How prove $\lim x_n = \sqrt{\frac{a}{b}}$ if $b \neq 0$ for $x_1=c$ and $x_{n+1}=\frac{x_n^2 + \frac{a}{b}}{2x_n}$? Let $x_1=c$ and $x_{n+1}=\frac{x_n^2 + \frac{a}{b}}{2x_n}$. How prove  $\lim x_n = \sqrt{\frac{a}{b}}$ if $b \neq 0$ ?

For convenience, let $s:=\sqrt{\dfrac ab}$.
Then,
$$\frac{x_{n+1}-s}{x_{n+1}+s}=\frac{\dfrac{x_n^2+s^2}{2x_n}-s}{\dfrac{x_n^2+s^2}{2x_n}+s}=\left(\frac{x_n-s}{x_n+s}\right)^2,$$
and by recurrence
$$\frac{x_n-s}{x_n+s}=\left(\frac{c-s}{c+s}\right)^{2^n}.$$
This gives us the explicit formula
$$\color{green}{x_n=s\frac{1+r^{2^n}}{1-r^{2^n}}},$$
where $$r:=\frac{c-s}{c+s}.$$
It (quickly) converges to $s$ when $$|r|=\left|\frac{c-s}{c+s}\right|<1,$$
which is true for any
$$c>0.$$

Addendum for numericians:
Choose the integer $k$ such that $2^k\le s^2<2^{k+1}$ ($k$ is the rank of the most significant bit in $s^2$) and let $$c=\frac1{\sqrt[4]2}\cdot2^{k/2}\text{ or }c=\sqrt[4]2\cdot2^{(k-1)/2}$$ depending on the parity of $k$. Then
$$\frac 1{\sqrt[4]2}\le\frac cs<\sqrt[4]2.$$
This makes an excellent starting approximation, as at worse $r=0.086427\cdots=2^{-3.5323\cdots}$ and the iterates give at least $3.5,7,28,56\cdots$ exact bits, which are enough for the single and double precision floating-point representations.
