Answer
$\dfrac{1}{120}$
Work Step by Step
Taylor series for $\sin \theta $ can be defined as: $\sin \theta= \theta-\dfrac{ \theta^3}{3!}+\dfrac{ \theta^5}{5!}-....$
Now, $\lim\limits_{\theta \to 0}\dfrac{\sin \theta-\theta+\dfrac{\theta^3}{6}}{\theta^5}=\lim\limits_{\theta \to 0} \dfrac{-\theta+\dfrac{\theta^3}{6}+\sin \theta}{\theta^5}$
or, $=\lim\limits_{\theta \to 0} \dfrac{\dfrac{\theta^5}{5!}- \dfrac{\theta^7}{7}+...}{\theta^5}$
or, $=\lim\limits_{\theta \to 0} [\dfrac{1}{5!}-\dfrac{\theta^2}{7!}+\dfrac{\theta^4}{9!}-....]$
or, $=\dfrac{1}{120}$
