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I'm stuck on proving the limit of the sequence defined recursively as $x_{n+1}= \frac{x_{n}}{2} + \frac{1}{x_{n}}$, $x_{1} = 2$.

I have proved that the sequence is decreasing and bounded below by $\sqrt{2}$, and found that the limit of the sequence is $\sqrt{2}$ (by letting n go to $\infty$ and getting $L = \frac{L}{2} + \frac{1}{L}$, solving to find $L=\sqrt{2}$) but I'm not satisfied with this solution. Is there a way using the definition of the limit of a sequence (i.e if $x$ is the limit of the sequence, $\forall \epsilon > 0 , \exists N(\epsilon), \forall n \geq N(\epsilon), |x_{n} - x| < \epsilon $) to prove that this is in fact the limit of the sequence? Moreover, how can we use the definition of the limit to prove limits for recursive sequences?

Thank you for your time.

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  • $\begingroup$ do you mean $x_{n+1}=\frac{1}{2}\left(x_n+\frac{1}{x_n}\right)$? $\endgroup$ Sep 27 '14 at 18:17
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    $\begingroup$ No, the problem is stated exactly as I have done so in the post. $\endgroup$
    – arcbloom
    Sep 27 '14 at 18:21
  • $\begingroup$ why have i got $-1$ my solution is also possible $\endgroup$ Sep 27 '14 at 18:24
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You might start with $$\left|x_{n+1}-\sqrt2\right|=\frac12\cdot\left|1-\frac{\sqrt2}{x_n}\right|\cdot\left|x_n-\sqrt2\right|,$$ and show that, in the regime one is interested in, $$\left|1-\frac{\sqrt2}{x_n}\right|\leqslant1,$$ for every $n$, leading to $$\left|x_{n+1}-\sqrt2\right|\leqslant\frac1{2^n}\cdot\left|x_1-\sqrt2\right|.$$

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  • $\begingroup$ Shouldn't it be in the factorization $|1 - \frac{\sqrt{2}}{x_{n}}|$? Thanks for the answer. $\endgroup$
    – arcbloom
    Sep 27 '14 at 18:28
  • $\begingroup$ Yes. You are welcome. $\endgroup$
    – Did
    Sep 28 '14 at 9:23

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