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Let $t\in (0,\pi)$ and $n$ change in natural numbers. I am wondering what is the answer to the following limit. $$\lim_{n\rightarrow \infty} \frac{\sin(nt)}{\sin(t)}.$$

Thank you.

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Raymond and I interpreted the question differently in our answers. Since you started out with "Let $t\in(0,\pi)$", I thought that $t$ is fixed; but the question makes a lot more sense if it's intended to be about convergence of distributions as Raymond interpreted it. If so, I think you should clarify it. – joriki Jul 18 '12 at 6:46
up vote 6 down vote accepted

Well $\ \frac {\sin(nt)}{\pi t}\to \delta(t)\ $ as $n\to\infty\ $ (equation (9)) so that (since $\frac t{\sin(t)}\to1$ as $t\to0$) : $$\lim_{n\to \infty} \frac{\sin(nt)}{\sin(t)}=\pi \frac t{\sin(t)}\delta(t)=\pi\delta(t)$$

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Your answer should be $\pi \frac t{\sin(t)}\delta(t)$, since you are not asked to take the limit as t goes to zero. – Mhenni Benghorbal Jul 18 '12 at 7:16
By the way, $\delta(t)$ is the Dirac delta function. – Mhenni Benghorbal Jul 18 '12 at 7:17
@Mhenni: $f(t)\delta(t)=f(0)\delta(t)$. – joriki Jul 18 '12 at 7:19
You are right. Thanks. – Mhenni Benghorbal Jul 18 '12 at 7:25
@MhenniBenghorbal: this has to be considered from the point of view of distributions. From a distribution point of view the value at a point ($0$ here) is unimportant since we are considering families of equivalent functions. The 'extended' continue function with value $0$ is from this point of view 'equal' to $\frac {\sin(t)}t$. You may too replace $f(0)$ by the limit as $t\to 0$ of $f(t)$ if you prefer. – Raymond Manzoni Jul 18 '12 at 7:57

$\sin t$ doesn't depend on $n$, so you can pull it out; so basically you're asking for the limit of $\sin(nt)$ for $n\to\infty$. Since $\sin x$ oscillates, there's no such thing.

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