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I've been busy trying to answer this question for several hours now and I've thought of looking for when it is eventually increasing or decreasing, but it doesn't seem to ever be either.

It came with the hint that $x_n\geq 0$ and $1\leq x_n^2\leq 3$ for all natural numbers n.

I understand that the limit would be $\sqrt{\frac{3}{2}}$, but showing that the sequence is Cauchy is the part I'm struggling with.

Please help.

Daniel Schurbohm
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    Its limit $L$ is the solution to $L=\frac13(L+\frac3L)$ – J. W. Tanner Nov 03 '20 at 17:24
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    ${x_n}$ is neither increasing nor decreasing, but the alternating terms ${x_{2n}}$ and ${x_{2n+1}}$ are (one increasing, the other decreasing); you should be able to show that. Alternately, you can find the limit (hint: if $x$ is the limit, it satisfies $x=\frac13(x+\frac3x)$) and then look at the behavior of the sequence of differences $\delta_n=x_n-x$. – Steven Stadnicki Nov 03 '20 at 17:25
  • Or this. There are many variants of this question around. Approach0 – Jyrki Lahtonen Nov 04 '20 at 13:29

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