= Solution
Write $r=\sqrt a>0$ and subtract it from the iteration:
$$
e_{n+1}=\frac{x_n^2+a-2rx_n}{2x_n}
=\frac{(x_n-r)^2}{2x_n}
=\boxed{\frac{e_n^2}{2x_n}}.
$$
This is an <Exact Newton error identity for a quadratic>. If $x_n>r$, then $x_n>0$ and $e_n>0$, so $e_{n+1}>0$. Induction from $x_0>r$ proves \b[$x_n>r$ for every $n$]. Also $x_{n+1}-x_n=(a-x_n^2)/(2x_n)<0$, so the iterates decrease while remaining above the root. Their limit must satisfy $L=(L+a/L)/2$, hence $L=r$. The squared error numerator explains the ensuing <quadratic convergence>.
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