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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!

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    $\begingroup$ What have you tried? $\endgroup$ Oct 22 at 1:45
  • $\begingroup$ 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.. $\endgroup$
    – namphamduc
    Oct 22 at 1:57
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    $\begingroup$ 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 $\endgroup$ Oct 22 at 2:11
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    $\begingroup$ 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. $\endgroup$ Oct 22 at 2:16
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    $\begingroup$ 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. $\endgroup$ Oct 22 at 2:28
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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.

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    $\begingroup$ Sweet and elegant answer! $\endgroup$ Oct 22 at 7:34
  • $\begingroup$ Yes, indeed. At the beginning i have misdirected, so i tried another approaches and got the answer... $\endgroup$
    – namphamduc
    Oct 22 at 7:41
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    $\begingroup$ There's a gap in this argument: how do you know that the limit $L$ exists? $\endgroup$ Oct 22 at 9:24

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