Answer
$$\lim _{\theta \rightarrow 0} f(\theta)=-0.16\bar{6}=-\frac{1}{6} $$
Work Step by Step
Given
$$f(\theta)=\frac{\sin \theta-\theta}{\theta^{3}}$$
Since
\begin{array}{|c|c|c|c|c|}
\hline
\theta & \pm0.002 & \pm0.0001 & \pm0.00005 & \pm0.00001 \\ \hline
f(\theta) & -0.166666633 & -0.166666661 & -0.166666648 & -0.166667285 \\ \hline
\end{array}
Then
$$\lim _{\theta \rightarrow 0} f(\theta)=-0.16\bar{6}=-\frac{1}{6} $$
