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?

You can use this way to get the formula. Note
\begin{eqnarray}
y_n-\sqrt{x}&=&\frac12\left(y_{n-1}-2\sqrt{x}+\frac{x}{y_{n-1}}\right)\\
&=&\frac12\left(\sqrt{y_{n-1}}-\frac{\sqrt{x}}{\sqrt{y_{n-1}}}\right)^2\\
&=&\frac12\left(\frac{y_{n-1}-\sqrt{x}}{\sqrt{y_{n-1}}}\right)^2.
\end{eqnarray}
Similary
\begin{eqnarray}
y_n+\sqrt{x}=\frac12\left(\frac{y_{n-1}+\sqrt{x}}{\sqrt{y_{n-1}}}\right)^2.
\end{eqnarray}
Now I think you can get the answer.
