We all knew, with $$\lim_{x\to 0}\frac{\sin x  x}{x(1\cos x)}$$ we can use L'Hôpital's rule or Taylor series to eliminate undefined form. But without all tools, by only using high school knowledge, how can we evaluate this limit? It seems difficult to transform numerator, any idea? Thank you!

2What have you tried? – ARNIE BEBITADRIS Oct 22 '21 at 01:45

I tried to transform the numerator, but it seems impossible, so i changed $$1\cos x$$ to $$2(\sin\frac{x}{2})^2$$ but cant find some common factors. May be i am misdirecting.. – MathChill Oct 22 '21 at 01:57

3Surprised that anyone would downvote such a question. In my opinion, this is actually a very deep question. First of all, the question only has meaning if the domain of the trig functions are real numbers, rather than angles. So, Analytical Geometry definitions must be excluded, and *some* definition of the sine and cosine functions must be given. Further, per the OP's question, these trig functions must not be defined in terms of a Taylor series, and no Taylor series can be involved in any way. ...see next comment – user2661923 Oct 22 '21 at 02:11

2The difficulty here is that *normally* in a *problemsolving* oriented high school Calculus class, you aren't given a formal definition of the sine and cosine functions, but are instead simply told that the domain of these functions is (dimensionless) arc lengths of a unit circle, rather than angles. One approach would be to somehow derive $\frac{d}{dx}\sin(x) = \cos(x)$ and $\frac{d}{dx}\cos(x) = \sin(x)$. Then, you would have to *reinvent the wheel*, walking in the path of the proof of L'Hopital's rule, and applying this *reinvention* to the specific problem. – user2661923 Oct 22 '21 at 02:16

Thank you for your idea, maybe it is the only way so far. It is a very hard problem to high school, i keep digging in to find some clues from another sources... – MathChill Oct 22 '21 at 02:24

1If you can afford to buy it, one approach is that described in Volume 1 of "Calculus : 2nd Edition" [Tom Apostol, 1966, 2 volume work]. Whatever *prooforiented* Calculus (AKA Real Analysis) text that you consult, you will need a formal definition of the sine and cosine functions. – user2661923 Oct 22 '21 at 02:28

There are plenty of similar questions already, where you can find the answer: https://math.stackexchange.com/questions/157903/evaluationoflimlimitsxrightarrow0fractanxxx3, https://math.stackexchange.com/questions/217081/determinelimxto0fracxsinxx3frac16withoutlhospita, https://math.stackexchange.com/questions/387333/arealllimitssolvablewithoutlh%c3%b4pitalruleorseriesexpansion, etc. – Hans Lundmark Oct 22 '21 at 09:26

Thank you, it is my fault when dont make a search before posting... – MathChill Oct 22 '21 at 09:40
1 Answers
I want to thank everyone for your help, after some hard work, i found the answer, i post it here for all of you (in case you all need reference later): $$\lim_{x\to 0}\frac{\sin x  x}{x(1\cos x)}=\lim_{x\to 0}\frac{\sin x  x}{2x\sin^2{\frac{x}{2}}}=\lim_{x\to 0}\frac{2(\sin x  x)}{x^3}\frac{\frac{x^2}{4}}{\sin^2{\frac{x}{2}}}=\lim_{x\to 0}\frac{2(\sin x  x)}{x^3}$$because $$\lim_{x\to 0}\frac{\frac{x^2}{4}}{\sin^2{\frac{x}{2}}}=1$$ when x approach to 0. Now, let $$L=\lim_{x\to 0}\frac{2(\sin x  x)}{x^3}(*)$$ use this indentity $$\sin x = 3\sin \frac{x}{3}  4\sin^3\frac{x}{3}$$ $$=>L=\lim_{x\to 0}\frac{2(3\sin \frac{x}{3}  4\sin^3\frac{x}{3}  x)}{x^3}=\lim_{x\to 0}\left(\frac{6}{27}\left(\frac{\sin\frac{x}{3}\frac{x}{3}}{(\frac{x}{3})^3}\right)\frac{8}{27}\right)$$ Now we can replace $$\lim_{x\to 0}\frac{\sin\frac{x}{3}\frac{x}{3}}{(\frac{x}{3})^3}=\frac{L}{2}$$by comparing to (*), finally we have $$L=\frac{6}{27}\frac{L}{2}\frac{8}{27}$$ solve this easy equation we conclude $$\lim_{x\to 0}\frac{\sin x  x}{x(1\cos x)}=L=\frac{1}{3}$$ And we have solved this limit without using L'Hopital's rule or Taylor series.
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Yes, indeed. At the beginning i have misdirected, so i tried another approaches and got the answer... – MathChill Oct 22 '21 at 07:41

3There's a gap in this argument: how do you know that the limit $L$ exists? – Hans Lundmark Oct 22 '21 at 09:24