Uniform Continuity of $x \sin x$

Let $f(x)=x\sin x$ and $x_n=\pi n$ and $y_n=\pi n+\frac{1}{n}$ then $\displaystyle\lim_{n\to\infty}( y_n-x_n)=0$ and

$f(y_n)-f(x_n)=(\pi n+\frac{1}{n})\sin(\pi n+\frac{1}{n})-\pi n\sin(n\pi)=_{n\longrightarrow\infty} \pi(-1)^n+o(1)\not \to 0$

Elaborating on David Mitra's Comment:

From continuity of $x\sin(x)$, you have that $$\forall \epsilon >0, \forall y, \exists \delta: |x-y|<\delta\implies |x\sin(x)-y\sin(y)|<\epsilon$$

But define $$p=x+2n\pi;\;\;\;q=p=y+2n\pi$$

Then you have $|p-q|<\delta$ but $$|(x+2n\pi)\sin(x+2n\pi)-(y+2n\pi)\sin(y+2n\pi)|$$ $$=|(x\sin(x)-y\sin(y))+2n\pi(\sin(x)-\sin(y))|>\epsilon$$ for sufficiently large $n$. An undesirable result for uniform continuity.