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Show that $f(x) = \text{cos}x\text{cos}\frac{\pi}{x}$, $x\in(0,1)$ is not uniformly continuous while $g(x) = \text{sin}x\text{sin}\frac{\pi}{x}$, $x\in(0,1)$ is uniformly continuous on the given intervals.

I have referred this solution https://math.stackexchange.com/a/3597074/697936. why does the solution for $f(x)$ we just have to see the cos($\frac{\pi}{x})$ part. Why does this conclude that $f(x)$ is not uniformly continuous.

And after going through the comments , I can also say $lim_{x \to 0} \text{sin}(\frac{\pi}{x})$ also does not have limit at 0. Since if you pick sequences $s_n=\frac{2}{4n+1}$ and $x_n=\frac{1}{n}$, both $s_n,x_n \to 0$ as $n \to \infty$. But $\text{sin}(\frac{\pi}{x})=1$ for $s_n$ and $\text{sin}(\frac{\pi}{x})=0$ for $x_n$. But does this conclude anything about $g(x)$?

I would also like to know how does this statement: 'Note that $|\sin x \sin (\frac {\pi} x)| \leq |\sin x| \to 0$ as $x \to 0$.' help us. (say, (*)) in determining whether limit exists.

Please help. Thanks in advance. Any other answer that proves the claim will also do help. (Please try to elaborate your answer)

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  • $\begingroup$ The point is that in the cosine case, nothing dampens down the oscillation near x=0, so it is like looking at just $\cos(\pi/x)$. In the sine case, the other sine dampens down the oscillation so it is like looking at $x \sin(\pi/x)$. $\endgroup$ – Ian Mar 27 at 10:36
  • $\begingroup$ As for this point about the limit, the point is that in general if $f(x) \to L$ as $x \to x_0$ then you can get a bound on $|f(x)-f(y)|$ for $x,y$ near $x_0$, namely $|f(x)-L|+|f(y)-L|$. $\endgroup$ – Ian Mar 27 at 10:43
  • $\begingroup$ Correct me if am wrong: you are assuming as $x \t0 0$ sin$x$ behaves like$x$ and cos$x$ doesn't affect. (its like for small $\theta$ , sin($\theta$)=$\theta$). But still , the solution is not clear. $\endgroup$ – user715501 Mar 27 at 11:01
  • $\begingroup$ @Ian Any other answer that proves the claim will also do help $\endgroup$ – user715501 Mar 27 at 11:05
  • $\begingroup$ @Ian what about my argument using $x_n$ and $s_n$? $\endgroup$ – user715501 Mar 27 at 11:24
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For the cosine case, say $\pi/x_n=n\pi/2$ i.e. $x_n=2/n$ (I guess $n$ must begin at $3$ or higher but this is not important). Then $x_n$ is Cauchy and $f(x_n)$ is not (because it oscillates between $0$, nearly $1$, and nearly $-1$), so $f$ is not uniformly continuous. (It is a good exercise to show that uniformly continuous functions map any Cauchy sequence in the domain to a Cauchy sequence in the codomain, even when the Cauchy sequence in the domain isn't convergent in the domain).

For the sine case this doesn't break anything, because $g(x_n)$ actually is Cauchy since $|g(x)| \leq |x|$, so instead it's just a single "witness" of continuity and is thus not particularly useful. Instead in the sine case you can simply note that extending the definition of $g$ using the formula at $x=1$ and as a limit at $x=0$ (that limit being $0$, since $|g(x)| \leq |x|$) gives a continuous function on a compact interval so you can apply Heine-Cantor.

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