limits - 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?

Answer

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

$$x=\frac12\Bigl(x+\frac ax\Bigr)\quad\iff\quad x=\sqrt a.$$