Answer
$$\frac{3}{2}$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{t \to 2} \frac{{{t^3} - {t^2} - 2t}}{{{t^2} - 4}} \cr
& {\text{evaluating the limit}} \cr
& = \frac{{8 - 4 - 4}}{{4 - 4}} \cr
& = \frac{0}{0} \cr
& \cr
& {\text{applying l'Hopital's rule}} \cr
& = \mathop {\lim }\limits_{t \to 2} \frac{{\frac{d}{{dt}}\left[ {{t^3} - {t^2} - 2t} \right]}}{{\frac{d}{{dt}}\left[ {{t^2} - 4} \right]}} \cr
& = \mathop {\lim }\limits_{t \to 2} \frac{{3{t^2} - 2t - 2}}{{2t}} \cr
& {\text{evaluating the limit}} \cr
& = \mathop {\lim }\limits_{t \to 2} \frac{{3{t^2} - 2t - 2}}{{2t}} = \frac{{3{{\left( 2 \right)}^2} - 2\left( 2 \right) - 2}}{{2\left( 2 \right)}} \cr
& = \frac{6}{4} \cr
& = \frac{3}{2} \cr} $$
