I have a task in which I need to show that some improper integral does not exist. This is one limit that has come up when trying to do so:
$$\lim_{\delta\to 0^{-}} \frac{1}{\sqrt 3}\cdot\ln \biggl(\frac{\sin\frac{\pi}{6}-(\sin(\frac{\pi}{6}-\delta))}{1-\cos(\frac{\pi}{6}-(\frac{\pi}{6}-\delta))}\biggr) $$
If I am not mistaken this evaluates to:
$$\lim_{\delta\to 0^{-}} \frac{1}{\sqrt 3} \ln \frac{0}{0} $$ which is an ideterminate form. I now am not sure, whether I am done, by saying that this limit does not exist(therefore the improper integral does not exist), or if I still need to apply L'Hopital's rule.
I dont know if that is even possible in this case with the $\frac{1}{\sqrt 3}\ln$ present.
So that is my question. Can I apply L'Hôpital's rule to functions of the kind $$ \ln\biggl(\frac{f(x)}{g(x)}\biggr)$$ ?
Thanks in advance to anyone taking the time to answer some novice's question.