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Evaluate the following limit for all real values of $x$.

$$\lim_{n \rightarrow \infty} \lim_{m \rightarrow \infty}{\cos^m(n!2\pi x)}$$

My claim is that the limit is $1$ when $x=\frac{k}{n!}$ where $k\in \mathbb{Z}$, $0$ when $x=\frac{1+2k}{n!4}$, limit does not exist when $x=\frac{1+2k}{n!2}$.

When $x$ is not of the above values, limit is $0$.

My reason is that I notice when $\cos(n!2\pi x)=1$, limit will always be one and when $\cos(n!2\pi x)=-1$, we have $\lim(-1)^m$ does not exist. When $-1<\cos(n!2\pi x)<1$, the limit of $m$ will make $\cos(n!2\pi x)=0$ for any fixed $n$.

Is my claim true?

Idonknow
  • 15,643

2 Answers2

5

We quite obviously have: $$\lim_{m\to\infty}\cos^m(n!2\pi x)=\cases{1&if $n!x\in\mathbb{Z}$\\0&if $2n!x\notin\mathbb{Z}$\\\mathrm{doesn't\ exist}&else}$$ And thus: $$\lim_{\mathbb{N}\ni n\to\infty}\left(\lim_{m\to\infty}\cos^m(n!2\pi x)\right)=\cases{1&if $x\in\mathbb{Q}$\\0&else}$$ I'll leave it to you to check the details.

Daniel Robert-Nicoud
  • 29,795
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If $x\in\mathbb{Q}$ then $x=\frac{a}{b}$ for some $a,b\in\mathbb{Z}$ so we have for $$n>b$$ $n!x\in\mathbb{Z}$ and therefore $$\lim_{m\to \infty}{\cos^m{(n!2\pi x)}}=\lim_{m\to \infty}{1^m}=1$$ Thus $\forall \epsilon>0$ we can take $n>N=b$ to get $$\left|\lim_{m\to \infty}{\cos^m{(n!2\pi x)}}-1\right|=|1-1|=0<\epsilon$$ so the limit is $1$.

If $x$ is irrational then $$\lim_{m\to \infty}{\cos^m{(n!2\pi x)}}=0$$ regardless of the value of $n$, so we have $$\lim_{n\to\infty}{0}=0$$