Question: Please help with my homework question:

Define $\{a\_n\}\_{n=1}^{\infty}$, $\{b\_n\}\_{n=1}^{\infty}$ as:

$b\_{n+1}=\frac{a\_n+b\_n}{2}, a\_{n+1}=\frac{a\_n^2+b\_n^2}{a\_n+b\_n}, 0<b\_1<a\_1$

**Prove the limits converge and that $\lim \limits\_{n \to \infty}a\_n = \lim \limits\_{n \to \infty}b\_n $**.

I have tried to show that they are monotone and bounded, tried looking for a property of arithmetic mean (related to $b\_n$) but I'm completely stuck as none of these yielded anything useful.

Any help would be appreciated

Answer: 
We have that

$$\frac{\sin(xy)}{|x-y|}=\frac{\sin(xy)}{xy}\frac{xy}{|x-y|}$$

and $\frac{\sin(xy)}{xy}\to 1$ but for $x=2y$ 

$$\frac{xy}{|x-y|}=\frac{2y^2}{|y|} \to 0$$

and for $x=t+t^2$ and $y=t\to 0$ 

$$\frac{xy}{|x-y|}=\frac{t^2+t^3}{t^2} \to 1$$

therefore the given limit doesn't exist.
Alternatively, if you apply user's method, for $\lim\limits\_{(x,y)\to(0,0)}\frac{xy}{|x-y|}:$ try a sequence similar to the one in [this](https://math.stackexchange.com/a/1228063/721644) answer.

Let $(x\_n,y\_n)=\left(\frac1n+e^{-n},\frac1n\right).$
Then, $$\frac{x\_ny\_n}{|x\_n-y\_n|}=\frac{\frac1{n^2}+\frac{e^{-n}}n}{e^{-n}}=\frac{e^n}{n^2}+\frac1n\to+\infty\text{ as } n\to+\infty$$