A student showed me the following exercise:
Is $h(x) = x \sin\left(\frac{1}{x}\right)$ uniformly continuous on $(0,1)$?
Admittedly, it has been a while since I have looked at problems involving uniform continuity and so I am not sure how to go about this problem.
My guess is that $h(x)$ is uniformly continuous on this interval but I am not sure how to go about picking an appropriate $\delta > 0$ such that $|x-c|< \delta$ implies $|h(x) – h(c)| < \epsilon$ for $x,c \in (0,1)$.
The student and I looked at $|x \sin(x) – c \sin(c)| \leq |x \sin(x)| + |c \sin(c)| \leq |x| + |c|$ although this feel like the wrong path to go down.
I appreciate any help! Thanks in advance!
Best Answer
$h(x)=x\sin(\frac{1}{x})$ can be extended to a continuous function on $[0,1]$ by setting $h(0)=0$. As $[0,1]$ is compact it follows that $h(x)$ is uniformly continuous on $[0,1]$, and therefore is uniformly continuous on $(0,1)$.