Prove $ \lim\limits_{x\to\infty}y_{n}=\sqrt{x}$ if $y_{n}=\frac{1}{2}\left(y_{n-1}+\frac{x}{y_{n-1}}\right),n\in \mathbb{N},x>0,y_{0}>0$ Can someone say how to solve this problem? In solution, it says that it stars with $$\frac{y_{n}-\sqrt{x}}{y_{n}+\sqrt{x}}=\left(\frac{y_{n-1}-\sqrt{x}}{y_{n-1}+\sqrt{x}}\right)^2,n\ge 1$$
How to get to this formula?

The function $y \to 1/2(y+ x/x)$ is continous for $y > 0$. So the limit of your sequence can only be $\sqrt{x}$, i.e. the positive root of $y=f(y)$.
Then you should study the monotonicy of $(y_n)$.
Your sequence is an example of the Newton's method applied to the function $y \to y-x^2$.
