= Solution
Take $x_n=2\pi n$ and $y_n=2\pi n+1/n$. Then $|y_n-x_n|=1/n\to0$, while
$$
f(x_n)=0,\qquad f(y_n)=\left(2\pi n+\frac1n\right)\sin\frac1n\longrightarrow2\pi.
$$
The <sequential criterion for uniform continuity> would force $f(y_n)-f(x_n)\to0$. These pairs contradict it: for example, for all sufficiently large $n$ their output separation is greater than $\pi$, despite arbitrarily small input separation. Thus \b[$x\sin x$ is not uniformly continuous on $\mathbb R$].
Back to article page
