# Evaluating the double limit $\lim_{m \to \infty} \lim_{n \to \infty} \cos^{2m}(n! \pi x)$

I have to find out the following limit $$\lim_{m\to\infty}\lim_{n\to\infty}[\cos(n!πx)^{2m}]$$ for $x$ rational and irrational. for $x$ rational $x$ can be written as $\frac{p}{q}$ and as $n!$ will have $q$ as its factor the limit should be equal to 1. the second part of irrational is giving me problems. I first thought that limit should be zero as absolute value of cosine term is less than 1 and power it to infinity you should get $0$. But then I realised that it was wrong. I brought the limit down to this form. $$e^{-\sin^2(n!πx)m}$$ after this I find the question quite ambiguous as they have just said $x$ is irrational. If I take $x$ as $\frac{1}{n!π\sqrt{m}}$ I get the limit as $\frac{1}{e}$ but if I take $x$ as $\frac{2}{n!π\sqrt{m}}$ I get the limit as $\frac{1}{e^4}$. please help me and tell me where have I gone wrong?

• It doesn't make sense to "take $x$ as $\frac{1}{n! x \sqrt{m}}$---by definition, you're evaluating a limit in $m$ and $n$, i.e., studying the behavior for fixed $x$ as $m, n$ vary. Anyway, I don't think the limit exists for irrational $x$, at least not for this ordering of the limits. For fixed $m$, $\cos(n! \pi x)$ with vary in $(-1 , 1)$ without converging, and thus so will $\cos^{2m}(n! \pi x)$. – Travis Oct 16 '14 at 7:02
• related discussion – Petite Etincelle Oct 16 '14 at 7:30
• See this and this and other questions shown there among linked questions. – Martin Sleziak Oct 16 '14 at 8:03
• It says it is 0, haven't I just shown that is wrong @martin – avz2611 Oct 16 '14 at 8:24
• By the "related discussion" link, the inner limit for $x=e$ is $1$, and so is the outer limit. Presumably, you can extend that to all integer powers of $e$ and many other reals, using series with a faster decrease than $1/i!$. – Yves Daoust Feb 20 '15 at 9:23

## 1 Answer

For $m>0$ the limit of $cos^{2m}(n!\pi.x)$ as $n$ goes to infinity does not exist for some $x$. For example, for natural number k, let $f(k)=1/2$ if $k$ is an integer power of $2$, otherwise let $f(k)=0$. Let $x=f(0)/0!+f(1)/1!+f(2)/2!+...$