 Newton iteration for a square root

For $c>0$ and $a_0>0$, the displayed recurrence relation stays positive and satisfiesAfter the first step, $a_n^2\ge c$, so $a_{n+1}-a_n=(c-a_n^2)/(2a_n)\le 0$. The bounded monotone sequence theorem gives a limit of a sequence $L\ge 0$; the inequality $L^2\ge c>0$ makes $L$ positive, and passage to the limit of a sequence in the recurrence relation gives $L^2=c$. This proves existence of the positive square root without assuming it in the construction. Once $s=\sqrt{c}$ is known, the exact error identity $a_{n+1}-s=(a_n-s{)}^2/(2a_n)$ explains the eventual quadratic convergence.