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The proof of $\lim\limits_{x\to 0}\frac{\sin x}{x}=1$ has been asked before and numerous answers were given.

Most proofs uses the idea of comparing the area of one sector and two triangle (or two sectors and one triangle) as is shown below:

enter image description here

My question is, how does one prove that in the graph above, the sector $ABC$ is contained in the triangle $ABD$? Namely, given any point $P$ which is contained in the sector $ABC$, how do we prove that $P$ is always contained in the triangle $ABD$?

It seems to me that in most of Calculus books, this fact is assumed as intuitively clear but never strictly proved.

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While a lot about this relies on fairly fuzzy intuition, that part at least is fairly clear. If $E$ is a point on the segment $BD$, then $AE =\sqrt{AB^2+BE^2}>AB$ by the Pythagorean theorem, and the intersection $F$ of line $AE$ with the circle lies inside the segment $AE$. Then $AF\subset AE$ (as sets of points), and every point inside the circular wedge in that direction from $A$ is inside the triangle. Take the union over all possible $E$, and the circular wedge is a subset of the triangle.

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  • $\begingroup$ Thanks! This is a paragraph that is missing from most calculus books. $\endgroup$ – Zuriel Feb 11 at 11:43

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