# Show $f(x)=x\ln x$ is not uniformly continuous

Show $f:(0,\infty)\rightarrow \mathbb{R}, f(x)=x\ln x$ on $(0,\infty)$ is not uniformly continuous.

I think that the general way to prove that something is not continuous in a metric space is to let $\epsilon=...$ and show that $\forall\delta>0$, $d'(f(x)-f(y))>\epsilon$. I can't use the Mean Value Theorem because we haven't gone over it yet.

Here's an attempt:
Let $\epsilon=1$. Without loss of generality, let $x>y>0$. $|x\ln x-y\ln y|<|(x-y)(\ln x)|\leq|x-y||\ln x|$...$\delta=\frac{\epsilon}{\ln x}$ and therefore doesn't work for all $x$?

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What is the domain of the function? – azarel Feb 19 '12 at 18:38
$X=(0,\infty)$. – Emir Feb 19 '12 at 18:42

You are not proving that $f$ is not continuous, you are proving that $f$ (which is, in fact, continuous) is not uniformly continuous. So you want to show that for some $\epsilon$ there is no $\delta$ that works for all $x$. In fact you can take $\epsilon = 1$. Hint: if $y > x > 0$, $y \ln y - x \ln x > (y - x) \ln x$.
I doubt that there's a general way of doing it "by inspection". Basically you want to see if you can find points $x, y$ that are arbitrarily close ($d(x,y) < \delta$) but where $f(x)$ and $f(y)$ are not close ($d'(f(x),f(y)) \ge \epsilon$). Since a continuous function on a compact metric space is uniformly continuous, this often involves $x$ and $y$ going off "to infinity" in some sense. – Robert Israel Feb 19 '12 at 20:00