Prove that : $\lim_{n\to +\infty}\left(x_{0}x_{1}...x_{n-1}\right)^{1/2^{n}}=\frac{3+\sqrt{5}}{2}$ 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 
 A: 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 $$\log\left(\alpha^{2^k}+\alpha^{-2^k}\right)=
2^k\log\alpha+\log\left(1+\alpha^{-2^{k+1}}\right)$$
A: Let $L$ be the limit to be computed. The purely computational path is as follows:
$$
\begin{aligned}
\ln L 
&=
\lim_{n\to \infty}
\ln \left(x_0x_1\dots x_{n-1}\right)^{\frac1{2^n}}
\\
&=
\lim_{n\to \infty}
\frac{\ln x_0+\ln x_1+\dots+ \ln x_{n-1}}{2^n}
\\
&=
\lim_{n\to \infty}
\frac{\ln x_{n-1}}{2^n-2^{n-1}}
=
\lim_{n\to \infty}
\frac{\ln x_{n-1}}{2^{n-1}}
=
\lim_{n\to \infty}
\frac{\ln x_n}{2^n}
\\
&=
\lim_{n\to \infty}
\ln x_n^{1/2^n}
=
\ln 
\lim_{n\to \infty}
x_n^{1/2^n}
\ .
\end{aligned}
$$
Now observe that we have an explicit formula:
Let $a,b$ be the numbers $(3\pm\sqrt 5)/2$ with $a>1$, $b=1/a\in(0,1)$, $ab=1$. Then we have inductively:
$$
\begin{aligned}
x_0 &= a+b\ ,\\
x_1 &= (a+b)^2-2=a^2+b^2+2-2=a^2+b^2\ ,\\
x_2 &= (a^2+b^2)^2-2=a^4+b^4+2-2=a^4+b^4\ ,\\
&\vdots\qquad
\vdots\qquad 
\vdots\qquad 
\vdots\qquad 
\vdots\qquad 
\vdots\qquad 
\vdots\qquad 
\\
x_n &= a^{2^n}+b^{2^n}\ ,
\\
&\vdots\qquad 
\vdots\qquad 
\vdots\qquad 
\vdots\qquad 
\vdots\qquad 
\vdots\qquad 
\vdots\qquad 
\end{aligned}
$$
So the limit $L$ is 
$$
L=\lim_{n\to \infty}
x_n^{1/2^n}
=a\ .
$$

Later EDIT: The one obscure step addressed in the comments uses Cesàro-Stolz - wiki link, the "$\infty:\infty$" case. This is the usual trick to digest limits of fractions where in numerator (and/or denominator) we have a sum of terms. (My hope was that the form of the denominator $2^n-2^{n-1}$ already rings the bell.) 
In the loc. cit (wiki link) consider $a_n$ to be the numerator $\ln x_0+\ln x_1+\dots+\ln x_{n-1}$, and $b_n$ to be the denominator $2^n$.
A: *

*Your limit is $\phi^2=\phi+1$ where $\phi$ is the golden ratio.  

*$x_n=\big\lceil \phi^{2^{n+1}}\big\rceil$  

*so the product $x_{0}x_{1}...x_{n-1}$ is a little larger than $\phi^{2^{n+1}-2}$ while less than $\phi^{2^{n+1}}$ 

*and the $2^n$th root of the product is thus between $\phi^{2-2^{-(n-1)}}$ and $\phi^2$ 

*converging to $\phi^2$ as $n$ increases

