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I have problem with calculating $\displaystyle \lim_{n\to \infty}(\lim_{k \to \infty}(\cos{(|n!\pi x|)^{2k}} )) $

I only get hint that it will be $1 \vee 0$ regarding to wheter $x \in \mathbb{Q}$ or $x \in (\mathbb{R}- \mathbb{Q})$ but don't have idea how to find it.

alex
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  • Intuitively, which do you think is zero and which $1$? – Simon S Dec 31 '14 at 21:28
  • Intuitevely $1$ for rational, but these two limits means the same as one limit with double index $n,k$ meaning $\displaystyle \lim_{n,k \to \infty}$ ?? – alex Dec 31 '14 at 21:35
  • Note that if $x$ is a rational number, then $n!x$ is an integer (in fact, an even integer) for all sufficiently large integral $n$. And if $x$ is not rational, then $n!x$ is not an integer for any positive integral $n$. – MPW Dec 31 '14 at 21:51
  • It is not a single limit. You must first take one limit, then the other (a so-called "iterated" limit). Order is important here. – MPW Dec 31 '14 at 21:52
  • If $x$ is rational then for large enough $n$ we have $n!x$ is an integer, so the limit is $1$. For fixed $n$, if $x$ is irrational, $(\cos(n!\pi x))^{2k}$ can be made arbitrarily close to $0$. – André Nicolas Dec 31 '14 at 21:53

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