I am self-learning Real Analysis from the text, Understanding Analysis by Stephen Abbott. I want to verify the claim: $h(x)=x \sin (1/x)$ is uniformly continuous on $(0,1)$. I'd like some hint (but not the complete proof/solution), on how to progress with this problem.
Intuitively, from the plot of $x \sin (1/x)$. If I fix an $\epsilon > 0$, and pick two points that are no more than a distance $\delta$ apart, we can always find points to their left, which require a smaller $\delta$-response. (Although, this could be wrong.)
I tried a couple of things. Let $\epsilon_0 > 0$. Let $(x_n)$ be the sequence defined as $x_n=\frac{1}{\pi/2+2n\pi}$ and $(y_n)$ be the sequence defined as $y_n = \frac{1}{3\pi/2+2n\pi}$. Now, $\lim |x_n - y_n|=0$. But, $|f(x_n)-f(y_n)|=\frac{4}{\pi}\left(\frac{4n+2}{(4n+1)(4n+3)}\right)$. But, this quantity can be made as small as we please.
So, I wonder, if I can prove that $x \sin (1/x)$ is perhaps, uniformly continuous. How do I simplify or prove the inequality $|x \sin (1/x) - y \sin (1/y)|<\epsilon$?
