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This limit is proposed to be solved without using the L'Hopital's rule or Taylor series: $$ \lim_{x\to0}\frac{\ln\sin^2(ax)}{\ln\sin^2(bx)}, $$ where $a,b=const$. I know how to calculate this limit using the L'Hopital's rule: $$ \lim_{x\to0}\frac{\ln\sin^2(ax)}{\ln\sin^2(bx)}= \lim_{x\to0}\frac{\frac{2a\sin (ax)\cos (ax)}{\sin^2(ax)}}{\frac{2b\sin (bx)\cos (bx)}{\sin^2(bx)}}= \lim_{x\to0}\frac{a}{b}\cdot\frac{\sin (ax)\cos (ax)}{\sin (bx)\cos (bx)}\cdot\frac{\sin^2(bx)}{\sin^2(ax)} $$ (using the asymptotic equivalence $\sin x\sim x$) $$ =\lim_{x\to0}\frac{a}{b}\cdot\frac{ax}{bx}\cdot\frac{(bx)^2}{(ax)^2}=1, $$ but I don't know to calculate this limit without derivatives.

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  • $\begingroup$ Off-topic: when $x\to 0$ then $\cos(ax)=\cos(bx)=1$ and not $ax/bx$. $\endgroup$
    – Sebastiano
    Oct 23, 2021 at 23:11
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    $\begingroup$ @Sebastiano Yes, I know. I mean $\sin (ax)\sim ax$, $\sin(bx)\sim bx$. $\endgroup$
    – mathstudent1
    Oct 23, 2021 at 23:16
  • $\begingroup$ What is the source of this problem? $\endgroup$
    – user2661923
    Oct 23, 2021 at 23:18
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    $\begingroup$ The reason that I asked is that normally, for a problem with the constraints that you have given (i.e. no L'Hopital's rule or Taylor series), the book will first prepare the student before presenting the problem. Typically, this preparation involves theorems, worked examples or previously solved problems. The idea is that then, the posed problem represents an application of the already presented theory. However, it is unclear what the phrase "problem book" signifies. If the book is only a presentation of problems, with no build-up, then the situation clearly becomes problematic. $\endgroup$
    – user2661923
    Oct 23, 2021 at 23:27
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    $\begingroup$ @Sebastiano I am not opposed to downvotes in general, but I think in that case I would be nice to explain the downvote (actually I did not downvote your answer) $\endgroup$
    – Severin Schraven
    Oct 23, 2021 at 23:53

2 Answers 2

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Since, $\sin^2(x)=\sin^2(-x)$, we can accept $a,b\in \mathbb R^{+}$. Thus we have,

$$\begin{align}\lim_{x\to0}\frac{\ln\sin^2(ax)}{\ln\sin^2(bx)}&=\lim_{x\to0}\frac{2\ln|\sin(ax)|}{2\ln|\sin(bx)|}\\ &=\lim_{x\to0}\frac{\ln|\sin(ax)|}{\ln|\sin(bx)|}\\ &=\lim_{x\to0^{+}}\frac{\ln\sin(ax)}{\ln\sin(bx)}\\ &=\lim_{x\to0^{+}}\left(\frac{\frac{\ln\sin(ax)}{\ln ax}}{\frac{\ln\sin(bx)}{\ln bx}}\times \frac{\ln ax}{\ln bx}\right)\\ &=1.\end{align}$$

Explanation:

Let $\alpha,\thinspace x\in\mathbb R^{+}$, then we can write the following limits:

  1. $\lim_{x\to0^{+}}\frac{\ln\sin(\alpha x)}{\ln\alpha x}=1$.

Use:

$\ln(\sin \alpha x) = \ln \alpha x + \ln\left(\frac {\sin \alpha x} {\alpha x}\right)$

  1. $\lim_{x\to 0^{+}}\frac{\ln ax}{\ln bx}=1$

Use:

