Uniform Continuity of $x \sin x$ 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 !
 A: 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.
A: Since $x\sin(x)$ is continuous, we won't be able to show discontinuity. It is the uniformity of the continuity that we have to consider. $f$ is uniform continuous if and only if

$$\forall\epsilon\gt0,\exists\,\delta\gt0:\forall x,y\in\mathbb{R},|x-y|\le\delta\implies|f(x)-f(y)|\le\epsilon\tag{1}$$

The inverse of $(1)$ is

$$\exists\,\epsilon\gt0:\forall\,\delta\gt0,\exists\,x,y\in\mathbb{R}:|x-y|\le\delta\wedge\,|f(x)-f(y)|\gt\epsilon\tag{2}$$

We can take $\epsilon=1$. Given $\delta>0$, let $\eta=\min(\delta,\pi/2)$ and $x=2\pi\left\lceil\dfrac1{2\pi\sin(\eta)}\right\rceil$ and $y=x+\eta$. Then $f(x)=0$ and
$$
\begin{align}
f(y)
&=(x+\eta)\sin(x+\eta)\\
&=(x+\eta)\sin(\eta)\\
&\gt x\sin(\eta)\\
&\ge1
\end{align}
$$
A: 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 $n$. 
A: 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 
$f(y_n)-f(x_n)=(\pi n+\frac{1}{n})\sin(\pi n+\frac{1}{n})-\pi n\sin(n\pi)=_{n\longrightarrow\infty} \pi(-1)^n+o(1)\not \to 0$
