$f(x)=x\cdot \ln(x)$ uniformly continuous in $(0, \infty)$

Question

I have to decide if the function $f(x)=x\cdot \ln(x)$ in the interval $(0,+\infty)$ is uniformly cotinuous .

Answer 1

$$f(x)=x\cdot \ln(x)$$ is not uniformly continuous on $(0,\infty)$

Note that $$f'(x)=\ln(x)+1$$ grows without bound so you can find two points $x$ and $y$ as close as you wish, where $|f(x)-f(y)|>1$

Let $\epsilon>0$ be given. Choose an $x> e^{2/{\epsilon}}$ and let $y=x+\epsilon /2$

According to the Mean Value Theorem there exist a $c$ between $x$and $y$ such that $$|f(x)-f(y)|=|f'(c)||x-y|= |1+\ln c|(\epsilon /2) =(1+2/{\epsilon})(\epsilon /2)>1$$

That contradicts the uniform continuity.

The trouble is infinity not zero.