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This is a problem from ISI MStat 2018 PSA Problem 10 based on Dirichlet Function.

Let \(x\) be a real number. Then \( \lim {m \rightarrow \infty}\left(\lim {n \rightarrow \infty} \cos ^{2 n}(m ! \pi x)\right) \)

- (A) does not exist for any x
- (B) exists for all x
- (C) exists if and only if x is irrational
- (D) exists if and only if x is rational

Limit

Sandwich Theorem

But try the problem first...

Answer: is (B)

Source

Suggested Reading

ISI MStat 2018 PSA Problem 10

Introduction to Real Analysis by Bertle Sherbert

First hint

Check two cases separately one when x is rational and other is when x is irrational.

Second Hint

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 \( \lim {m \rightarrow \infty}\left(\lim {n \rightarrow \infty} \cos ^{2 n}(m ! \pi x)\right) =1 \)

Final Step

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.

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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.