Show sequences converge to $\sqrt 2$ let $x_0$ = 2, $y_0 = 1$;  $x_{n+1} = \frac{x_n + y_n}{2}$, $y_{n+1} = \frac{2}{x_{n+1}}, n = 0, 1, 2...$.
Need to show the sequences converge to $\sqrt 2$. I wish I could add more in terms of my attempted solution but I have really gotten anywhere; I haven't worked on problems like this in a while, and after spending time reviewing and attempting I haven't made much progress! Hints/solution would be appreciated. 
 A: Note that you are attempting to calculate the limit of $x_{n}$ where $$x_{n}=\frac{1}{2} \left(x_{n-1}+\frac{2}{x_{n-1}} \right)$$
This is known as the Babylonian Method, and you can prove it yourself by proving that first, that is bounded below by $\sqrt{2}$, and then that $x_{n}$ is a decreasing sequence. This is relatively simple, and follows from $\text{AM-GM}$, so I leave to you. 
This gives us the sequence must converge. Thus, note that if the limit is $a$ , $a$ satisfies $$2a=a+\frac{2}{a}$$
And you can use $x_{n}$ to compute $y_{n}$. 
A: We have directly the following:
$$x_{n+1}=\frac{x_n+\frac2{x_n}}2$$
So $y_n$ is really irrelevant.  Now, if $L=\lim\limits_{n\to\infty}x_n$, then
$$L=\frac{L+\frac2L}2\implies L\stackrel?=\pm\sqrt2$$
Applying AM-GM inequality, we see that
$$x_{n+1}>\sqrt{x_n\frac2{x_n}}=\sqrt2$$
Thus, if $L$ exists, then it follows that it must be the square root of $2$.
To prove it exists, one may observe that if $x_n>\sqrt2$, then
$$\frac2{x_n}<\sqrt2$$
$$x_n+x_n>x_n+\sqrt2\implies x_n>\frac{x_n+\sqrt2}2$$
$$x_{n+1}=\frac{x_n+\frac2{x_n}}2<\frac{x_n+\sqrt2}2<x_n$$
Thus,
$$\sqrt2<x_{n+1}<x_n$$
So it converges, and to our desired value.
