Prove length of arc is the same as chord when $\theta$ tends to $0$

Question

I'm trying to prove that the length of an arc is the same as the length of a chord in a circle when $\theta$ tends to $0$. Let

$$ \begin{eqnarray} arc &=& \theta \\ chord &=& \sqrt{(1 - \cos \theta)^2 + \sin^2 \theta} \end{eqnarray} $$

If $arc = chord$ then their ratio should be $1$, so

$$ \begin{eqnarray} && \lim_{\theta \to 0} \frac{\sqrt{(1 - \cos \theta)^2 + \sin^2 \theta}}{\theta} \\ &=& \lim_{\theta \to 0} \frac{\sqrt{2 - 2 \cos \theta}}{\theta} \\ &=& 2 \lim_{\theta \to 0} \frac{1 - \cos \theta}{\theta \sqrt{2 - 2 \cos \theta}} \\ &=& ? \end{eqnarray} $$

I happen to recognize $lim_{\theta \to 0} \frac{1 - \cos \theta}{\theta} = 0$ but I don't think I can use that while there's still $\cos \theta$ in the denominator. So that's where I'm stuck, how can I proceed or where did I make a mistake?

Answer 1

Remember that $1-\cos(\theta) =2\sin^2(\theta/2) $

so that

$\begin{array}\\ \frac{\sqrt{2 - 2 \cos \theta}}{\theta} &= \sqrt{2}\frac{\sqrt{1 - \cos \theta}}{\theta}\\ &=\sqrt{2}\frac{\sqrt{2\sin^2(\theta/2)}}{\theta}\\ &=2\frac{\sin(\theta/2)}{\theta}\\ &=\frac{\sin(\theta/2)}{\theta/2}\\ \end{array} $

You should be able to take it from here.

Answer 2

Usin Taylor series, $\sqrt{(1 - \cos \theta)^2 + \sin^2 \theta}=\sqrt{2(1-\cos\theta)}\sim_0\sqrt{2(\frac{\theta^2}2)}=\theta$

I'll edit this if I find another way. (Edit : see marty's answer)