This is a problem from ISI MStat 2018 PSA Problem 10 based on Dirichlet Function.
Let \(x\) be a real number. Then \( \lim {m \rightarrow \infty}\left(\lim {n \rightarrow \infty} \cos ^{2 n}(m ! \pi x)\right) \)
Limit
Sandwich Theorem
Answer: is (B)
ISI MStat 2018 PSA Problem 10
Introduction to Real Analysis by Bertle Sherbert
Check two cases separately one when x is rational and other is when x is irrational.
If \( m!x \) is an integer, then \( cos ^{2 n}(m ! \pi x) =1 \)
If x is rational \( \frac{p}{q}\), then, eventually, for large enough m, m! will be divisible by q , so that \(m!x\) will be an integer, and we have \( \lim {m \rightarrow \infty}\left(\lim {n \rightarrow \infty} \cos ^{2 n}(m ! \pi x)\right) =1 \)
If x is irrational, \( m!x \) will never be an integer, and \( |cos(m! {\pi } x)|<1\) , so that \( \lim {m \rightarrow \infty}\left(\lim {n \rightarrow \infty} \cos ^{2 n}(m ! \pi x)\right) =0 \) for all m>0 by Sandwich Theorem.
What I can see that the limit exists for all real $x.$ If $x$ is rational the limit evaluates to $1$ and if $x$ is irrational the limit evaluates to $0.$ Then why do say that the limit doesn't exist for any real $x$? The reasoning is not quite clear to me. Can you please explain?
Don't worry. It will be (B) i.e limit exists for all real x we have also shown that . I mistakenly put A as option . Between thanks for your valuable reply.