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If $x \in \mathbb{Q}$ then what will be the value of $\lim_{m \to \infty} \lim_{n \to \infty}(1+\cos^{2m}(n!\pi x))$ ?

Approach: Well the expression $\cos^{2m}(n!\pi x)$ would basically mean $(\pm1)^{\infty}$ which is not defined. Then it is possible to calculate this limit somehow?

The answer was given as 2.

Arturo Magidin
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marks_404
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  • the $2m$ makes it always even powers, which makes it $1$ – Alan Aug 16 '21 at 04:22
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    For sufficiently large $n$, $n!\pi x$ is an integer multiple of $\pi$, so $\cos(n!\pi x)$ is $1$ or $-1$. But since you are calculating the $m$th power of $\cos^2(n!\pi x)$, this is going to be $1$ for sufficiently large $n$, for all $m$. So the inside limit is $2$ for any $m\geq 1$, and then you are taking the outside limit of a constant function. – Arturo Magidin Aug 16 '21 at 04:23
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    Look at https://math.stackexchange.com/questions/264889/double-limit-of-cos2nm-pi-x-at-rationals-and-irrationals – DSD Aug 16 '21 at 04:25
  • @ArturoMagidin do we implicitly assume here that $n$ is always an integer? – DatBoi Aug 16 '21 at 04:46
  • @DatBoi: The use of $n$ and $m$ strongly suggest that, yes. The fact that we are using $n!$ (rather than a $\Gamma$-function) also tells us that it is almost certain $n$ is supposed to be an integer, and so that $m$ is as well. – Arturo Magidin Aug 16 '21 at 04:52
  • @Alan indeed, no idea why I missed that, Thanks a lot! – marks_404 Aug 16 '21 at 06:19

2 Answers2

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You can break down the $\cos^{2m}(n!\pi x)$ into $(\cos^{2}(n!\pi x))^m$. Then as soon as $n!$ starts to kill the denominator of $x$, you get $(1)^m=1$, so yes the limit is 2

Alan
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For any finite $x$, $n!*x$ is always an even number if $n \to \infty$.

Thus, $(\cos{(\text{even multiple of}\, \pi)})^{even}=1$

Nandeesh Bhatrai
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