Let $$f(x)=\limsup_{n\rightarrow \infty, n\in\mathbb{N}} \sin(n)^{n^x}.$$ Compute $f(1)$ and $f(2)$.
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In view of a result by Khinchine (see e.g. Theorem 3.6(a)), there exist infinitely many pairs $(p,q)$ of natural numbers such that $|2\pi q+\tfrac\pi2-p|<1/q$. For such $p$ and $q$, letting $q\to\infty$, we have $\sin p=1-O(1/q^2)=1-O(1/p^2)$ and hence $\sin^p p\to1$. So, $f(1)=1$.
As noted by Gerald Edgar, the same argument shows that $f(x)=1$ for all $x\in(0,2)$. For $x=2$, it seems we have to deal with Diophantine approximation to $\pi$, rather than that to a arbitrary irrational number. This seems to boil down to an open problem; cf. this answer.
answered Jul 14, 2023 at 14:56
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$\begingroup$ Same argument for $1\le x < 2$. But $x=2$ is on the border, this computation does not answer that case. $\endgroup$ Commented Jul 14, 2023 at 15:43
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$\begingroup$ @GeraldEdgar : You are quite right. For $x=2$, it seems we have to deal with Diophantine approximation to $\pi$, rather than that to a arbitrary irrational number. This seems to be an open problem. $\endgroup$ Commented Jul 14, 2023 at 16:13
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$\begingroup$ Thanks a lot for the answer! I found this problem on one of the t-shirts that were dustributed in a math competition about 10 years ago. Unfortunately, I don't have an answer to the second part of the problem either. $\endgroup$ Commented Jul 16, 2023 at 16:22
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$\begingroup$ By the same argument with specifying the constant in $O(1/q^2)$, we may obtain a positive lower bound of $f(2)$. $\endgroup$ Commented Aug 3, 2023 at 17:04
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$\begingroup$ @SungjinKim : Yes, you are right. $\endgroup$ Commented Aug 3, 2023 at 17:56