<s>[INST] Equation with limit $\lim\limits_{n\to \infty}\sqrt{1+\sqrt{x+\sqrt{x^2+...+\sqrt{x^n}}}}=2$
I had never faced with problems like this. Give me, please, a little hint.
 [/INST] We are trying to solve for $x$ where $\lim_{n\to\infty}a_n=2$ and $a_n=\sqrt{1+\sqrt{x+\sqrt{x^2+\sqrt{\dots x^n}}}}$
The trivial possible solutions are $x=0,1$, which do not work.
For $x$
$$\sqrt{1+\sqrt{x+\sqrt{x^2+\sqrt{\dots x^n}}}}=\sqrt{1+\sqrt{x}\sqrt{1+\sqrt{1+\sqrt{\frac1x+\sqrt{\frac1{x^4}+\sqrt{\dots x^n}}}}}}$$
$$<\sqrt{1+\sqrt{x}\sqrt{1+\sqrt{1+\sqrt{\dots+1}}}}\tag{If $x>1$}$$
$$=\sqrt{1+\frac{1+\sqrt{5}}{2}\sqrt{x}}$$
So if we set that equal to $2$, we have
$$x>\left(\frac6{1+\sqrt{5}}\right)^2=\frac{27-9\sqrt5}2$$
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