# Find the limit of a sequence defined recursively

$$\displaystyle a_{n+1} = \frac{1}{2}\left(a_n+\frac{x}{a_n}\right) \quad \mathrm{with} \quad a_0 = c$$ such that $x > c > 1 \quad \text{and} x,c \in \mathbb{R}$. Find the limit $\displaystyle \lim_{n\to\infty} a_n$ in terms of $c$ and $x$.

I said that, assume the limit is $\ell$ then we have $$\ell = \frac{1}{2}\left(\ell+\frac{x}{\ell}\right) \iff \ell^2 = x \implies \ell = \sqrt{x}.$$

I don't see any $c$ in my evaluation nor do I know how to get a $c$ term in the limit, is what I've done correct?

Edit: The question specifies that the limit exists, so there is no need to prove that the limit does exist - however, for future knowledge, if I wanted to prove that the limit did exist, how would I go about it? Would I use the epsilon-delta proof?

• How do you know that the limit $l$ exists? Oct 22 '15 at 10:39
• @Hetebrij in this case we know it because it's the squareroot computation limit :P Oct 22 '15 at 10:40
• If limit exist your calculation is right but can you prove the limit exists? Oct 22 '15 at 10:41
• I haven't included the full text of the question, but it specifies that $(a_n)_{n \in \mathbb{N}}$ is a convergent sequence and the limit exists. Oct 22 '15 at 10:42

What you've done is correct, the $c$ of choice is irrelevant for it because it'll always go for the positive solution by how it is structured. This is the old way to calculate squareroots since the babylonians if I recall correctly. $c$ is just a start value which ultimately means nothing in analysis but does change how long it takes to compute.
• A starting value ($c$ in this case) is needed for the recursive process to be well defined, whether or not it appears in the expression for the limit. Oct 22 '15 at 11:12
Consider that you need to find the zero of $$f(a)=a^2-x$$ Newton formula is $$a_{n+1}=a_n-\frac{f(a_n)}{f'(a_n)}=\frac{1}{2}\left(a_n+\frac{x}{a_n}\right)$$ and $a_0=c$ is your starting guess.