Past exam of the mathematics course of the University of Cambridge / 2007 / ib / Paper 3 / 3H / i / Solution

Take $x_n=2\pi n$ and $y_n=2\pi n+1/n$. Then $|y_n-x_n|=1/n\to 0$, whileThe sequential criterion for uniform continuity would force $f(y_n)-f(x_n)\to 0$. These pairs contradict it: for example, for all sufficiently large $n$ their output separation is greater than $\pi$, despite arbitrarily small input separation. Thus $x\sin x$ is not uniformly continuous on $\mathbb{R}$.