Let $x>0$. Let $a_1=\sqrt{x}, a_{n+1}=\sqrt{x+a_n}$. So $a_2=\sqrt{x+\sqrt{x}}, a_3=\sqrt{x+\sqrt{x+\sqrt{x}}}$ and so on.
Let $t:=\frac{1+\sqrt{1+4x}}2$. It's easy to show that:
$$(1) \lim_{n \rightarrow \infty} a_n = t$$
$$(2) \lim_{n \rightarrow \infty} \frac{t-a_n}{t-a_{n+1}}=2t$$
So one might try calculating:
$$f(x):=\lim_{n \rightarrow \infty} (2t)^n (t-a_n)$$
I took interest in this kind of limits since $f(2)=\frac{\pi^2}4$ via Viète's formula. I was able to approximate some other limits, but found nothing meaningful so far.
Has this function been considered before? Do you have any hints on how to analyze it?
EDIT: It may be helpful to look at $f(x)$ as a result of a single recurrency:
$$b_1=(2t)(t-\sqrt{x})$$
$$b_{n+1}=(2t)^{n+1}(t-\sqrt{x+t-\frac{b_n}{(2t)^n}})$$
$$f(x)=\lim_{n \rightarrow \infty } b_n$$
EDIT2:
Please note that $(t+\sqrt{x+t-\frac{b_n}{(2t)^n}})b_{n+1}=(2t)b_n$. Since $\sqrt{x+t-\frac{b_n}{(2t)^n}} < t$, we know that $b_{n+1}>b_n$.
Furthermore, $(t+\sqrt{x+t-\frac{b_n}{(2t)^n}})(b_{n+1}-b_n)=(2t)b_n-(t+\sqrt{x+t-\frac{b_n}{(2t)^n}}) b_n=(t-\sqrt{x+t-\frac{b_n}{(2t)^n}})b_{n}=\frac{b_nb_{n+1}}{(2t)^{n+1}}$
This can be weakened to:
$b_{n+1}-b_n \leq \frac{f^2(x)}{(2t)^{n+2}}$
Which means that:
$$f(x) \leq b_1 + \sum_{i=1}^n \frac{f^2(x)}{(2t)^{n+2}}=b_1+\frac{f^2(x)}{4t^2-2t}$$
EDIT3: The determinant of this inequality is:
$$\Delta:=1-4\frac{b_1}{4t^2-2t}=1-4\frac{2t(t-\sqrt{x})}{4t^2-2t}=1-4\frac{t-\sqrt{x}}{2t-1}$$
$\Delta \leq 0$ for $x \leq \frac{23+8\sqrt{7}}{36}$. For $\Delta >0$, this gives us two options:
$f(x) \geq (4x+2t)(1+\Delta)$
$f(x) \leq (4x+2t)(1-\Delta)$
We already know that $f(2)=\frac{\pi^2}{4}$ -- this satisfies the second inequality. Since $f(x)$ is quite clearly continuous, we know that:
$$f(x) \leq (2x+t)(1-\Delta) \text{ for } x\geq\frac{23+8\sqrt{7}}{36}$$
$\lim_{x \rightarrow \infty} \frac{b_1}{\sqrt{x}} = \lim_{x \rightarrow \infty} (2x+t)(1-\Delta) = 1$, so $\lim_{x \rightarrow \infty}\frac{f(x)}{\sqrt{x}}=1$.
