My main claim is to prove $\lim_{x\to 0} \sin x/x = 1$. So I drew a circular sector and some triangles so I could get the following formula from its areas: $\sin (x / 2) < x / 2 < \tan (x / 2)$. But I think I can prove that the area of a circular sector is $x / 2$ only from $\lim_{x\to 0} \sin x/x = 1$ or integrals. Am I wrong? How can I prove that the area of a circle is $\pi r^2$ without integrals and $\lim_{x\to 0} \sin x/x = 1$?
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Alternatively, you could try to prove it using curve lengths and perimeters instead of areas, see for example my answer here for a sketch of such proof. – dxiv Aug 12 '21 at 05:26
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Here and here are summaries of one way to do it. – David Mitra Aug 12 '21 at 05:34
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To deal with area you need to use some limit based definition of area which mostly uses integrals. The same goes for arc-length also. – Paramanand Singh Aug 12 '21 at 20:53
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@ParamanandSingh Though in cases where geometry is introduced before calculus, one of length or area may be considered a primitive notion. Most often the curve length, for example that's how the measure of an angle is usually defined. – dxiv Aug 13 '21 at 07:36
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@dxiv: I prefer to make following assumptions for a beginner: an arc of a circle has a well defined length, a sector of a circle has a well defined area and finally if $A$ is area of a sector of circle of radius $R$ and $L$ is length of corresponding arc then $A=RL/2$. With these assumptions one may prove the limit in question. – Paramanand Singh Aug 13 '21 at 11:42