# Limit and existence of this function

I have the following function f(x) = $$\lim_{m \to \infty} \lim_{n \to \infty} (\cos(\pi m!x))^{2n}$$

I have to prove that f(x) exists and calculate its value. I was wondering if using $$e^{i\pi x}$$ to direct calculate it was enough or I need the definition with $$\epsilon$$.

After that, I need, as well, to find {x$$\in\mathbb{R}$$| f is continuous at x}.

I decided to consider the série $$e^u = \sum_{k=0}^{n}\frac{u^k}{k!}$$. Here, u = $$i\pi x m!$$ I choose r > 0 and x$$\in$$[-r,r], so we can $$i\pi x m! \leq i \pi r m!$$ and using the Abel's theorem (don't know how it's called in english) then the serie is normally convergent.

Thus, f is continuous when x $$\in \mathbb{R}$$

• I am not sure why you need to go to $e^{i\pi x}$. Just consider rational and irrational points separately. Once you have the limit, continuity should be more clear. Nov 7, 2020 at 16:49
• consider $m=2$ fixed. $0 \leq \cos(2\pi x)^2 \leq 1$. Only the points where the value is 1 will converge to 1 as $n\rightarrow \infty$. At all other points, it converges to 0. For all values of $m$, there is a similar form. You then need to look at convergence of these functions at rational and irrational points. Nov 7, 2020 at 19:50
• @Dunham Re($(e^{i x \pi})^{m!}$) = cos($\pi x m!$). We know that $-1 \leq e^{I x \pi} \leq 1$ so f(x) = 1. Am I wrong? Nov 7, 2020 at 19:52
• $e^{i \pi x}$ is not real valued, and $f(x)$ is not identically 1. Nov 7, 2020 at 19:52
• I don't think that expanding the exponential is a good idea.
– user65203
Nov 9, 2020 at 20:27

Define $$f_m(x) = \lim_{n\to \infty} (\cos(\pi m! x))^{2n}$$.
If $$x$$ is rational, then $$x=p/q$$ for some integers $$p,q$$. If $$2q$$ divides $$m!$$, then $$(\cos(\pi m! x))^2 = 1$$, which will be the case for $$m$$ sufficiently large. Hence $$f_m(x)=1$$ eventually for all $$m$$ after some point.
If $$x$$ is irrational, then $$0<(\cos(\pi m! x))^2<1$$, so $$f_m(x)=0$$ for all $$m$$.
Therefore $$f(x) = 0$$ for irrational $$x$$ and $$f(x)=1$$ for rational $$x$$. This function is not continuous anywhere.
• Thanks! It helps verify what I have done. Actually, when I understood that $\cos(m!\pi x)$ was like $\cos(kx)$ here $k = \pi m!$. It's very easy to do after that thinking. Nov 9, 2020 at 20:59