<s>[INST] 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. 
 [/INST] $$A = \sqrt{1+\sqrt{x+\sqrt {x^2+\sqrt{x^3+\sqrt{x^4+\sqrt{x^5+...}}}}}} = 2$$
$$A = \sqrt{1+\sqrt{x\left(1+\sqrt {1+\sqrt{x^1+\sqrt{x^2+\sqrt{x^3+...}}}}\right)}} = 2$$
$$A = \sqrt{1+\sqrt{x(1+A)}} = 2,A =2$$
$$ \sqrt{1+\sqrt{3x}} = 2$$
$$ 1+\sqrt{3x} = 4$$
$$ \sqrt{3x} = 3$$
$$ x = 3$$
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