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?

Working with $\epsilon$ and $\delta$ quickly becomes tedious and annoying, it is thus better to learn more convenient and powerful techniques. Remember that $f$ is uniformly continuous on $S$ if and only if for every sequences $(x_n), (y_n) \subseteq S$ with $d(x_n, y_n) \to 0$ we have that $d(f(x_n), f(y_n)) \to 0$ (with $d$ denoting the distances in the domain the definition and the range of $f$).
In our case, we suspect that $\ln$ fails to be uniformly continuous towards of $0$. Choose, therefore, $x_n = \textrm e^{-n}$ and $y_n = \textrm e^{-n + 1}$. Notice that $|x_n - y_n| \to |0 - 0| = 0$, but $$| \ln x_n - \ln y_n | = | \ln \textrm e^{-n} - \ln \textrm e^{-n + 1} | = | -n - (-n + 1)| = 1 \not\to 0 ,$$
which shows that $\ln$ is not uniformly continuous on $(0,\infty)$. (It is, though, on every interval of the form $[a,b)$ with $a > 0$ and $b$ possibly infinite.)
