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Identify the function $f(x) = \displaystyle \lim_{n \to \infty}(\lim_{k \to \infty}(\cos(n!\pi x)^{2k}))$.

I was confused how to find this limit because cosine oscillates, so I didn't think it was possible to find the limit.

Puzzled417
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    I haven't thought very hard about it but it might be good to break it into the cases of $x$ rational and $x$ irrational and consider sufficiently large (fixed) $n$ for the $k$ limit. – Cameron Williams Aug 01 '16 at 17:00
  • (So, it's the indicator function of the rationals?) – user66081 Aug 01 '16 at 17:02
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    If $x$ is rational then eventually $n!x$ is an integer. If $x$ is irrational then $n!x$ is never an integer. From this you can figure out the properties of $|\cos(n! \pi x)|$ which give the result. – Ian Aug 01 '16 at 17:04
  • It's the characteristic function for the set of rational numbers. – Asinomás Aug 01 '16 at 17:09
  • This one is more interesting: $f(x) = \displaystyle \lim_{k \to \infty}(\lim_{n \to \infty}(\cos(n!\pi x)^{2k}))$ – Asinomás Aug 01 '16 at 17:12
  • nevermind, the first limit doesn't exist. – Asinomás Aug 01 '16 at 17:13
  • @CarryonSmiling The first limit does exist if $x$ is rational. I don't think it exists if $x$ is irrational though. – Ian Aug 01 '16 at 19:10
  • either way $$ f(x) = \begin{cases}0 \quad & \text{if } x \text{ is irrational} \ 1 \quad & \text{if } x \text{ is rational} \\end{cases}$$ – robert bristow-johnson Feb 03 '17 at 17:20
  • https://math.stackexchange.com/questions/264889/double-limit-of-cos2nm-pi-x-at-rationals-and-irrationals?noredirect=1&lq=1 – StubbornAtom Jan 04 '19 at 19:44

2 Answers2

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Set $$D_n = \{ x \in \mathbb{Q} \, : \, x=p/q \text{ with } q \text{ integer divisor of } n! \}. $$It is enough to notice that $$\lim_{m \to \infty} [\cos^2 (n! \pi x)]^m=\begin{cases} 1 & \text{if } x \in D_n \\ 0 & \text{if } x \in \mathbb{R} \setminus D_n \end{cases}=f_n(x).$$Since $$\bigcup_{n=1}^\infty D_n = \mathbb{Q}$$we have that $$\lim_n f_n(x) = \chi_{\mathbb{Q}} (x)= \begin{cases} 1 & \text{if } x \in \mathbb{Q} \\ 0 & \text{if } x \in \mathbb{R} \setminus \mathbb{Q} \end{cases}$$

gangrene
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if x \in Q then x = p/q then for some very large n n!/q is even number therefore making cos(n! \pi x) = 1 hence making the limit 1

else if x is irrational then (n!\pi x) is never \pi hence Z = (cos(n! \pi x) )^2 <1 implies Z^k=1 for all k

Karmanya GB
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