Answer
$\dfrac{1}{120}$
Work Step by Step
$\lim\limits_{\theta \to 0}\dfrac{\sin \theta-\theta+\dfrac{\theta^3}{6}}{\theta^5}=\lim\limits_{\theta \to 0} \dfrac{-\theta+\dfrac{\theta^3}{6}+\sin \theta}{\theta^5}\\=\lim\limits_{\theta \to 0} \dfrac{\dfrac{\theta^5}{5!}- \dfrac{\theta^7}{7}+...}{\theta^5}\\=\lim\limits_{\theta \to 0} \dfrac{1}{5!}- \lim\limits_{\theta \to 0} \dfrac{\theta^2}{7!}+\lim\limits_{\theta \to 0} \dfrac{\theta^4}{9!}-...$
So, $\lim\limits_{\theta \to 0}\dfrac{\sin \theta-\theta+\dfrac{\theta^3}{6}}{\theta^5}=\dfrac{1}{120}$
