Show that $f(x)=\ln(x)$ is not uniformly continuous on $(0,\infty)$ I'm trying to show that $f(x)=\ln(x)$ is not uniformly continuous on the interval $(0,\infty).$
This is what I have so far:
Let $\epsilon=1.$
Choose $\delta=$
if $x,y\in(0,\infty)$ with $|y-x|<\delta$ then $|f(y)-f(x)|=|\ln\left(\frac{y}{x}\right)|$
I'm stuck at this point though, are there any well known inequalities I can use here?

Let $\epsilon=1$ be like you've chosen. Then for all $\delta>0$, just pick $y=\frac{3}{4}\delta>0$ and $x=\frac{1}{4}\delta>0$. This works because
$$
|y-x|=\frac{1}{2}\delta<\delta\quad\text{but}\quad|\log(y)-\log(x)|=\log(3)>1=\epsilon.
$$
