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$x_1 = \sin x_0 > 0$

$x_{n+1} = \sin x_n$

Prove

$\lim_{x \to \infty }$ $\sqrt{\frac{n}{3}} $ $x_n = 1$

having problem of trying to figure out what value for the $x_0$ starts at.

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    @RagibZaman your good, I tried checking before I posted. Now I feel bad. Thanks2012-04-23

1 Answers 1

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Use Stolz theorem:

$nx_n^2=\frac{n}{\frac{1}{x_n^2}}\to\frac{n+1-n}{\frac{1}{x_{n+1}^2}-\frac{1}{x_{n}^2}}=\frac{x_{n+1}^2x_{n}^2}{x_{n}^2-x_{n+1}^2}=\frac{x_n^2\sin x_n^2}{x_n^2-\sin^2x_n}=\frac{\sin^2x_n}{1-\frac{\sin^2x_n}{x_n^2}}$

By $\sin x\sim x,\displaystyle\frac{\sin x}{x}\sim 1-\frac{x^2}{3!}$ and $x_n\to 0$, you can obtain $nx_n^2\to3$

I assume $x\to\infty$ is a typo, which should be $n\to\infty$

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    @ziyuang thanks. that really makes it easier.2012-04-27