How do i prove that x*sin(x) is not uniformly continuous?
$\displaystyle x \sin(x)-(x+\delta) \sin(x+\delta)=x \delta \cos(x) + xO(\delta^2)$
and so for $\displaystyle \delta$ small enough and $\displaystyle x \ne n \pi/2, \ n\in \mathbb{Z}$ there is a $\displaystyle k>0$ independent of $\displaystyle x$ such that:
$\displaystyle x \sin(x)-(x+\delta) \sin(x+\delta)<kx \delta \cos(x)$
So however small we make $\displaystyle |\delta|>0$ there is an $\displaystyle x$ which makes $\displaystyle |x \sin(x)-(x+\delta) \sin(x+\delta)|$ as large as we like.
Note: The $\displaystyle O(\delta^2)$ term can be made independent of $\displaystyle x$ as it stands in place of a remainder term of a Taylor series which is the product of a constant $\displaystyle \kappa$, $\displaystyle \delta^2$ and a circular trig function at a point in $\displaystyle [x,x+\delta]$ all of which which is bounded by $\displaystyle |\kappa \delta^2|$. If the reader needs the detail they can develop it them selves.
CB