TITLE: Prove that : $\lim_{n\to +\infty}\left(x_{0}x_{1}...x_{n-1}\right)^{1/2^{n}}=\frac{3+\sqrt{5}}{2}$
QUESTION [5 upvotes]: Let sequence $x_{n+1}=x_{n}^{2}-2$ with $x_{0}=3$ 
Then prove that : 

$$\displaystyle\lim_{n\to +\infty}\left(x_{0}x_{1}...x_{n-1}\right)^{\frac{1}{2^{n}}}=\frac{3+\sqrt{5}}{2}$$

I don't know how I started but my result in try is : 
I see $\frac{3+\sqrt{5}}{2}+\frac{2}{3+\sqrt{5}}=3=x_{0}$ 
I don't know  where this can help me 
I have already to see your hints or ideas to approach it

REPLY [2 votes]: As I said in a comment, your observation is the key to the problem.
HINTS:
Prove by induction that $$x_k=\alpha^{2^{k+1}}+\alpha^{-2^{k+1}},$$ where $\alpha=\frac{3+\sqrt5}{2}.$
Take logarithms.
Use $$\log\left(\alpha^{2^k}+\alpha^{-2^k}\right)=
2^k\log\alpha+\log\left(1+\alpha^{-2^{k+1}}\right)$$