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Let $0<x_0<1$ if $x_{n+1} = sin(x_n)$ show that $\lim_{n\to\infty} \frac{x_n}{\sqrt{3}/n} = 1$

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  • $\begingroup$ thanks, I already solved the problem with Stolz–Cesàro theorem. . $\endgroup$
    – Sergio MNZ
    Nov 18, 2018 at 4:39
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    $\begingroup$ The first appearance of this question seems to be MSE question "Convergence of $\sqrt{n}x_{n}$ where $x_{n+1} = \sin(x_{n})$" from 2010. $\endgroup$
    – Somos
    Nov 18, 2018 at 5:29

1 Answer 1

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We have that:

$$\frac{x_n}{(\frac {\sqrt3}{n})}=\frac{nx_n}{\sqrt3}$$

Also, since for $0 <x <1$ we have that $x >\sin x$, $$\lim_{n\to \infty}{x_n}=0$$

You can do the rest.

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