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The question says that if $\frac{\cos x}{\sin ax}$ is a periodic function then find the find the value of $$\lim _{m \rightarrow \infty} \lim_{n \rightarrow \infty} \left(1+ \cos^{2m} n! \; \pi a\right)$$ I really don't have any clue on how to begin the question. I have never seen a question like this. The only thing I could think of since is that since $\frac{\cos x}{\sin ax}$ is periodic then we can assume $a$ to be $1$ since $\cot x$ is periodic. So then we have to find the limit of the function, $$\lim _{m \rightarrow \infty} \lim_{n \rightarrow \infty} \left(1+ \cos^{2m} n! \; \pi \right)$$ But even this is based on an assumption and I have no idea on how to go from here. Any clue or hint would be much appreciated.

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... we can assume $a$ to be 1...

We can (must) assume $a$ to be rational.

In your case (the general case is almost the same) the inner limit $$\lim_{n\rightarrow \infty}\left(1+\cos^{2m}n!\pi\right)$$ is trivial because for $k\in\Bbb Z$, $\cos(k\pi) = \cdots$

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  • $\begingroup$ Alright so we can just say that since anything of the form $\cos kx = 1$ then we can say the limit of the function is $1+1=2$? $\endgroup$ Jan 28 '19 at 14:39
  • $\begingroup$ I did actually think of the problem in that way but I just wasn't convinced by that argument $\endgroup$ Jan 28 '19 at 14:39
  • $\begingroup$ @PrakharNagpal, yes. In the general case, $n!a$ is integer for $n$ large enough. $\endgroup$ Jan 28 '19 at 14:41
  • $\begingroup$ Alright, thank you so much! $\endgroup$ Jan 28 '19 at 14:42
  • $\begingroup$ @PrakharNagpal, $\cos k\pi = \pm 1$. $\endgroup$ Jan 28 '19 at 14:42

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