# Clarity on uniform continuity of functions

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.

• 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 $x \sin(\pi/x)$. – Ian Mar 27 at 10:36
• 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 $|f(x)-L|+|f(y)-L|$. – Ian Mar 27 at 10:43
• 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. – user715501 Mar 27 at 11:01
• @Ian what about my argument using $x_n$ and $s_n$? – user715501 Mar 27 at 11:24
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.