TITLE: Uniform continuity of $\ln(x)$
QUESTION [9 upvotes]: Is $f(x)=\ln(x)$ uniformly continuous on $(1,+\infty)$? If so, how to show it?
I know how to show that it is not uniformly continuous on $(0,1)$, by taking $x=\frac{1}{\exp(n)}$ and $y = \frac{1}{\exp(n+1)}$.
Also, on which interval does $\ln(x)$ satisfy the Lipschitz condition?

REPLY [5 votes]: Let $x,y\in(1,+\infty)$  then by mean value theorem there's $z$ such that
$$\log(x)-\log(y)=\frac{1}{z}(x-y)$$
hence we have
\begin{array}\\
|\log(x)-\log(y) |&\leq\sup_{z>1}\frac{1}{z}|x-y|\\&\leq|x-y|
\end{array}
so  the function $\log$ is uniformly continuous on the interval $(1,+\infty)$ since it satisfies the lipschitz condition.