Finding $\lim_{n \to \infty} x_n$ Let $x$ be a positive real number. A sequence $x_n$ of real numbers is defined as follows:
$$x_1=\frac 12\left(x+\frac5x\right)$$ and $$x_{n+1}=\frac 12\left(x_n+\frac 5{x_n}\right) \quad\forall n\geq 1$$
The question is to show that $$\frac{x_n-\sqrt 5}{x_n+\sqrt 5}=\left(\frac{x-\sqrt 5}{x+\sqrt 5}\right)^{2^{n}}$$ and hence or otherwise evaluating $$\lim_{n \to \infty} x_n$$

To show first part I tried using induction. It is clear that $n=1$ is true. Let it be true for $n=k$. So $$\frac{x_n-\sqrt 5}{x_n+\sqrt 5}\left(\frac{x-\sqrt 5}{x+\sqrt 5}\right)^2=\left(\frac{x-\sqrt 5}{x+\sqrt 5}\right)^{2^{n+1}}$$ 
I tried manipulating this but could not proceed. Any ideas? Thanks.
 A: If it true for $n=k$ then we have $\frac{x_k - \sqrt{5}}{x_k + \sqrt{5}} = \left(\frac{x-\sqrt{5}}{x+\sqrt{5}}\right)^{2^k}$.
Then $$\frac{x_{k+1} - \sqrt{5}}{x_{k+1} + \sqrt{5}} = \frac{x_k + \frac{5}{x_k} - 2\sqrt{5}}{x_k + \frac{5}{x_k} + 2\sqrt{5}} = \frac{x_k^2 - 2x_k\sqrt{5} +5}{x_k^2 + 2x_k\sqrt{5}+5}$$
which is a square $$\frac{(x_k-\sqrt{5})^2}{(x_k+\sqrt{5})^2} = \left(\left(\frac{x-\sqrt{5}}{x+\sqrt{5}}\right)^{2^k}\right)^2 = \left(\frac{x-\sqrt{5}}{x+\sqrt{5}}\right)^{2^{k+1}}$$
as required. 
A: The limit, if it exists, should be a fixed point of $f(x) = \frac{1}{2} \left( x + \frac{5}{x} \right)$. Rearranging $x = \frac{1}{2} (x + \frac{5}{x})$ gives $x^2 - 5 = 0$ or $x = \pm \sqrt{5}$, and a negative limit is impossible as the series is manifestly positive. (And it's easy to show that the limit exists: note that if $x < \sqrt{5}$ then $x < f(x) < \sqrt{5}$, and vice versa if $x > \sqrt{5}$, so the sequence is monotonic and bounded and therefore has a limit.)
For the induction step of the proof of the closed form, you just have to prove that $$\frac{x_{n+1} - \sqrt{5}}{x_{n+1} + \sqrt{5}} = \left( \frac{x_n - \sqrt{5}}{x_n + \sqrt{5}} \right)^2.$$ Manipulating the RHS of this, using the fact that $2 x_n x_{n+1} = x_n^2 + 5$ (a rearrangement of the recursive relation that defines the sequence):
$$\left( \frac{x_n - \sqrt{5}}{x_n + \sqrt{5}} \right)^2 = \frac{x_n^2 - 2\sqrt{5} x_n + 5}{x_n^2 + 2\sqrt{5} x_n + 5} = \frac{2 x_n x_{n+1} - 2\sqrt{5}{x_n}}{2 x_n x_{n+1} + 2\sqrt{5} x_n}$$
which gives the desired result after cancelling $2 x_n$ from numerator and denominator.
Now to show the limit using the closed form: if $x$ is positive, then $$\left| \frac{x - \sqrt{5}}{x + \sqrt{5}}\right| < 1 $$ and so
$$0 =  \lim_{n \to \infty} \left( \frac{x - \sqrt{5}}{x + \sqrt{5}} \right)^{2^n} = \lim_{n \to \infty} \frac{x_n - \sqrt{5}}{x_n + \sqrt{5}} = \frac{ \lim_{n \to \infty} x_n - \sqrt{5}}{\lim_{n \to \infty} x_n + \sqrt{5}}$$
which gives you $\lim_{n \to \infty} x_n = \sqrt{5}$.
