<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] Hint.
$$f(x) = \sqrt{x+\sqrt{x^2+\sqrt{x^3+\sqrt{x^4+\ldots}}}} $$
is a continuous and increasing function on $\mathbb{R}^+$, as well as:
$$ f_k(x) = \sqrt{x^k+\sqrt{x^{k+1}+\sqrt{x^{k+2}+\ldots}}}$$
It is enough to check that $f(4)=3$, or:
$$ f_1(4) = 2\cdot 4^0+1, $$
or:
$$ f_2(4) = f_1(4)^2-4 = (2\cdot 4^0+1)^2-2^2=4+1 $$
or:
$$ f_3(4)=f_2(4)^2-4^2 = (4+1)^2-4^2 = 2\cdot 4+1 $$
or:
$$ f_4(4)=f_3(4)^2-4^3 = (2\cdot 4+1)^2 - 4^3 = 4^2+1 $$
or:
$$ f_5(4)=f_4(4)^2-4^4 = (4^2+1)^2-(4^2)^2 = 2\cdot 4^2+1 $$
$\ldots$ A pattern emerged. Notice that

$$ f_{2n+1}(4) = 2\cdot 4^n+1 \qquad f_{2n}(4) = 4^n+1 \tag{1}$$

are equivalent, since
$$ f_{2n+2}(x)=f_{2n+1}(x)^2-x^{2n+1}, \qquad f_{2n+1}(x)=f_{2n}(x)^2-x^{2n}. $$
So it is enough to prove that

$$ \lim_{n\to +\infty}\frac{f_{2n}(4)}{4^n+1}=\lim_{n\to +\infty}\frac{f_{2n+1}(4)}{2\cdot 4^n+1}=1 \tag{2}$$

hold by squeezing.
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