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)
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.