with $M$ negative, which means that the root $x$ would be complex, which is not what you expect I assume). We show further that $x_n \ge x_{n+1} \ge M$. Now we can see that 
\begin{align}
x_n^2 - 2 \sqrt M x_n + M = (x_n-\sqrt M)^2 \ge 0  \quad & \Longrightarrow \quad x_n + \frac{M}{x_n} \ge 2 \sqrt M \\
& \Longrightarrow \quad \frac 12 \left( x_n + \frac {M}{x_n} \right) \ge \sqrt M \\
\end{align}
and therefore we know that after the first iteration, $x_n$ will always be above $\sqrt M$, since the only thing I assumed here was that $x_0$ was positive (strict).
Therefore the sequence is converging because it decreases and it is bounded below by  $\sqrt M$. To find the limit, we have 
$$
x_n \to L \qquad \Longrightarrow \qquad x_{n+1}  \to L = \frac 12 \left( L + \frac ML \right) \qquad \Longrightarrow \qquad L = \sqrt M
$$
by knowing the limit has to be positive (you'll find a quadratic with two roots in there).
To explain the first iteration thing, you have to note that 
$$
x_0 = \frac M2 \ge \sqrt M \qquad \Longleftrightarrow \qquad (\sqrt M)^2 - 2 \sqrt M \ge 0
$$
which in turn is equivalent to $\sqrt M \ge 2$, which means $M \ge 4$ and $x_0 \ge 2$. 
Hope that helps,
A: if the limit $x$ exists, then $x=(1/2)(x+M/x)$ and $x=\pm\sqrt{M}$.  see
newton's method applied to $f(x)=x^2-M$
A: $\newcommand{\+}{^{\dagger}}
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$\ds{x_{0}\equiv{M \over 2},\quad
     x_{n+1}\equiv\half\,\pars{x_{n} + {M \over x_{n}}}.\qquad M \geq 0.}$

Let's $\ds{y_{n} = {x_{n} \over \root{M}}\quad\imp\quad
y_{n + 1} = \half\pars{y_{n} + {1 \over y_{n}}}\,,\quad y_{0} = {\root{M} \over 2}}$
  \begin{align}
{y_{n + 1} + 1 \over y_{n + 1} - 1}&
={\half\pars{y_{n} + 1/y_{n}} + 1 \over \half\pars{y_{n} + 1/y_{n}} - 1}
={y_{n}^{2} + 2y_{n} + 1 \over y_{n}^{2} - 2y_{n} + 1}
=\pars{y_{n} + 1 \over y_{n} - 1}^{2}
\end{align}

\begin{align}
{y_{n + 1} + 1 \over y_{n + 1} - 1}&
=\pars{y_{n - 1} + 1 \over y_{n - 1} - 1}^{4}
=\pars{y_{n - 2} + 1 \over y_{n - 2} - 1}^{8}=\cdots
=\pars{y_{0} + 1 \over y_{0} - 1}^{2^{n + 1}}
\end{align}

$$
y_{n + 1} = {\pars{y_{0} + 1}^{2^{n + 1}} + \pars{y_{0} - 1}^{2^{n + 1}}
\over \pars{y_{0} + 1}^{2^{n + 1}} - \pars{y_{0} - 1}^{2^{n + 1}}}
$$
  Now, you can take from here.

