Does $\lim_{x\rightarrow 0}\frac{\ln(x)}{\cot(x)}$ exist or not?

Question

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 1

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.

Answer 2

Needless to call L'Hospital for that: the answer is obvious with equivalents: near $0$, $\sin x \sim x$, $\cos x\sim 1$, so $$\frac{\ln x}{\cot x}=\frac{\ln x\sin x}{\cos x}\sim_{0}\frac{x\ln x}1=x\ln x, $$ and it's a result from high school that $$\lim_{x\to 0^+}x\ln x=0.$$

Answer 3

Sometimes some tacit assumptions are omitted by authors, and sometimes there are oversights and typos. It should have said $\lim_{x\to 0^+}\;.$

We have $\lim_{x\to 0^+}(x\ln x)=0$ and $\lim_{x\to 0^+}\frac {\tan x}{x}=1.$

For $\pi /2>x>0$ we have $\cot x \ne 0$ and $$\frac {\ln x}{\cot x}=(\ln x)(\tan x)=(x\ln x)\cdot\frac {\tan x}{x}.$$

So $\lim_{x\to 0^+} \frac {\ln x}{\cot x}=0\cdot 1=0.$