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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?

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    $\begingroup$ Nothing. You're right: the limit must be one sided, otherwise it isn't even well defined. $\endgroup$ – DonAntonio Feb 10 '18 at 23:05
  • $\begingroup$ It is preferable do not use the term "exist" or "do not exist" to express the fact that the limit is or not meaningless. It is more correct to say that the bilateral limit or left limit are meaningless and only the right limit can be considered. $\endgroup$ – gimusi Feb 10 '18 at 23:18
  • $\begingroup$ @gimusi I see what you mean but could you be more specific as to when we should use the vocabulary of "meaningless" or "not well-defined" versus "does not exist". I would expect a calculus student for exemple to simply write down $/\exists$. Should we agree that all "meaningless" limits do not exist but not necessarily the converse? (Hope that's clear...) $\endgroup$ – orion2112 Feb 10 '18 at 23:24
  • $\begingroup$ Also, (not relevant to the question itself but still curious), can anyone explain why Mathematica doesn't see this and why it sees "meaning" in a limit that is now established as meaningless? $\endgroup$ – orion2112 Feb 10 '18 at 23:27
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    $\begingroup$ @orion2112 We say that a "limit does not exist" when the limit expression is well defined but it is not $L\in \mathbb{R}$ nor $+\infty$ nor $-\infty$. For example $lim_{x\to +\infty} \sin x$ is a well defined expression but the limit does not exist and $lim_{x\to +\infty} \arcsin x$ in not well defined. $\endgroup$ – gimusi Feb 10 '18 at 23:37
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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.

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

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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 $$\lim_{x\to 0^+}x\ln x=0.$$

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    $\begingroup$ Neat, but I found the limit in an exercise set made for practicing l'Hôpital's rule. The question was also more about the existence of the limit or not (with $x\rightarrow 0$) rather than how to solve it. $\endgroup$ – orion2112 Feb 10 '18 at 23:30
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    $\begingroup$ “Is a result from high school” is a dangerous phrase! $\endgroup$ – JavaMan Feb 10 '18 at 23:52
  • $\begingroup$ Why, dangerous? $\endgroup$ – Bernard Feb 11 '18 at 0:47
  • $\begingroup$ @Bernard Not every country has the same school system. The result you are using was probably in the same exercise set as the limit I was talking about, so it's not as though it was an obvious, intuitive result everyone has in mind when learning to do limits. $\endgroup$ – orion2112 Feb 11 '18 at 1:05

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