How can i prove sin(x)/x tends to 1 without circular unit circle reasoning?

[Isabella.L]

I keep getting stuck when I try to prove that the limit of sin(x)/x as x approaches 0 is 1. I understand the squeeze theorem conceptually, but when I try to set up the inequalities using the unit circle and areas, my geometric argument for why sin(x) < x < tan(x) feels circular. I'm not sure if my justification for the area of a sector is actually relying on the limit I'm trying to prove.

[Addison.W]

That sector triangle setup can work if you pin down the definitions cleanly. On a unit circle the area of the sector with angle x is x/2 because area scales with fraction of the circle, and radians tell you arc length equals x. The inscribed triangle has area 1/2 sin x and the tangent triangle has area 1/2 tan x, so for small positive x the three areas sit in order around the sector. The leap to the limit happens later when you turn those inequalities into a bound on sin x over x, but the geometry itself doesn’t depend on the limit.

[Thomas_H]

I tried the same image and kept worrying I was already assuming what I’m trying to prove. I doodled the circle, drew the wedge, and labeled the three areas, but every time I tried to assert the inequality I felt like I was assuming something about limits I shouldn’t. It left me unsettled.

[PatrickL]

I did a quick reality check by eye and noticed sin grows a touch slower than the line and tan grows faster, but that’s not a proof. The clean part for me is trusting radians so that sector area is x/2; once you have that, the comparison is geometrical rather than calculus heavy. Still, I’m not fully confident about all the steps.

[DennisR]

Is the sticking point just keeping track of which shape bounds which, or is there a hidden circularity in using the sector area formula itself?