$\ln (\alpha x)=\ln \alpha+\ln x.$

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  • $\begingroup$ Interesting approach, in that the result that $\displaystyle \lim_{x \to 0^+} \frac{\sin(x)}{x} = 1$ can be interpreted as a consequence of applying the squeeze theorem against the axioms used to define the sine and cosine functions (re Tom Apostol's sine/cosine axioms). This indicates that you are not (inadvertently) applying L'Hopital's rule. Just out of curiosity, re the comments that I left following the question, is it your opinion that your approach may have been intended by the problem composer? $\endgroup$
    – user2661923
    Oct 24, 2021 at 0:02
  • $\begingroup$ @user2661923 I fixed some of my mistakes. $\endgroup$
    – lone student
    Oct 24, 2021 at 0:30
  • $\begingroup$ @user2661923 I'm not sure about the answer to your question and I have not a good calculus experience. Please, feel free to show my mistakes. $\endgroup$
    – lone student
    Oct 24, 2021 at 0:36
  • $\begingroup$ You misread my comment. I didn't scrutinize your answer that closely, because I have an aversion to this type of problem. That is, it is problems like this that L'Hopital's rule was designed for. Anyway, I am unaware of any mistakes in your answer. $\endgroup$
    – user2661923
    Oct 24, 2021 at 0:49
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    $\begingroup$ @Sebastiano Thank you for reviewing my answer ( ◜‿◝ ) $\endgroup$
    – lone student
    Oct 25, 2021 at 4:39
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Can someone tell me what is wrong in what I'm doing? Assuming $ \lim_{x\to0}\frac{\sin\left(\alpha x\right)}{\alpha x}=1 $ $$ \require{cancel} \lim_{x\to0}\frac{\ln\left(\sin^{2}\left(ax\right)\right)}{\ln\left(\sin^{2}\left(bx\right)\right)}=\frac{\ln\left(\frac{\sin^{2}\left(ax\right)}{\left(ax\right)^{2}}\cdot\left(ax^{2}\right)\right)}{\ln\left(\frac{\sin^{2}\left(bx\right)}{\left(bx\right)^{2}}\cdot\left(bx^{2}\right)\right)}=\frac{\ln\left(\cancelto{1}{\frac{\sin^{2}\left(ax\right)}{\left(ax\right)^{2}}}\right)+\ln\left(\left(ax\right)^{2}\right)}{\ln\left(\cancelto{1}{\frac{\sin^{2}\left(bx\right)}{\left(bx\right)^{2}}}\right)+\ln\left(\left(bx\right)^{2}\right)}=\frac{\ln\left(1\right)+\ln\left(\left(ax\right)^{2}\right)}{\ln\left(1\right)+\ln\left(\left(bx\right)^{2}\right)}=\frac{\ln\left(\left(ax\right)^{2}\right)}{\ln\left(\left(bx\right)^{2}\right)}\Rightarrow \\ \frac{2\ln\left(ax\right)}{2\ln\left(bx\right)}=\frac{\ln\left(ax\right)}{\ln\left(bx\right)}=\frac{\ln\left(a\right)+\ln\left(x\right)}{\ln\left(b\right)+\ln\left(x\right)} $$ Dividing both sides by $\ln(x) $ will bring us to $ \lim_{x\to0}\frac{\ln\left(a\right)+\ln\left(x\right)}{\ln\left(b\right)+\ln\left(x\right)}=\lim_{x\to0}\frac{\frac{\ln\left(a\right)}{\ln\left(x\right)}+\frac{\ln\left(x\right)}{\ln\left(x\right)}}{\frac{\ln\left(b\right)}{\ln\left(x\right)}+\frac{\ln\left(x\right)}{\ln\left(x\right)}}=\frac{0+1}{0+1}=1 $

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  • $\begingroup$ You need to keep your limits. What is the computation in the very end? $\endgroup$
    – Severin Schraven
    Oct 23, 2021 at 23:50
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    $\begingroup$ Now that you figured it out, good work. Thanks for the edit! $\endgroup$
    – amWhy
    Oct 23, 2021 at 23:51
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    $\begingroup$ @Danny Blozrov This is undeniable. The question is different: why it is possible to take a part of a function and replace it with the limit of this part. I was taught that, in general, this is wrong. $\endgroup$
    – mathstudent1
    Oct 24, 2021 at 0:06
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    $\begingroup$ @Danny Blozrov I think you are wrong. We can't in the limit $\lim_{x\to 0}\frac{\frac{\tan x}x-\frac{\sin x}x}{x^2}$ replace $\frac{\tan x}x$ or $\frac{\sin x}x$ with 1. This gives us some wrong result, although the identical function is continuous $\endgroup$
    – mathstudent1
    Oct 24, 2021 at 0:48
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    $\begingroup$ @Danny Blozrov The theorem from the video applies to the function under the limit sign as a whole, and not to a separate part of it. I'm sorry, but I downvoted your answer because I am convinced that it is wrong $\endgroup$
    – mathstudent1
    Oct 24, 2021 at 0:54

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