# Does this double limit of $\cos^{2n}(m! \pi x)$ exist?

I've figured out that the repeated limit exists:

$$\lim_{m \to \infty} \lim_{n \to \infty} \cos^{2n}(m! \pi x) = \begin{cases} 1,&x\text{ is rational}\\ 0,&x\text{ is irrational}\end{cases}$$

but does the double limit also exist?

as in: $$\lim_{m,n \to \infty}\cos^{2n}(m! \pi x)$$

If x is rational then for a large enough m the value will be $$1^n$$ constantly so that's not an issue.

But if x is irrational I'm having a hard time justifying why the limit should be zero. It seems that I can't rule out the possibility that $$m! * \pi * x$$ could somehow converge on to a multiple of $$\pi$$ fast enough to also be 1 or 0 depending on the subsequence of indexes thereby making the limit nonexistent.

• See principal of mathematical analysis by rudin – Cloud JR Nov 20 '19 at 13:27
• When cos attain +1 or -1? If abs(cos x) <1 and applying limit as n tends to infinite what happen ( of course we fix the x and take limit ) – Cloud JR Nov 20 '19 at 13:31

$$m! \cdot x$$ will always be irrational when x is irrational, since $$m!$$ is always rational. Therefore, cosx always lies between 1 and -1 for any irrational multiple of $$\pi$$. Taking $$cos^{2n}(m!\pi x)$$ in this case would mean that the limit is zero, which proves that the double limit exists.
• But m is also changing so how can we ensure that in fact $m! * \pi * x$ doesn't somehow approach a multiple of pi very quickly? – davegri Nov 20 '19 at 13:39
• Exactly. It's the same kind of wrong argument that $(1-\frac1n)^{n}$ converges to $0$ for $n \to \infty$, "because the base is always smaller than $1$, and taken to a power that gets bigger and bigger". – Ingix Nov 20 '19 at 14:45
• Because since x is irrational, $m! \cdot x$ will also be irrational! it's not about reaching multiples of pi, it's about reaching an integral multiple, since only integral multiples will be equal to either 1 or -1 – Aniruddha Deb Nov 20 '19 at 15:13
• @AniruddhaDeb For $x$ irrational, find $m_n(x)$ such that $\cos(m_n(x)! \pi x)>1-1/n$, then the limit of $f_{m_n(x),n}(x)$ exceeds $e^{-2}$ if it exists. One can do this over to show the limit along another sequence is less than that if it exists. This means the double limit cannot exist. – Ian Nov 20 '19 at 18:42