Can we identify the limit of this arithmetic/geometric mean like iteration? Let $a_0 = 1$ and $b_0 = x \ge 1$. Let
$$
a_{n+1} = (a_n+\sqrt{a_n b_n})/2, \qquad b_{n+1} = (b_n + \sqrt{a_{n+1} b_n})/2.
$$
Numeric computation suggests that regardless of the choice of $x$, $a_n$ and $b_n$ always converge to the same value. Can we prove this?
Moreover, if we assume this is true and define $f(x) = \lim_{n \to \infty} a_n$, then numeric computation shows that
$$
f'(1) \approx 0.57142857142857 \approx 4/7 \\
f''(1) \approx -8/49 \\
f^{(3)}(1) \approx 1056/4459 \\
f^{(4)}(1) \approx -65664/111475.
$$
There seems to be some patterns here.
Can we actually find the limit with these clues?

I was actually trying to check this recursion
$$
a_{n+1} = (a_n+\sqrt{a_n b_n})/2, \qquad b_{n+1} = (b_n + \sqrt{a_{n} b_n})/2.
$$
I made a mistaked in my code and computed the one at the top. See The Computer as Crucible, pp 130.
 A: We have $1\le a_0\le b_0\le x$.
We can prove that $1\le a_n\le a_{n+1}\le b_{n+1}\le b_n\le x$ by mathematical induction.
$\displaystyle a_1=\frac{1+\sqrt{x}}{2}$ and $\displaystyle b_1=\frac{x+\sqrt{a_1x}}{2}$.
Obviously, $a_1\ge 1$. Hence $\displaystyle b_1-a_1=\frac{(x-1)+(\sqrt{a_1}-1)\sqrt{x}}{2}\ge0$ 
Note that
$$a_1-x=\frac{1+\sqrt{x}-2x}{2}=\frac{(1-\sqrt{x})(1+2\sqrt{x})}{2}\le0$$
and 
$$b_1\le\frac{x+\sqrt{x\cdot x}}{2}=x$$
Therefore, $1\le a_0\le a_{1}\le b_{1}\le b_0\le x$.
Suppose that $1\le a_k\le a_{k+1}\le b_{k+1}\le b_k\le x$. Then 
$$a_{k+2}=\frac{a_{k+1}+\sqrt{a_{k+1}b_{k+1}}}{2}\ge\frac{a_{k+1}+\sqrt{a_{k+1}a_{k+1}}}{2}=a_{k+1}$$
$$b_{k+2}=\frac{b_{k+1}+\sqrt{a_{k+2}b_{k+1}}}{2}\ge\frac{a_{k+1}+\sqrt{a_{k+1}b_{k+1}}}{2}=a_{k+2}$$
and
$$b_{k+2}=\frac{b_{k+1}+\sqrt{a_{k+2}b_{k+1}}}{2}\le\frac{b_{k+1}+\sqrt{b_{k+2}b_{k+1}}}{2}$$
So,
\begin{align}
b_{k+2}&\le\frac{b_{k+1}+\frac{b_{k+2}+b_{k+1}}{2}}{2}\\
4b_{k+2}&\le3b_{k+2}+b_{k+1}\\
b_{k+2}&\le b_{k+1}
\end{align}
This complete the induction proof.
So, $\{a_n\}$ is increasing and bounded above by $x$. $\{b_n\}$ is decreasing and bounded below by $1$. Both the sequences converge.
Let $\displaystyle \lim_{n\in\infty} a_n=a$ and $\displaystyle \lim_{n\in\infty} b_n=b$. Then we have
$$a=\frac{a+\sqrt{ab}}{2}$$
This implies that $a=b$.
Unsolved: How to find the value of the common limit?
