Finding the convergent value of a recursion similar to Arithmetic-Geometric Mean recursion The sequence is defined as follows :
Start : $(x_0,y_0)$ with $ 0 < x_0 < y_0 $
Step  : $x_{n+1} = \frac {x_n+y_n} {2}$ ,  $y_{n+1}= \sqrt{x_{n+1}y_n} $
Find  $\lim_{n\to \infty}(x_n,y_n)$ .

I have proved that the sequences do converge by showing that the sequences are monotone bounded sequences. The prove goes as follows ,
First we will prove that: If $0<x_k<y_k<a$ then $0<x_{k} < x_{k+1} < y_{k+1} < y_k < a$.
$0<\frac {x_k+x_k} 2 = x_k <x_{k+1} = \frac {x_k+y_k} 2 < \frac {y_k+y_k} {2}=y_k$  and $x_{k+1}=\sqrt{x_{k+1}x_{k+1}} <y_{k+1} =\sqrt{x_{k+1}y_k} < \sqrt{y_ky_k}=y_k<a$
Thus   $\,0<x_k<x_{k+1}<y_{k+1}<y_k<a$. It follows that $0<x_n<x_{n+1}<y_{n+1}<y_n<y_0$. Hence the $x_n$ and $y_n$ are monotone bounded sequence. Hence the sequence $(x_n,y_n)$ converges.
Let
$\lim_{x \to \infty } x_n=\lim_{x\to \infty} x_{n+1}=l$
$\lim_{x \to \infty } y_n=\lim_{x\to \infty} y_{n+1}=w$
$l=\lim_{x\to \infty}x_{n+1}=\lim_{x\to \infty} \frac {x_n+y_n} {2}=\frac{\lim_{x\to \infty} x_n+\lim_{x\to \infty} y_n} {2}=0.5l+0.5w$. Hence $l=w$.I am unable to find the convergent value. I would like some assistance on this.
I think  $\lim_{x\to \infty}(x_n,y_n)$ exists for a more general recursion
Start : $(x_0,y_0)$ with $ 0 < x_0 < y_0 $
Step  : $x_{n+1}=tx_n+(1-t)y_n $ with $t\in(0,1)$  ,  $y_{n+1}= \sqrt{x_{n+1}y_n} $
In this case what is the value of $\lim_{x\to \infty}(x_n,y_n)$ ?
Any insight /answer is greatly appreciated !
Thanks
 A: A partial answer (for $t=\frac{1}{2}$): notice that from
$$ y_{n+1}=\sqrt{\frac{x_n+y_n}{2}y_n}$$
it follows that:
$$ y_{n+1}^2 - x_{n+1}^2 = \frac{1}{4}(y_n^2-x_n^2).\tag{1} $$
Now notice that if
$$ x = \frac{2}{2^m \tan\frac{\pi}{2^m}},\quad y = \frac{2}{2^m \sin\frac{\pi}{2^m}}$$
then:
$$\frac{x+y}{2}=\frac{2}{2^{m+1}\tan\frac{\pi}{2^{m+1}}},\quad \sqrt{\frac{x+y}{2}y}=\frac{2}{2^{m+1}\sin\frac{\pi}{2^{m+1}}},\tag{2}$$
so starting from $x_0=0$ and $y_0=1$ we have convergence towards $\frac{2}{\pi}$. In general, if:
$$ x_0 = \frac{\alpha}{\tan\theta},\qquad y_0=\frac{\alpha}{\sin\theta} $$
with $\alpha\in\mathbb{R}^+$ and $\theta\in(0,\pi/2]$, we have convergence towards $\frac{\alpha}{\theta}$, so:

$$\lim_{n\to +\infty} x_n = \lim_{n\to +\infty} y_n= \frac{\sqrt{y_0^2-x_0^2}}{\arccos\frac{x_0}{y_0}}.$$

Notice that this is just the Archimedean algorithm for finding approximation of $\pi$ by computing the perimeters of the inscribed and circumscribed regular $2^n$-agons in the unit circle.
A: since note
$$\left(\dfrac{x_{n+1}}{y_{n+1}}\right)^2=\dfrac{x^2_{n+1}}{x_{n+1}y_{n}}=\dfrac{x_{n+1}}{y_{n}}=\dfrac{1}{2}\left(\dfrac{x_{n}}{y_{n}}+1\right)$$
so let $\dfrac{x_{n}}{y_{n}}=\cos{c_{n}},c_{n}\in(0,\dfrac{\pi}{2})$
use this $$\dfrac{1}{2}\left(\cos{2x}+1\right)=\cos^2{x}$$
so
$$\cos{c_{n+1}}=\cos{\dfrac{c_{n}}{2}}\Longrightarrow c_{n+1}=\dfrac{c_{n}}{2}$$
so
$$c_{n}=\dfrac{c_{0}}{2^n}\Longrightarrow \dfrac{x_{n}}{y_{n}}=\cos{\dfrac{c_{0}}{2^n}}$$
then use following well know identity 
$$\cos{\dfrac{x}{2}}\cdots\cdots\cos{\dfrac{x}{2^n}}=\dfrac{sin{x}}{2^n\sin{\dfrac{x}{2^n}}}$$
then easy have limits
A: I just saw this problem in Arthur Engel's book on page 3 and the solution goes like this:
$$ \frac{x_{n+1}}{y_{n+1}}=\frac{x_{n+1}}{\sqrt{x_{n+1}y_{n}}}=\sqrt{\frac{x_{n+1}}{y_n}}=\sqrt{\frac{1+x_{n}/y_{n}}{2}}.\tag{1}\label{eq1}$$
this reminds us of the half-angle relation
$$\cos\frac{\alpha}{2}=\sqrt{\frac{1+\cos\alpha}{2}}.$$
since we have $ 0<x_n /y_n<1,$ we may set $x_n/y_n=\cos\alpha_n$ then\eqref{eq1} becomes
$$ \cos\alpha_{n+1}=\cos\frac{\alpha_n}{2}\implies\alpha_n=\frac{\alpha_0}{2^n}\implies 2^n\alpha_n=\alpha_0$$
which is equivalent to
$$2^n\arccos\frac{x_n}{y_n}=\arccos\frac{x_0}{y_0}.\tag{2}\label{eq2} $$
to avoid square roots, we consider $y_n^2-x^2_n$ instead of $y_n-x_n$ and get $$ y^2_{n+1}-x^2_{n+1}=\frac{y^2_n-x^2_n}{4}\implies 2\sqrt{y^2_{n+1}-x^2_{n+1}}=\sqrt{y^2_{n}-x^2_{n}}  $$
or $$2^n\sqrt{y^2_{n}-x^2_{n}}=\sqrt{y^2_{0}-x^2_{0}}.\tag{3}\label{eq3} $$ from \eqref{eq2},\eqref{eq3} we get
$$ \arccos\frac{x_0}{y_0}=2^n\arccos\frac{x_n}{y_n}=2^n\arcsin\frac{\sqrt{y^2_n - x^2_n}}{y_n}=2^n\arcsin\frac{\sqrt{y^2_0 - x^2_0}}{2^ny_n}. $$
the right-hand side converges to $\sqrt{y^2_0-x^2_0}/y$ for $ n\to\infty $ finally we get
$$ x = y = \frac{ \sqrt{y^2_0 - x^2_0}}{ \arccos (x_0/y_0)}. $$
