<s>[INST] how can one solve for $x$, $x =\sqrt{2+\sqrt{2+\sqrt{2\cdots }}}$ 
Possible Duplicate:
Limit of the nested radical $\sqrt{7+\sqrt{7+\sqrt{7+\cdots}}}$ 

how can one solve for $x$, $x =\sqrt[]{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2\cdots }}}}}}$
we know, if $x=\sqrt[]{2+\sqrt{2}}$, then, $x^2=2+\sqrt{2}$
now, if $x=\sqrt[]{2+\sqrt{2}}$, then, $(x-\sqrt{2})(x+\sqrt{2})=\sqrt{2}$
 [/INST] In general, the function $f(x)=\displaystyle\bigg(\small\sqrt{\normalsize x+\small\sqrt{\normalsize x+\sqrt{ x+\sqrt{x}}}}\;\normalsize\bigg)$ is:
$0.5(1+\sqrt[]{1+4x})$
Ref: Math-Integration of nested square roots of x
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