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Consider the limit:

$$\lim_{x\to0^+}x\ln(\sin x)$$

I wish to evaluate this limit without L'Hôpital's Rule and without using the result $\lim_{x\to0^+}x\ln x=0$(say (*)) since I can evaluate this limit using this result(see below):

$$\lim_{x\to0^+}x\ln(\sin x)$$ $$=\lim_{x\to0^+}x\ln\left(\frac{\sin x}{x}\right)+x\ln x$$ The first term is of the form $0×0$, hence its limit is $0$ and by the result (*), the second term is also $0$. Hence, the given limit is $0$.

I tried many times but am not able to get a different idea. How do I approach this problem? Any help would be highly appreciated.

Integreek
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  • Are you approaching this limit from the right? Otherwise, if $x$ is just a bit negative then you're taking the natural log of a negative number, which is not defined (for the purpose of an intro to Calc course, at least!). – Benjamin Dickman Sep 15 '24 at 15:35
  • Yes, I meant from the right! – Integreek Sep 15 '24 at 15:36
  • @BenjaminDickman , But, if we were to approach from the left, how would we evaluate the limit? – Integreek Sep 15 '24 at 15:37
  • The limit doesn't exist when you approach from the left; you can see this by just inspecting a graph e.g. in Desmos. – Benjamin Dickman Sep 15 '24 at 15:38
  • Yes, it doesn't exist in the reals, but what if we consider complex numbers? – Integreek Sep 15 '24 at 15:39
  • @BenjaminDickman I suppose it would involve complex analysis, but I am not very familiar with it, all I know is about the basic definition of the complex logarithm. Can you please explain how to evaluate the LHL considering complex numbers? – Integreek Sep 15 '24 at 17:01
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    @BenjaminDickman According to the more general definition of limit, in real analysis, when we indicate $\lim_{x\to 0}\log x$ or similar it is implicit the assumption that $x\to 0^+$. We can indicate it or not, this is not so important. Of course we can't indicate or assume that $x\to 0^-$. – user Sep 16 '24 at 14:06
  • What is your motivation for forbidding the (very natural) use of $\lim_{x\to0^+}x\ln x=0$? – Anne Bauval Sep 25 '24 at 14:37
  • @AnneBauval since there is already a post on this site that uses that approach. – Integreek Sep 25 '24 at 15:21
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    This does not answer my question. Since that approach is natural, the 3 answers below are but variants of it. They don't avoid to (reprove and) use in disguise your forbidden result. – Anne Bauval Sep 25 '24 at 15:32
  • @AnneBauval yes, I agree that in a way they can be used to prove the limit I don’t want to use explicitly. My aim was that I wanted some more strategies to evaluate this limit like using inequalities(I am not an expert in this, other than simple ones like $0<\sin^2 x<1$) rather than just simply making use of $\lim_{x\to0^+}x\ln x=0$. – Integreek Sep 25 '24 at 15:52
  • This is because before asking this question I only knew how to prove $\lim_{x\to0^+}x\ln x=0$ and $\lim_{x\to0^+}x\ln \sin x=0$ by L-Hopital’s rule which I did not find very rigorous. – Integreek Sep 25 '24 at 15:58
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    I personally dislike the systematic use of L'Hôpital's rule, which seems to be a habit on this site (maybe a habit in US or more widely english-speaking countries), but it is rigorous. – Anne Bauval Sep 25 '24 at 15:59
  • @AnneBauval Thanks for correcting me, but I feel like using a shortcut or missing out on some beautiful insights whenever I use L-Hôpital’s rule in limits. So, whenever I do, I try to look for an alternative method. Am I thinking correct or is only using L- Hôpital’s rule fine? – Integreek Sep 25 '24 at 16:09
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    @MathGuy The problem with L'Hôpital's rule is that it gives a "black box" solution without giving any useful insight on what is really going on in the limit. Algebraic manipulation, Squeeze theorem, Standard limits and Taylor's series (when strictly needed) are preferable. In some cases L'Hôpital's rule is the best option, as for example for limits involving integrals. – user Sep 25 '24 at 16:22
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    This question is essentially (or should be) a duplicate of $\lim_{x\to0^{+}} x \ln x$ without l'Hopital's rule, which is more reasonable and easier to answer honestly. – Anne Bauval Sep 25 '24 at 18:19
  • @AnneBauval It's not a duplicate since I don't wish to use the result $\lim_{x\to0^+}x\ln x=0$ in a ready-made way as some of the comments in the answers have already mentioned. – Integreek Sep 26 '24 at 04:03
  • I know, but the only reason you gave for this "wish" is that you don't know proofs of the fact $\lim_{x\to0^+}x\ln x=0 $ without L'Hopital. So, better remedy this directly than ask for a contrived avoidance of an explicit use of this fact, while reproving it in disguise, like in all the answers below. – Anne Bauval Sep 26 '24 at 04:56
  • @AnneBauval Ok, I get your point. But what about the answers here using inequalities that are not there in the original question? – Integreek Sep 26 '24 at 05:40
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    I am not sure to understand this question. I commented each of these answers, in the same spirit as my last comment here. The inequalities they used are but those commonly used for proofs without l'hôpital of $\lim_{x\to0^+}x\ln x=0$ – Anne Bauval Sep 26 '24 at 05:43

