limit of recursive function I've been struggling with this for a while now and can't seem to get it right.
Let $f(x)= \frac{x^2+3}{2x}$ from $0 $ to $\infty $. Show that $0\lt x \lt \sqrt{3}$ implies $\sqrt{3} \lt f(x)$ 
and show if we define $x_{n+1} = f(x_n)$ then 
$$ \lim_{n\to\infty}x_n =\sqrt{3}$$ when $x_0 \gt 0$
It's easy to check that $f(\sqrt{3}) = \sqrt{3}$ but limits is not one of my strong points. I've also shown that $f(x) < 3/x$.

Note that $f(x)-\sqrt{3} = {(x-\sqrt{3})^2 \over 2x} \ge 0$. So we have $x_k \ge \sqrt{3}$ for $k \ge 1$.
Since $f(x)-\sqrt{3} = {x-\sqrt{3} \over 2x} (x-\sqrt{3})$, this suggests that we look at $\phi(x) = {x-\sqrt{3} \over 2x}$. It is straightforward to show that $\phi$ is strictly increasing for $x >0$, and $\lim_{x \to \infty} \phi(x) = {1 \over 2}$. Furthermore, $\phi(x) \ge 0$ for $x \ge \sqrt{3}$.
Hence we have $f(x) -\sqrt{3} \le {1 \over 2} (x-\sqrt{3})$ for $x \ge \sqrt{3}$.
In particular, this gives $0 \le x_{n+1}-\sqrt{3} \le {1 \over 2} (x_n-\sqrt{3})$ for $n \ge 1$ and so
$x_n-\sqrt{3} \le {1 \over 2^{k-1}} (x_1-\sqrt{3})$.
