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If we choose two points on a circle, they form an arc and a chord. We know that if the angle subtended at the centre of the circle by the two points is small, the arc length is very close to the chord length. I wanted to prove/formalise this idea using limits. I think I’ve found a satisfactory way to do this, and I’ll be posting my method as an answer to this question. Any other approaches/ideas would be appreciated.

Heisenberg
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Consider the circle $C(O,r)$. Take any two points $A$, and $B$ on the circle. The straight line distance $AB$ is the length of the chord in question, and the arc length of the arc $AB$ is the arc length in question. I’ve named them $h$, and $s$, respectively. Let $\angle{AOB}=\theta$. Now, using the law of cosines, $h^2=r^2+r^2-2r^2cos(\theta)$. Simplifying this, $h^2=2r^2(1-cos(\theta)$. This gives, $h(\theta)=r\sqrt{2(1-cos(\theta)}$. Also, $s(\theta)=r\theta$. Now, consider $lim_{\theta \to 0}\frac{h(\theta)}{s(\theta)}$. By cancelling out $r$, we get $\frac{h(\theta)}{s(\theta)}=\frac{\sqrt{2(1-cos(\theta)}}{{\theta}}$. To evaluate the limit of this as $\theta$ tends to $0$, we make use of the fact that $1-cos{\theta}=2sin^2{\frac{\theta}{2}}$. Substituting that in, we get that the limit is equal to $2lim_{\theta \to 0}{\frac{sin(\frac{\theta}{2})}{\theta}}$. This easily evaluates to $1$, and the proposition that as the central angle approaches $0$, the chord length approaches the arc length is proved.

Heisenberg
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    How do you evaluate the last limit in your working, without using the result that you're trying to prove? – Dan Jun 20 '22 at 00:27
  • @Dan I divided the denominator by $2$, and multiplied the fraction by $\frac{1}{2}$. This is the same as multiplying the expression with $1$. Then, I pulled out $\frac{1}{2}$, and evaluated the limit of $\frac{sin(\frac{\theta}{2})}{\frac{\theta}{2}}$ as ${\theta}$ tends to $0$. – Heisenberg Jun 20 '22 at 15:37
  • And how did you evaluate that limit? – Dan Jun 20 '22 at 15:41
  • @Dan I used the Taylor Expansion of $sin(x)$. Is my argument circular? – Heisenberg Jun 21 '22 at 13:39
  • The Taylor expansion requires the derivative of $\sin{x}$. How do you differentiate $\sin{x}$ from first principles without ever using $\lim_{n\to0}\frac{\sin{x}}{x}=1$? – Dan Jun 21 '22 at 13:42
  • @Dan I’m not trying to prove that the limit of $\frac{sin(x)}{x}$ as $x$ tends to $0$ is 1. I’m taking it as a standard result, and using it to prove a fact about circles. I don’t see how what I’m doing is circular. – Heisenberg Jun 21 '22 at 13:46
  • The "standard result" that you are assuming is pretty much what you're trying to prove in the OP, isn't it? On the unit circle, $\sin{x}$ is the half the length of a chord, and $x$ is half the length of the corresponding arc. – Dan Jun 21 '22 at 13:50
  • Oh yeah, you’re right. I just didn’t think of it that way. – Heisenberg Jun 21 '22 at 13:54
  • Possibly useful: https://math.stackexchange.com/questions/75130/how-to-prove-that-lim-limits-x-to0-frac-sin-xx-1?noredirect=1&lq=1 – Dan Jun 21 '22 at 13:59