Determine $x$ such that $\lim\limits_{n\to\infty} \sqrt{1+\sqrt{x+\sqrt{x^2…+\sqrt{x^n}}}} = 2$ Find the value of $x$ such that $\lim\limits_{n\to\infty} \sqrt{1+\sqrt{x+\sqrt{x^2…+\sqrt{x^n}}}} = 2$  
I tried getting rid of square roots and got $(...((9-x)^2-x^2)^2-...)^2-x^n = 0$ which I don't think helped. Please point me in the right direction. 
 A: Let me describe a sketch of proof that $x=4$. 
A. Observe that if $f(x)=\lim_{n\to\infty}\sqrt{1+\sqrt{x+\sqrt{x^2+\cdots\sqrt{x^n}}}}$, then $f$ is strictly increasing.
B. We shall show that $f(4)=2$, and hence $x=4$ is the unique answer.
$B_1.$ Fix $m\in\mathbb N$ and show that, for $n=m,m-1,m-2,\cdots$ (induction backwards) 
$$
2^n<\sqrt{4^n+\sqrt{4^{n+1}+\cdots\sqrt{4^{m-1}+\sqrt{4^m}}}}<2^n+1,
$$
while
$$
\sqrt{4^n+\sqrt{4^{n+1}+\cdots\sqrt{4^{m-1}+\sqrt{4^m}+1}}}=2^n+1.
$$
$B_2.$ Next estimate the difference
$$
(2^n+1)-
\sqrt{4^n+\sqrt{4^{n+1}+\cdots\sqrt{4^{m-1}+\sqrt{4^m}}}} \\
=\sqrt{4^n+\sqrt{4^{n+1}+\cdots\sqrt{4^{m-1}+\sqrt{4^m}+1}}}-
\sqrt{4^n+\sqrt{4^{n+1}+\cdots\sqrt{4^{m-1}+\sqrt{4^m}}}} \\
=\frac{\sqrt{4^{n+1}+\cdots\sqrt{4^{m-1}+\sqrt{4^m}+1}}-
\sqrt{4^{n+1}+\cdots\sqrt{4^{m-1}+\sqrt{4^m}}}}{\sqrt{4^n+\sqrt{4^{n+1}+\cdots\sqrt{4^{m-1}+\sqrt{4^m}+1}}}+
\sqrt{4^n+\sqrt{4^{n+1}+\cdots\sqrt{4^{m-1}+\sqrt{4^m}}}}} \\
<\frac{{\sqrt{4^{n+1}+\cdots\sqrt{4^{m-1}+\sqrt{4^m}+1}}-
\sqrt{4^{n+1}+\cdots\sqrt{4^{m-1}+\sqrt{4^m}}}}}{2\cdot 2^n} \\
<\cdots<\frac{(\sqrt{4^m}+1)-\sqrt{4^m}}{2^{m-n}\cdots 2^{n+(n+1)+\cdots+(m-1)}}=2^{-\frac{(m-n)(n+m+1)}{2}}
$$
Thus
$$
\lim_{m\to\infty}\sqrt{4^n+\sqrt{4^{n+1}+\cdots\sqrt{4^{m-1}+\sqrt{4^m}}}}=2^n+1.
$$
For $n=0$ we have
$$
\lim_{m\to\infty}\sqrt{1+\sqrt{4+\cdots\sqrt{4^{m-1}+\sqrt{4^m}}}}=2^0+1=2.
$$
A: Hint: Pretend that there's an extra $+1$ at the next-to-last level. Then $$4^{n-1}+\sqrt{4^n}+1~=~2^{2(n-1)}+2^n+1~=~2^{2(n-1)}+2\cdot2^{n-1}+1~=~(2^{n-1}+1)^2.$$ Can you see what happens ? :-$)~$ Now, as $n\to\infty,$ the numerical influence gained by adding that extra $+1$ at the top level tends towards $0.$
A: $$A = \sqrt{1+\sqrt{x+\sqrt {x^2+\sqrt{x^3+\sqrt{x^4+\sqrt{x^5+...}}}}}} = 2$$
$$A = \sqrt{1+\sqrt{x\left(1+\sqrt {1+\sqrt{x^1+\sqrt{x^2+\sqrt{x^3+...}}}}\right)}} = 2$$
$$A = \sqrt{1+\sqrt{x(1+A)}} = 2,A =2$$
$$ \sqrt{1+\sqrt{3x}} = 2$$
$$ 1+\sqrt{3x} = 4$$
$$ \sqrt{3x} = 3$$
$$ x = 3$$
