Find value of $x$:
$x=\sqrt[8]{2207-\frac{1}{2207-\frac{1}{2207-....and\,so\, on}}}$
On solving ,we have $x^8=2207-\frac{1}{x^8}$
$x^8+\frac{1}{x^8}=2207$
$x^4+\frac{1}{x^4}=47$
$x^2+\frac{1}{x^2}=7$
$x+\frac{1}{x}=3$
$x=\frac{3+\sqrt{5}}{2}\,,\frac{3-\sqrt{5}}{2}$
But $x$ can take only one value.
The problem is which of these two values is to be accepted/rejected and why.
On
You want the value which is greater than $1$ - clearly by estimating.
From $x^8+\frac 1{x^8}=2207$ you know that if $x$ satisfies that equation so will $\frac 1x$. Put $\frac 1x$ in the original equation and modify the eighth root accordingly and you will see where the alternative answer comes from and that this is clearly less than $1$.
The 8th root is not pertinent. You are asking how to assign a value to the continued fraction inside the 8th root. As the equation for $x^8$ shows, there are two solutions of $f = 2207 - \frac{1}{f}$.
My first thought was that the even and odd "levels" of the continued fraction converge to the two different solutions (making it hard to distinguish the two solutions in principle) but I realized later that it is also possible that one of the solutions is attracting and the other repelling. In this case the value of the attracting solution is the one that makes sense and the only one to which convergence will occur.
The graph of $y = 2207 - 1/x$ lies above the line $y=x$ between the two solutions. This means the smaller solution is repelling and the larger one, attracting. So the convergents of the continued fraction are pushed closer and closer to the larger value as long as one starts above the small solution. The fraction is thus $(\frac{3 + \sqrt{5}}{2})^8$.