3 Answers3

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Recall the inequality

$$1-\frac{1}{t}\leq\ln t\leq t-1$$

for all $t>0$. From the first inequality we also get that

$$\frac{1}{2}\ln t=\ln\sqrt t\geq1-\frac{1}{\sqrt{t}}$$

for all $t>0$. Consequently we have the inequality

$$2-\frac{2}{\sqrt{t}}\leq\ln t\leq t-1.$$

In particular, taking $t=\sin x$ and multiplying by $x>0$ we have that

$$2x-\frac{2x}{\sqrt{\sin x}}\leq x\ln(\sin x)\leq x(\sin x-1).$$

The limit of the right-hand side expression is clearly zero as $x\downarrow0$, and for the left-hand side expression we have that

$$2x-\frac{2x}{\sqrt{\sin x}}=2x-2\sqrt{x}\underbrace{\sqrt{\frac{x}{\sin x}}}_{\to1}\to0$$

as $x\downarrow 0$. It follows by the squeeze theorem that the limit is zero.

Lorago
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  • (+1) Wow, that's a nice answer! Thank you for your help! I am able to do it now. – Integreek Sep 15 '24 at 16:02
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    The right handside is useless since $\forall x\in(0,\pi/2)\quad x\ln(\sin x)\le0$. – Anne Bauval Sep 25 '24 at 15:04
  • @AnneBauval good observation, the important part of the proof is more so the other inequality – Lorago Sep 25 '24 at 15:10
  • And that important part, $t\ln t\ge2(t-\sqrt t)$, is essentially a (re)proof of the (accordingly weirdly) forbidden result $\lim_{t\to0^+}t\ln t=0$. Replacing $t$ by $\sin x$ before or after the use of the squeeze theorem doesn't make a big difference, so that imo, none of the answers here really adresses the question. It might be interesting but I doubt it is possible. – Anne Bauval Sep 25 '24 at 15:28
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We have that as $y\to \infty$

$$0\le \log y \le \sqrt y \implies 0\le \frac{\log y}y \le \frac1{\sqrt y} $$

then by $\sin x=\frac1y \to 0^+$

$$0\le -\sin x \log (\sin x) \le \sqrt {\sin x}$$

therefore

$$0\le -x\log(\sin x)=-\frac x{\sin x}\sin x \log (\sin x)\le \frac x{\sin x}\sqrt {\sin x} \to 1 \cdot 0=0$$

then by squeeze theorem the given limit is equal to zero.

user
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  • (+1) This inequality is even more intuitive! – Integreek Sep 16 '24 at 15:10
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    @RakshithPL Thanks, of course the first inequality implies also the limit you don't want use esplicitely. – user Sep 16 '24 at 15:52
  • (-1) This answer is essentially tacitely reproving the $\lim_{y\to0^+}y\ln y=0$, and using this (accordingly weirdly) forbidden result (plus $\sin x\sim_{(x\to0)}x$). – Anne Bauval Sep 25 '24 at 15:02
  • @AnneBauval Yes I understand you observation and indeed I've commented here above that this result also implies the limit but for the proof here I'm using the latter inequality by squeeze theorem without referring at that standard limit. – user Sep 25 '24 at 15:05
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    @AnneBauval Note that the question is requiring "without using the result" not without using any result which implies that limit. I understand that the difference is subtle but there is a difference. – user Sep 25 '24 at 15:07
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You already know (by squeezing) that

$$\lim_{x\to0^+}x\ln(\sin x)=\lim_{x\to0^+}x\ln(x).$$

Now by a change of variable $x=e^{-t}$,

$$\lim_{x\to0^+}x\ln(x)=-\lim_{t\to\infty}e^{-t}t.$$

And with $a_t=te^{-t}$,

$$\frac{a_{t+1}}{a_t}=\frac{(t+1)e^{-t-1}}{te^{-t}}=\left(1+\frac1t\right)e^{-1}<\frac2e$$ for $t>1$ and the infinite product of such terms tends to zero.

  • $(+1)$ I like these interesting alternative approaches to evaluate $\lim_{x\to0}x\ln x$ and $\lim_{x\to \infty}xe^{-x}$! – Integreek Sep 25 '24 at 09:26
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    @MathGuy: this allow you to settle the case of $x^a b^x$ for any $a,b$. –  Sep 25 '24 at 10:09
  • (-1) Since you are (reproving and) using $\lim_{x\to0^+}x\ln x=0$, this answer does not adress the (accordingly weird) question. – Anne Bauval Sep 25 '24 at 14:39
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    @AnneBauval: this is not how I understand it. I am not using the result ready-made, I am proving it. –  Sep 25 '24 at 15:13