Limit that is able to count a square root.

Question

Can anyone explain me why does that limit is actually solving square root :

$\lim_{n\rightarrow \infty}{x}_{n} = \sqrt{a} \\{x}_{0} = \frac{a}{2}; \quad {x}_{n} = 0.5*({x}_{n-1} + \frac{a}{{x}_{n-1}}); \\n = 1, 2, ...$

I was not able to find anything else except binary search for square root problem, so I have no idea how the limit will turn out to be the right answer.

Answer 1

note that $$\frac{1}{2}\left(x_{n-1}+\frac{a}{x_{n-1}}\right)\geq \sqrt{a}$$

Answer 2

Here is the easiest approach I found to get it:

1. If $\lim_{x\rightarrow \infty}{x}_{n}=b$ then $\lim_{x\rightarrow \infty}{x}_{n-1}=b$ as well.
2. Now let us notice that the limit actually exists.

Having 1. and 2. we infer that $2b = b + a/b \implies b^2 = a.$

So, the limit really counts square root.