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.

  • $\begingroup$ Note that the $\lim_{x \to 0} \ln{x} = -\infty$ $\endgroup$
    – Joseph Eck
    May 1, 2018 at 23:24

2 Answers 2


Are you sure you have transcribed the problem correctly. The denominator looks suspicious.

Anyway you can factor the constant out to the front.

i.e. $\lim_\limits{x\to a} K f(x) = K\lim_\limits {x\to a} f(x)$

Composition of functions:

If $f(x)$ is continuous.

and $\lim_\limits{x\to a} g(x) = L$ then $\lim_\limits{x\to a} (f\circ g)(x) = f(L)$

$\lim_\limits{\delta\to 0} \frac {1}{\sqrt 3} \ln\left( \frac {\sin \frac {\pi}{6} - \sin (\frac {\pi}{6}-\delta)}{-(\cos \frac {\pi}{6} - \cos (\frac {\pi}{6}-\delta))}\right) = \frac {1}{\sqrt 3}\ln\left(\lim_\limits{\delta\to 0} \frac {\sin \frac {\pi}{6} - \sin (\frac {\pi}{6}-\delta)}{-(\cos \frac {\pi}{6} - \cos (\frac {\pi}{6}-\delta))}\right) = \frac {1}{\sqrt 3}\ln\left(\frac {\cos \frac {\pi}{6}}{\sin\frac {\pi}{6}}\right) = \frac {1}{\sqrt 3}\ln (\sqrt 3)$

Now I am not sure that is actually the limit you are trying to solve, but it should point you in the right direction.

  • $\begingroup$ It does help and I'm pretty sure I have it correct up until now, as in this task we got given $ F(x) = \frac{1}{\sqrt 3} \ln \biggl(\frac{sin\frac{\pi}{6}-(sin(\frac{\pi}{6}-x))}{1-cos(\frac{\pi}{6}-x)}\biggr) $ and all I did is split the integral from 0 to $\frac{\pi}{6}$ and $\frac{\pi}{6}$ to $ \frac{\pi}{2}$ and plug in values, to check if the limit exists (it shouldn't) $\endgroup$
    – Viceh
    May 1, 2018 at 23:46

You can rewrite it like this since the log is continuous in this interval until $\delta= 0$

$\displaystyle\lim_{\delta \to 0^-}\frac{1}{\sqrt{3}}\ln{\frac{\sin{\frac{\pi}{6}}-\sin{(\frac{\pi}{6}}-\delta)}{1-\cos{\delta}}} = \frac{1}{\sqrt{3}}\ln{\displaystyle\lim_{\delta \to 0^-}\frac{\sin{\frac{\pi}{6}}-\sin{(\frac{\pi}{6}}-\delta)}{1-\cos{\delta}}}$

$\displaystyle\lim_{\delta \to 0^-}\frac{\sin{\frac{\pi}{6}}-\sin{(\frac{\pi}{6}}-\delta)}{1-\cos{\delta}}=\lim_{\delta \to 0^-}\frac{\cos{(\frac{\pi}{6}}-\delta)}{\sin{\delta}}\to-\infty$


$Re(\ln{(-\infty)})=Re(\ln{((-1)\infty)}) = Re(\ln{(-1)})+\ln{\infty}$

$Re(\ln{(-1)}) = Re(\ln{1}+i(\pi+2n\pi))=0$

$\frac{\ln{(\infty)}}{\sqrt{3}} =\infty$


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