Limit with Circle

Suppose the chord subtends an angle $\theta$ . Then for a unit circle $c = 2\sin(\frac{\theta}{2})$ and $s(c) = \theta$ . Now $\theta \to 0$ as $c \to 0$ , so

$\displaystyle \lim_{c \to 0} \dfrac{s(c) - c}{c^{3}} = \lim_{\theta \to 0} \dfrac{\theta - 2 \sin(\frac{\theta}{2})}{8 \sin^{3}(\frac{\theta}{2})} = \lim_{x \to 0} \dfrac{x - \sin(x)}{4 \sin^{3}(x)}$ , where $x = \frac{\theta}{2} \to 0$ as $\theta \to 0$ .

Now as the limit is of the indeterminate form $0/0$ we could employ L'Hopital's rule twice to find the answer, but for sake of variety note that the Maclaurin series for $\sin(x)$ is $x - \dfrac{x^{3}}{3!} + O(x^{5})$ , so as $x \to 0$ we see that $x - \sin(x) \to \dfrac{x^{3}}{6}$ , and thus our limit becomes $\displaystyle \lim_{x \to 0} \dfrac{x^{3}}{24 \sin^{3}(x)} = \dfrac{1}{24} \approx \boxed{0.04167}$ ,

where we have used the fact that $\displaystyle \lim_{x \to 0} \dfrac{\sin(x)}{x} = 1$ .