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

1 Answers1

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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.

Rhys Hughes
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