Answer
diverges toward $-\infty$
Work Step by Step
In the proof of Th.5.1. we noted that
as $t\rightarrow 0^{+}$ (t approaches 0 from the right)
then ln(t)$\rightarrow-\infty$ (diverges toward $-\infty$).
(see fig.5.3)
Here, if we set t=x-3, and
when $x\rightarrow 3^{+},$then $t\rightarrow 0^{+}$, so we can write
$\displaystyle \lim_{x\rightarrow 3^{+}}\ln(x-3)=\lim_{t\rightarrow 0^{+}}\ln t=-\infty$
