Show that $x\sin(x)$ is not uniformly continuous on $\Bbb R$ For one of my homework problems, I'm supposed to show that $f(x)=x\sin(x)$ is not uniformly continuous on the real numbers, but this doesn't seem right to me. 
I know that if a function is continuous on a closed and bounded interval then it is uniformly continuous on that interval. I also know that f is continuous on $\Bbb R$. It seems like a simple induction would show that f is UC on $\Bbb R$, so what's going on? 
Please don't actually tell me how to solve the problem, but if you can tell me what I'm missing I would appreciate it. 

Any continuous function on a compact interval is uniformly continuous.
This does not extend to $\mathbb{R}$ by induction, as $\mathbb{R}$ is not compact.
If it is uniformly continuous then for all $\epsilon>0$ there is some $\delta >0$ such that $|f(x)-f(y)| < \epsilon$ whenever $|x-y| < \delta$.
So, you need to show that there is some $\epsilon >0$ such that for all $\delta>0$ there is some $x,y$ with $|x-y| < \delta$ such that $|f(x)-f(y)| \ge \epsilon$.
In your particular example, you can pick an arbitrary $\epsilon>0$.
Hint: Look at how quickly the function changes in the neighbourhood of a zero crossing.
