Given the following function: $x\sin\bigg(\frac{1}{x}\bigg)$
How do I show that this is uniformly continuous on $(0, \infty)$?
The reason why I am having trouble is that I am use to finding pathologies in functions which causes them to not have a certain property.
Furthermore, is there a more "analysis" like way of doing this.
(1) $x\sin(\frac{1}{x})=\frac{\sin(\frac{1}{x})}{\frac{1}{x}}$
(2) L'Hospital's Rule: $\sin(\frac{1}{x})-\frac{\cos(\frac{1}{x})}{x}$. Taking all appropriate limits I get $1$.
This is about where I'm stuck.