TITLE: Uniform Continuity of $x \sin x$
QUESTION [12 upvotes]: How does one go about proving that the function $x \sin x$ is not uniformly continuous on the set of real numbers ?
Any method that uses the sequential criterion for discontinuity would be preferred.
Thanks in advance !

REPLY [2 votes]: Let $(x_n)_{n\ge1}$ and $(y_n)_{n\ge1}$ be any two sequences from the domain such that $x_n=2n^2\pi+\dfrac{1}{n}$ and $y_n=2n^2\pi$.
Now, 
\begin{align}\left\lvert x_n\sin x_n-y_n\sin y_n\right\rvert&=\left\lvert \left(2n^2\pi+\dfrac{1}{n}\right)\sin \left(\dfrac{1}{n}\right)\right\rvert\\&=\left(2n^2\pi+\dfrac{1}{n}\right)\sin \left(\dfrac{1}{n}\right)\\&>2n^2\pi\sin\left(\dfrac{1}{n}\right)\\&>1\end{align} for all sufficiently large $n$.