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I've worked my way through the geometric Squeeze Theorem proof of $\lim\limits_{x\to 0}\frac{\sin{x}}{x} $.

I was wondering how the same could be done for:

$$\lim\limits_{x\to 0}\frac{1 - \cos{x}}{x} $$

I know that we could just solve using the previous limit via multiplying by $1 + \cos(x)$ and substituting. But I'd like to be able to prove this limit with geometric intuition like we did the first.

I'm unclear how to geometrically see the initial inequality for this one. E.g. this proof begins with the geometric intuition:

$$\cos{x}<\frac{\sin{x}}{x}<1$$

and proceeds from there. Can someone explain what I should look at geometrically on the unit circle to get this or other appropriate inequality intuitions for use in the squeeze theorem? A picture of a triangle would be great, but words work just as well if you can explain the geometric intuition that sets us on our way to the squeeze theorem.

Ben G
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  • $\cos(\alpha)$ is the $x$ coordinate of a point on the unit circle such that the vector from the origin to this point forms an angle of $\alpha$ with the positive $x$ axis (positive angle means a counterclockwise rotation). So $1-\cos(\alpha)$ is just the complement. Draw a circle, pick a point on it, draw the projection of the point on the $x$ axis. The complement is $1-\cos(\alpha)$, i.e. the distance from the projection of this point on the $x$ axis to the point $(1,0)$. The fraction $\frac{1-\cos(\alpha)}{\alpha}$ then measures the ratio of this small distance to the angle (arc) itself. – Snaw Dec 11 '21 at 22:17
  • "draw the projection of the point" can you explain this. also which inequalities are you using here for the squeeze proof? – Ben G Dec 11 '21 at 22:53
  • I didn't comment on how to give a geometric proof, I only mentioned what the geometric meaning of $1-\cos(x)$ is (and of $\frac{1-\cos(x)}{x}$). The projection of the point $(x,y)$ onto the $x$ axis is the point $(x,0)$. – Snaw Dec 11 '21 at 23:15
  • @bgcode $1-\cos x = 2 \sin^2 x/2$ so this is just a manipulation of the $\sin(x)/x$ limit. You could probably straitjacket it into a geometric construction if you really wanted to, but it doesn't add much. – dxiv Dec 11 '21 at 23:47
  • @dxiv Is the sin(x)/x squeeze approach for geometrically obvious than this one? I'm just wondering if we build on this one instead of that one for particular reason, and also trying to build geometric intuition and proving skills. – Ben G Dec 12 '21 at 00:25
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    @bgcode It is straightforward to prove $,1 - \cos x = 2 \sin^2 x/2,$ geometrically, then $,(1-\cos x) / x$ $= \sin x/2 \cdot \frac{\sin x/2}{x/2},$ and the limit follows. I guess you could translate that into an entirely geometric argument from end to end, but it would be hard to beat the original intuition for $,\sin{\theta} \leq \theta \leq \tan{\theta},$. – dxiv Dec 12 '21 at 03:52
  • It seems to be zero. – Moti Dec 12 '21 at 04:47

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