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? 

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 $$ \sqrt{a} $$
