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I'm doing some textbook problems in Burden and Faires Numerical Analysis, when I encountered this question that I did not understand how to do.

How would I go about finding the rate of convergence for

$$\lim_{n\to\infty}\sin\dfrac{3}{n}=0$$

I know that it is true, because $\dfrac{3}{n}$ goes toward $0,$ and $\sin0=0$, but I don't know the method/ procedure in order to calculate the order of convergence for this. Any help or tips would be appreciated. Thanks!

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    $\begingroup$ Perhaps use $\lim_{x \to 0} \frac{\sin x}{x} = 1$? What definition of rate of convergence do you use? $\endgroup$ – copper.hat Sep 30 '13 at 23:14
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Let $x_n = \sin \frac{3}{n}$. We know $x_n \to 0$.

One approach is to calculate $\mu = \lim_n \frac{|x_{n+1}-L|}{|x_{n}-L|}$.

Here we have $\mu = \lim_n \left| \frac{ \sin\frac{3}{n+1}}{\sin \frac{3}{n}} \right| = \lim_n \left| \frac{ \frac{3}{n+1}}{ \frac{3}{n}} \right| = 1$, hence we have what is known as sublinear convergence.

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As you say, for small $x$ we have $\sin x \approx x$, so for large $n$ we have $\sin \frac 3n \approx \frac 3n$. Then whatever terminology you have for $\frac 3n$ approaching zero as $n$ goes to infinity.

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