Algorithm for computing square roots. Fix a positive number $\alpha$. Choose $x_1>\sqrt{\alpha}$ and define $x_2, x_3, x_4, \dots$ by the recursion formula $$x_{n+1}=\frac{1}{2}\left(x_n+\frac{\alpha}{x_n}\right).$$It's easy to check that $\{x_n\}$ decreases monotonically and $x_n\to\sqrt{\alpha}$.
Put $\varepsilon_n=x_n-\sqrt{\alpha}$ and we get $\varepsilon_{n+1}=\dfrac{\varepsilon_n^2}{2x_n}<\dfrac{\varepsilon_n^2}{2\sqrt{\alpha}}$ so that, setting $\beta=2\alpha,$ $$\varepsilon_{n+1}<\beta\left(\frac{\varepsilon_1}{\beta}\right)^{2^n} \qquad(*)$$
Am I true that formula $(*)$ shows the rapidity of convergence of above prove?
What about if $x_1>3\sqrt{\alpha}$? Because in this case $$\frac{\varepsilon_1}{\beta}=\frac{x_1-\sqrt{\alpha}}{2\sqrt{\alpha}}>\frac{2\sqrt{\alpha}}{2\sqrt{\alpha}}=1$$

Your (*) inequality is not contradictory with $\frac{\epsilon_1}{\beta}>1$ for $x_1 > 3\sqrt{\alpha}$.
It just means that the convergence is as fast as (*) for $x_n$ close enough to $\sqrt{\alpha}$.
