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Write $r=\sqrt{a}>0$ and subtract it from the iteration: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 $x_n>r$ 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.

Use the Exact Newton error identity for a quadratic to expose the cancellation without a long quotient expansion. Put $e_k=x_k-x_{\ast }$. Since $f(x_{\ast })=0$, direct expansion givesFrom part (ii), $e_1=\frac{1}{2}{\epsilon }^2+O({\epsilon }^3)$ and $x_1-\epsilon \to 1$. It follows thatThus the stronger result isThe vanishing lower-order terms are a concrete instance of quadratic convergence: to leading order, this step squares the previous error.

After the first update, the sequence lies in $[\sqrt{a},x_1]$ and is nonincreasing. The monotone convergence theorem for sequences gives a limit $L\ge \sqrt{a}$. The update is continuous on this interval, so passing to the limit givesPositivity forces $\boxed{x_n\to \sqrt{a}\text{ for every }{x}_0>0}$. If an iterate reaches the root exactly, the constant tail gives the same conclusion. The Exact Newton error identity for a quadratic here is $x_n-\sqrt{a}=(x_{n-1}-\sqrt{a}{)}^2/(2x_{n-1})$, also showing quadratic convergence once the error is small.