TITLE: How to prove $\sin(1/x)$ is not uniformly continuous
QUESTION [17 upvotes]: How do I go about proving $f(x)=\sin(1/x)$ is not uniformly continuous?
(Or: different question, but same intention* how do I prove that $x\sin(x)$ is not uniformly continuous)
*I'm trying to grasp how one would prove $f$ is not uniformly continuous for functions other than the simple $x^n$. I have seen one technique being to set an $\epsilon$ and set $x, y$ in the form of $\delta$ (e.g. $\delta/2$, etc.) then subsequently proving that $f(x)-f(y)\ge\epsilon$

REPLY [13 votes]: Choose two sequences $T_n = \frac{1}{n}$ and $S_n = \frac{1}{n+\pi}$. Their difference goes to zero as $n$ goes to infinity. But $|f(S_n)-f(T_n)|=\left|2 \sin\left(\frac{n+\pi-n}{2}\right)\cos\left(\frac{n+\pi+n}{2}\right)\right|=\left|2 \cos\left(n+\frac{\pi}{2}\right)\right|=\left|2\sin(n)\right|>0$ for $n \in \mathbb{N}$, so the function is not uniformly continuous.