Hero's Method/Proof 2

Proof
Let $a > 0$.

We make no statement about $x_1$.

We specify that:
 * $x_{n + 1} = \dfrac {x_n + \dfrac a {x_n} } 2$

Now:

If we now assume that $x_1 \ge \sqrt a$, then it follows (as above) that $x_n \ge \sqrt a$.

So:

If $\size y < 1$, then $y^n \to 0$ as $n \to \infty$ from Sequence of Powers of Number less than One.

So, by Limit of Subsequence equals Limit of Real Sequence:
 * $y^{2^n} \to 0$ as $n \to \infty$

Thus we see that:
 * $x_n \to \sqrt a$ as $n \to \infty$

provided that:
 * $\dfrac {x_1 - \sqrt a} {2 \sqrt a} < 1$

that is, that:
 * $\sqrt a \le x_1 < 3 \sqrt a$

We assumed (above) that $x_1 \ge \sqrt a$.

Now we have shown that $x_n \to \sqrt a$ as $n \to \infty$ provided that $\sqrt a \le x_1 < 3 \sqrt a$.

However, we have already shown that $x_n \to \sqrt a$ as long as $x_1 \ge 0$.

The advantage to this analysis is that this gives us an opportunity to determine how close $x_n$ gets to $\sqrt a$.