Limit of the sequence $a_{n+1}=\frac{1}{2} (a_n+\sqrt{\frac{a_n^2+b_n^2}{2}})$ - can't recognize the pattern Consider the sequence:
$$a_0=x,~~~b_0=y$$
$$a_{n+1}=\frac{1}{2} \left(a_n+\sqrt{\frac{a_n^2+b_n^2}{2}} \right),~b_{n+1}=\frac{1}{2} \left(b_n+\sqrt{\frac{a_n^2+b_n^2}{2}}\right)$$
$$\lim_{n \to \infty} a_n=\lim_{n \to \infty} b_n=l(x,y)$$
I can't pin down the pattern for the limit. Numerically, I've got the following result:
$$\frac{1}{l(x,y)}=\arctan (f(x,y))$$

What is the explicit expression for $f(x,y)$?

Numerical examples:
$$\begin{array}( x & y & \frac{1}{l(x,y)} \\ 1 & 2 & \arctan \left( \frac{3}{4} \right) \\ 1 & 3 & \arctan \left( \frac{1}{2} \right) \\ 2 & 3 & \arctan \left( \frac{5}{12} \right) \\ 3 & 5 & \arctan \left( \frac{1}{4} \right) \\ 1 & 5 & \arctan \left( \frac{\sqrt{13}-3}{2} \right) \\ 4 & 5 & \arctan \left( \frac{9}{40} \right) \\ 3 & 4 & \arctan \left( \frac{7}{24} \right) \\ 3 & 7 & \arctan \left( \frac{\sqrt{29}-5}{2} \right) \end{array}$$

Edit
To summarize @IvanNeretin's comment (and adding my own numerical results), so far we have:

$$l(x,0)=\frac{2x}{\pi}$$

$$f(x,x+1)=\frac{2x+1}{2x(x+1)}$$
$$f(x,x+2)=\frac{1}{x+1}$$
$f(x,x+3)$ - see my answer below
$$f(x,x+4)=\frac{\sqrt{x^2+4x+8}-(x+2)}{2}$$

Here $x \in \mathbb{R}^{+}$, $x \neq 0$

Also by symmetry we have of course:
$$f(x,y)=f(y,x)$$

At this point it's obvious that no ordinary function works here.

My hope is to connect $f(x,y)$ to some special function or an integral.


Geometrically: 
$(a_{n+1}, b_{n+1})= \frac 12 (a_n,b_n) +  \frac 12 \sqrt{a_n^2+b_n^2} (\cos \pi/4, \sin \pi/4)$
So we draw a line between $(a_n,b_n)$ and connect it to a point equidistant from the origin on the line $x = y$, and find the mid-point.
$r_n = \sqrt{a_n^2 + b_n^2}\\
\phi_n = \tan^{-1}(\frac{b_n}{a_n})$
$\phi_{n+1} = \frac{\frac{\pi}{4} + \phi_n}{2}\\
r_{n+1} = r_n\cos(\frac{\frac{\pi}{4} - \phi_n}{2})$
