Originally Posted by

**sunmalus** Hello,

I want to show that

$\displaystyle f(x)=x \cdot \sin( \frac{1}{x})$

in the interval (0, infinity)

is uniformly continuous using the following definition:

$\displaystyle \text{Given }f \colon I \subset \mathbf{R} \longrightarrow \mathbf{R}. f \\

\text{ is uniformly continuous on I if }\forall \varepsilon >0, \exists \delta >0 \\

\text{ such that }\forall x,y \in I, | x - y| < \delta, |f(x)-f(y)|< \varepsilon$.

My problem is that I want to find something like:

$\displaystyle |f(x)-f(y)|\leq \underbrace{|x-y|}_{ \delta }\dots \leq \varepsilon$

but I've been thinking for a while and I cant find anything.