real analysis - Does $lim_{xrightarrow 0}frac{ln(x)}{cot(x)}$ exist or not?

I stumbled on the following limit in a calculus textbook today:

\begin{equation*}
\lim_{x\rightarrow 0}\frac{\ln(x)}{\cot(x)}
\end{equation*}

According to the book's solutions and Mathematica, this limit exists and is equal to 0. I can see why $0$ is obtained using l'Hôpital's rule twice:

\begin{equation*}
...=\lim_{x\rightarrow0}\frac{\left(\frac{1}{x}\right)}{-\csc^2(x)}=-\lim_{x\rightarrow0}\frac{\sin^2(x)}{x}=-\lim_{x\rightarrow0}\frac{2\sin(x)\cos(x)}{1}=2\sin(0)\cos(0)=0
\end{equation*}

If I recall correctly, l'Hôpital's rule is applicable when we have:
\begin{equation*}
\lim_{x\rightarrow a}\frac{f(x)}{g(x)}
\end{equation*}
even if $f$ and $g$ are not derivable at precisely $a$, so there should be no issue in using it on the above limit.

However, I can't reconcile the fact that $\ln(x)$ is defined over $]0,+\infty[$ (and usually, only $\lim_{x\rightarrow0^+}\ln(x)$ exists) with the fact that the above limit exists (both as $x\rightarrow0^+$ and as $x\rightarrow0^{-}$).

It seems to me that only

\begin{equation*}
\lim_{x\rightarrow 0^+}\frac{\ln(x)}{\cot(x)}
\end{equation*}

should exist and thus the "bilateral limit" (with $x\rightarrow 0$) does not exist since the limit with $x\rightarrow0^-$ doesn't.

Is there something I am missing?

Answer

Yes you are right, since $\ln x$ is defined for $x>0$ the limit for $x\to 0^-$ is meaningless and only

$$\begin{equation*}
\lim_{x\rightarrow 0^+}\frac{\ln x }{\cot x }
\end{equation*}$$

can be considered.

Note that this doesn't mean in general that the limit considered exists.

For example

\begin{equation*}
\lim_{x\rightarrow 0^+}\frac{\ln x }{\sin \frac1x}
\end{equation*}

can be considered but does not exist.