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$$\displaystyle a_{n+1} = \frac{1}{2}\left(a_n+\frac{x}{a_n}\right) \quad \mathrm{with} \quad a_0 = c$$ 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?

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    $\begingroup$ How do you know that the limit $l$ exists? $\endgroup$
    – Hetebrij
    Oct 22 '15 at 10:39
  • $\begingroup$ @Hetebrij in this case we know it because it's the squareroot computation limit :P $\endgroup$ Oct 22 '15 at 10:40
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    $\begingroup$ If limit exist your calculation is right but can you prove the limit exists? $\endgroup$
    – A.F.23
    Oct 22 '15 at 10:41
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    $\begingroup$ 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. $\endgroup$
    – Zain Patel
    Oct 22 '15 at 10:42
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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.

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  • $\begingroup$ The question must have been badly worded then, thanks! :) $\endgroup$
    – Zain Patel
    Oct 22 '15 at 10:39
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    $\begingroup$ It isn't badly worded as much as tricky :) It is meant to make you suspicious like you were and think it matters, but when you recognize the formula nad others you know it is just irrelevant garbage. $\endgroup$ Oct 22 '15 at 10:39
  • $\begingroup$ 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. $\endgroup$ Oct 22 '15 at 11:12
  • $\begingroup$ That is correct $\endgroup$ Oct 22 '15 at 11:16
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You could recognize that the sequence is the sequence of Newton iterates.

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.

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