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.

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. 
