Limit of = (sin nx) / (sin x) as n goes to infinity.

[uNmiN]

Hello everyone.
I need help trying to calculate/ trying to realize what the limit function of (sin nx)/(sin x) as n goes to infinity is.

from another topic here on MBH ("Show δn = (sin nx) / (pi x) is a delta distribution") and after research with Wolfram Alpha I know that the limit function of (sin nx)/(pi x) as n goes to infinity is the delta distribution δ(x). I am wondering now what happens if we exchange the "x" from the denominator to (sin x). Now the denominator is a periodic function of x and reaches zero periodically. By plotting the function (sin nx)/(sin x) with increasing values for n It seems that the function becomes something like a sum of various delta distributions at the zeros of (sin x) (which actually makes some sense..), with alternating signs. But I don't know how to tackle this problem more 'mathematically'. Is there a way we can reach that result and/or solve this limit?

Thanks in advance!
uNmiN.

 

[Sudharaka]

Hello everyone.
I need help trying to calculate/ trying to realize what the limit function of (sin nx)/(sin x) as n goes to infinity is.

from another topic here on MBH ("Show δn = (sin nx) / (pi x) is a delta distribution") and after research with Wolfram Alpha I know that the limit function of (sin nx)/(pi x) as n goes to infinity is the delta distribution δ(x). I am wondering now what happens if we exchange the "x" from the denominator to (sin x). Now the denominator is a periodic function of x and reaches zero periodically. By plotting the function (sin nx)/(sin x) with increasing values for n It seems that the function becomes something like a sum of various delta distributions at the zeros of (sin x) (which actually makes some sense..), with alternating signs. But I don't know how to tackle this problem more 'mathematically'. Is there a way we can reach that result and/or solve this limit?

Thanks in advance!
uNmiN.

Hi uNmiN, :)

$$\lim_{n\rightarrow\infty}\frac{\sin(nx)}{\sin x}=\frac{1}{\sin x}\lim_{n\rightarrow\infty}\sin(nx)$$

For a fixed value $x$, $\sin(nx)$ alternates sign as $n$ changes. Therefore the limit of $\sin(nx)$ as $n$ goes to infinity does not exist. Thus the limit $\lim_{n\rightarrow\infty}\frac{\sin(nx)}{\sin x}$ does not exist.

 

[uNmiN]

I don't think it does not exist, since the same can be said about sin(nx)/x and this limit exist being equal to the delta distribution

It's not the limit of the sequence sin(nx) I'm talking about, but the limit of sin(nx)/sin(x) as a function of x..