Convergence of the sequence $x_{k+1}=\frac{1}{2}(x_k+\frac{a}{x_k})$ 
Consider the sequence $x_{k+1}=\frac{1}{2}(x_k+\frac{a}{x_k}), a\gt 0, x\in\mathbb{R}$. Assume the sequence converges, what does it converge to?

I'm having trouble seeing how to start,
Any help would be appreciated
Thanks
 A: The number $x$ to which it converges should satisfy $x =\frac12\left(x+\frac a x\right)$.  Solve that equation for $x$.  (It may help to begin by multiplying both sides by $x$, thereby getting rid of the fraction.)
A: If $x_n \to g$, then, if $g \ne 0$, we have $\frac{1}{2}(x_n + \frac{a}{x_n}) \to \frac{1}{2}(g+\frac{a}{g})$. We also have $x_{n+1} \to g$, so since $x_{n+1} = \frac{1}{2}(x_n + \frac{a}{x_n})$, we have that $g = \frac{1}{2}(g + \frac{a}{g})$.
A: This sequence has a closed form.
First, if $x_0=\sqrt{a}$, $x_n=\sqrt{a}$ for all $n$, and the sequence is constant. We will thus assume $x_0\ne\sqrt{a}$ in the following.
Now, given $x_n\ne\sqrt{a}$ for some $n$, we have
$$x_{n+1}-\sqrt{a}=\frac{1}{2}\left(x_n+ \frac{a}{x_n} \right) - \sqrt{a}=\frac{1}{2}\left( \sqrt{x_n} - \frac{\sqrt{a}}{\sqrt{x_n}}\right)^2 > 0$$
Hence $x_n>\sqrt{a}$ for all $n>0$. To simplify a bit, we can safely assume that $x_0 > \sqrt{a}$ too.
We can thus write $x_n = \sqrt{a} \ \coth{t_n}$. Then
$$x_{n+1} = \frac{1}{2}\left( \sqrt{a} \ \coth{t_n} + \frac{a}{\sqrt{a} \ \coth{t_n}} \right) = \frac{1}{2}\sqrt{a} \left( \coth{t_n} + \mathrm{th}\ t_n\right) = \sqrt{a} \ \coth \ 2 t_n$$
We have thus $t_{n+1}=2\ t_n$, and $t_n= 2^n t_0$, then for $x_n$ :
$$x_n=\sqrt{a} \coth \left( 2^n \arg \coth \frac{x_0}{\sqrt{a}}\right)$$
And the value inside parentheses tends to infinity, so $\lim(x_n)=\sqrt{a}$.
