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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!

OnTheWay
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    What have you tried? – Jose Arnaldo Bebita Dris 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.. – OnTheWay Oct 22 '21 at 01:57
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    Surprised 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
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    The difficulty here is that normally in a problem-solving 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 re-invent the wheel, walking in the path of the proof of L'Hopital's rule, and applying this re-invention 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... – OnTheWay Oct 22 '21 at 02:24
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    If 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 proof-oriented 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/evaluation-of-lim-limits-x-rightarrow0-frac-tanx-xx3, https://math.stackexchange.com/questions/217081/determine-lim-x-to-0-fracx-sinxx3-frac16-without-lhospita, https://math.stackexchange.com/questions/387333/are-all-limits-solvable-without-lh%c3%b4pital-rule-or-series-expansion, etc. – Hans Lundmark Oct 22 '21 at 09:26
  • Thank you, it is my fault when dont make a search before posting... – OnTheWay Oct 22 '21 at 09:40

1 Answers1

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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.

OnTheWay
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