In https://proofwiki.org/wiki/Limit_of_Sine_of_X_over_X_at_Zero/Geometric_Proof and many other proofs, ($\sin x<x<\tan x(0<x<\pi/2)$) was used. And my concern is that ($\sin x<x<\tan x(0<x<\pi/2)$) is proved by the area formula with circle, which is proved by the limit of $\sin x/x$ as $x$ approaches 0. Is there a circular argument? If yes, could you please give me a rigorous proof? If no, could you please tell me what is wrong in my concern?
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There is a famously known proof for the $\pi r^2$ by splitting the into infinitesimal rings, which I believe does not require trigonometry.... – D S Nov 03 '23 at 16:29
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1It sounds like your question is: how can we prove that the area of a sector of the unit circle with angle $\theta$ in radians is $\theta/2$, without relying on this result? Is that correct? – Ian Nov 03 '23 at 16:44
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Another circular argument would be to use l'Hopital's Rule, as this limit is used to show that $\frac{d}{dx}\sin(x)=\cos(x)$. – Geoffrey Trang Nov 03 '23 at 16:48
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Which definition of sine are you using? – Deane Nov 04 '23 at 02:50
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2If you define the trigonometric functions by infinite series, then you can prove everything using these definitions, without appealing to trigonometry. – durianice Nov 04 '23 at 06:30
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You can also define the trigonometric functions by differential equations, which quickly leads to showing that they're analytic and gives an explicit form for their Taylor series. – anomaly Nov 04 '23 at 08:52
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Related: Non-Circular Proof of $\lim_{x \to 0} \frac{\sin x}{x} = 1$ – Adam Rubinson Nov 04 '23 at 11:01
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2This has been discussed in past (in particular see my answer in the thread referred by @AdamRubinson). The key to resolving such confusion is to first focus on definition of $\sin x$ for a given real number $x$. – Paramanand Singh Nov 04 '23 at 11:22
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@Adam Rubinson Thanks a lot – imaretard Nov 04 '23 at 15:00
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We can also avoid the use of the area formula entirely. Here is a presented a proof that $\sin(x)/x$ and $\tan(x)/x$ are squeezed to $1$ as $x\to0^+$, using the geometric fact that the shortest distance between two points is a straight line. The limit as $x\to0^+$ follows from the parity of the functions.
Begin with the unit circle centered at the origin $O$. Rendering $x$ as any positive number less than $\pi/2$, march $x$ units along the circle in both directions from the point $A=(1,0)$ to arrive at $B=(\cos(x),\sin(x))$ and $C=(\cos(x),-\sin(x))$. The arc length between these points is $2x$.
Draw the tangents at points $B$ and $C$ which intersect at $D$. From SOH CAH TOA applied to right triangles $OBD$ and $OBD$ we have $BD=CD=\tan(x)$. The sum of these distances is $2\tan(x)$.
Next draw the tangents through $A$ which intersects $\overline{BC}$ at $E$ and $\overline{BD}$ at $F$. $\overline{EF}$ is shorter than the sum of $\overline{BC}$ and $\overline{BD}$; and $\overline{OE}$ and $\overline{OF}$ respectively bisect $\angle AOB$ and $\angle AOC$. From these results we can infer that
$2\tan(x/2)<\tan(x);\quad 0<x<\pi/2$.
By iterating this bisection process the total lengths of the tangent segments is rendered as a decreasing sequence. Since the tangent segments approach the circular arc geometrically as this iteration proceeds, the decreasing sequence is bounded from below by the length of the circular arc and so
$\tan(x)>x,\,0<x<\pi/2.$
Thus $\tan(x)/x>1$.
The proof that $\sin(x)/x<1$ is simpler: simply draw $\overline{BC}$, whose length $2\sin(x)$ must be less than the arc length $2x$.
So as $x\to0^+$:
$\sin(x)<x<\tan(x)$
$\sin(x)/x<1<\tan(x)/x$
and yet both $\sin(x)$ and $\tan(x)/x$ must have the same limit because their ratio is $\cos(x)\to1$. That limit is therefore $1$.
Ian
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Oscar Lanzi
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