Calculating square roots using the recurrence $x_{n+1} = \frac12 \left(x_n + \frac2{x_n}\right)$ 
Let $x_1 = 2$, and define $$x_{n+1} = \frac{1}{2} \left(x_n + \frac{2}{x_n}\right).$$ Show that $x_n^2$ is always greater than or equal to $2$, and then use this to prove that $x_n − x_{n+1} ≥ 0$. Conclude that $\lim x_n =
\sqrt2$.

My question:
So I know how to do this problem but I don't know how to prove that $x_n >0$ for all $n\in \mathbb{N}$.
 A: As, @CiaPan said in comments, you need to prove that the limit to this sequence is exactly $\sqrt 2$.
So, here is a different approach to reach the goal.
Firstly, by induction, prove that $$\color{blue}{\frac {x_n-\sqrt 2}{x_n+\sqrt 2}=\left(\frac{x_1-\sqrt 2}{x_1+\sqrt 2}\right)^{2^n}}.$$
So, firstly, for $n=1$, we have, $$x_2=\frac 12\left(x_1+\frac 2{x_1}\right)\\\implies\frac {x_2}{\sqrt 2}=\frac {x_1^2+2}{2\sqrt2 x_1}.$$
Now, use inverse of componendo-dividendo, $$\frac {x_2-\sqrt 2}{x_2+\sqrt 2}=\left(\frac{x_1-\sqrt 2}{x_1+\sqrt 2}\right)^2.$$
So, for $n=1$, our claim holds.
Now, suppose, it is true for $n=k$, i.e. $$\frac {x_{k+1}-\sqrt 2}{x_{k+1}+\sqrt 2}=\left(\frac{x_1-\sqrt 2}{x_1+\sqrt 2}\right)^{2^{k+1}}.\tag 1$$
Let's see for $n=k+1$. So, we have $$x_{k+2}=\frac 12\left(x_{k+1}+\frac 2{x_{k+1}}\right),$$ and similar approach to the case $n=1$ takes us to $$\frac {x_{k+2}-\sqrt 2}{x_{k+2}+\sqrt 2}=\left(\frac{x_{k+1}-\sqrt 2}{x_{k+1}+\sqrt 2}\right)^2=\left(\frac{x_1-\sqrt 2}{x_1+\sqrt 2}\right)^{2^{k+2}}[from\;\;(1)].$$ 
So, for all $n$, we have $$\color{blue}{\frac {x_n-\sqrt 2}{x_n+\sqrt 2}}=\left(\frac{x_1-\sqrt 2}{x_1+\sqrt 2}\right)^{2^n}=\color{blue}{\left(\frac{2-\sqrt 2}{2+\sqrt 2}\right)^{2^n}}[\text{since }\, x_1=1].$$
So, as $\color{red}{n\to \infty}$, the right side, i.e. $\color{red}{\left(\frac{2-\sqrt 2}{2+\sqrt 2}\right)^{2^n}\to 0}$, so, $$\color{red}{\frac {x_n-\sqrt 2}{x_n+\sqrt 2}\to 0},$$ so $$\color{red}{x_n-\sqrt 2\to 0\\\implies x_n\to \sqrt 2}. $$
So, $${\color{fuchsia}{\lim x_n= \sqrt 2}}$$ 
A: If $x_n>0$ then $x_{n+1}$ cannot be negative. Since $x_1=2>0$, the claim follows.
A: From
$x_{n+1} 
= \frac{1}{2} \left(x_n + \frac{2}{x_n}\right)
$
we get, squaring,
$x_{n+1}^2 
= \frac{1}{4} \left(x_n^2 +4+ \frac{4}{x_n^2}\right)
$
so that
$\begin{array}\\
x_{n+1}^2 -2
&= \frac{1}{4} \left(x_n^2 -4+ \frac{4}{x_n^2}\right)\\
&= \frac{1}{4} \left(x_n - \frac{2}{x_n}\right)^2\\
&= \frac{1}{4} \left(\frac{x_n^2-2}{x_n}\right)^2\\
\end{array}
$
From this,
with the initial
$x_1$,
it is straightforward
to show that
$x_n \to \sqrt{2}$
and that the convergence is quadratic
(as it should be,
since this is Newton's iteration).

More generally,
using copy, paste, and edit:
From
$x_{n+1} 
= \frac{1}{2} \left(x_n + \frac{a}{x_n}\right)
$
we get, squaring,
$x_{n+1}^2 
= \frac{1}{4} \left(x_n^2 +2a+ \frac{a^2}{x_n^2}\right)
$
so that
$\begin{array}\\
x_{n+1}^2 -a
&= \frac{1}{4} \left(x_n^2 -2a+ \frac{a^2}{x_n^2}\right)\\
&= \frac{1}{4} \left(x_n - \frac{a}{x_n}\right)^2\\
&= \frac{1}{4} \left(\frac{x_n^2-a}{x_n}\right)^2\\
\end{array}
$
From this,
with the initial
$x_1$,
it is straightforward
to show that
$x_n \to \sqrt{a}$
and that the convergence is quadratic
(as it should be,
since this is Newton's iteration).
A: Once you know that the sequence is monotonically decreasing and bounded from below, it converges. 
Let $u$ be the limit.  Then $u=\dfrac 12(u+\dfrac 2u)\implies 2u=u+\dfrac 2u\implies u^2=2$.
Now use that $u\ge0$ (see the accepted answer).
