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?

We have that 
$$y_n=\frac12\left(y_{n-1}+\frac{x}{y_{n-1}}\right)\implies y_{n-1}^2-2y_ny_{n-1}+2x=0 \tag 1$$
Now, multiplying $(1)$ by $x^{1/2}$ yields the following obvious identity
$$y_{n-1}^2\sqrt{x}-2y_ny_{n-1}\sqrt{x}+2x\sqrt{x}=-(y_{n-1}^2\sqrt{x}-2y_ny_{n-1}\sqrt{x}+2x\sqrt{x}) \tag 2$$
Next, adding $y_ny_{n-1}^2+xy_n-2xy_{n-1}$ to both sides of $(2)$ reveals 
$$\begin{align}
&y_ny_{n-1}^2+xy_n-2xy_{n-1}+y_{n-1}^2\sqrt{x}-2y_ny_{n-1}\sqrt{x}+2x\sqrt{x}\\\\
&=y_ny_{n-1}^2+xy_n-2xy_{n-1}-(y_{n-1}^2\sqrt{x}-2y_ny_{n-1}\sqrt{x}+2x\sqrt{x}) \tag 3
\end{align}$$
Factoring both sides of $(3)$ we have 
$$(y_n-\sqrt{x})(y_{n-1}+\sqrt{x})^2=(y_n+\sqrt{x})(y_{n-1}-\sqrt{x})^2 \tag 4$$
whereupon rearranging $(4)$ gives the coveted identity
$$\frac{y_n-\sqrt{x}}{y_n+\sqrt{x}}=\left(\frac{y_{n-1}-\sqrt{x}}{y_{n-1}+\sqrt{x}}\right)^2$$

It is easy to see that the sequence 
$$\phi_n\equiv \frac{y_n-\sqrt{x}}{y_n+\sqrt{x}}$$ 
converges since we have $\phi_n=\phi_{n-1}^2$ and $\phi_n<1$ for all $n$.  
Thus, $\phi_n \to 0$ and thus $y_n\to \sqrt{x}$ as was to be shown!
