Showing the sequence converges to the square root For any $a > 0$, I have to show the sequence $x_{n+1}$ $=$  $ \frac 12$($x_n+ $ $ \frac {a} {x_n}$)
converges to the square root of $a$ for any $x_1>0$.
If I assume the limit exists ( denoted by $x$) then,
$x$ $=$  $ \frac 12$($x+ $ $ \frac {a} {x}$) can be solved to $x^2 = a$
How could I show that it does exist? 
 A: As mentioned in the comments, we need to show that the sequence is monotonic and bounded.
First, we observe that
$$
x_n-x_{n+1}=x_n-\frac12\Bigl(x_n+\frac a{x_n}\Bigr)=\frac1{2x_n}(x_n^2-a).
$$
Secondly, we obtain that
\begin{align*}
x_n^2-a
 &=\frac14\Bigl(x_{n-1}+\frac a{x_{n-1}}\Bigr)^2-a\\
 &=\frac{x_{n-1}^2}4-\frac a2+\frac{a^2}{4x_{n-1}^2}\\
 &=\frac14\Bigl(x_{n-1}^2-2a+\frac{a^2}{x_{n-1}^2}\Bigr)\\
 &=\frac{1}{4}\Bigl(x_{n-1}-\frac a{x_{n-1}}\Bigr)^2\\
 &\ge0.
\end{align*}
Hence, $x_n\ge x_{n+1}$ and $x_n$ is bounded from below since $x_n^2\ge a$ for each $n\ge2$.
Monotonic and bounded sequence converges. Denote the limit of the sequence $x=\lim_{n\to\infty}x_n$. Then we have that
$$
x=\frac12\Bigl(x+\frac ax\Bigr)\quad\iff\quad x=\sqrt a.
$$
A: That looks a lot like the well known method for computing square roots. It is derived by using Newton's on
$$ x^2 - a = 0 $$
$$ x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)} $$
$$=x_n-\frac{x_n^2-a}{2x_n}$$
$$=\frac{1}{2}\left(x_n+\frac{a}{x_n}\right) $$
If it converges it will converge to the $$ \sqrt{a} $$
