Chapter 2 - Limits - 2.2 Limits: A Numerical and Graphical Approach - Exercises - Page 54: 2

$$\lim_{t\to 0} h(t)=-0.5$$

Given $$ h(t)=\frac{\cos t-1}{t^{2}} $$ Since \begin{array}{|c|c|c|c|c|} \hline t & \pm0.002 & \pm0.0001 & \pm0.00005 & \pm0.00001 \\ \hline h(t) & -0.499999833 & -0.499999997 & -0.499999997 & -0.500000041 \\ \hline \end{array} Then $$\lim_{t\to 0} h(t)=-0.5$$