Answer
$6$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{t \to 4} \frac{{{t^2} - 2t - 8}}{{t - 4}} \cr
& {\text{Evaluate the limit by using the Direct Substitution Property}} \cr
& \mathop {\lim }\limits_{t \to a} f\left( t \right) = f\left( a \right) \cr
& \cr
& {\text{Therefore}}{\text{,}} \cr
& \mathop {\lim }\limits_{t \to 4} \frac{{{t^2} - 2t - 8}}{{t - 4}} = \frac{{{{\left( 4 \right)}^2} - 2\left( 4 \right) - 8}}{{\left( 4 \right) - 4}} \cr
& \cr
& {\text{Simplifying}} \cr
& \mathop {\lim }\limits_{t \to 4} \frac{{{t^2} - 2t - 8}}{{t - 4}} = \frac{0}{0}{\text{ Indeterminate}} \cr
& \cr
& {\text{Factoring the numerator}} \cr
& \mathop {\lim }\limits_{t \to 4} \frac{{{t^2} - 2t - 8}}{{t - 4}} = \mathop {\lim }\limits_{t \to 4} \frac{{\left( {t - 4} \right)\left( {t + 2} \right)}}{{t - 4}} \cr
& {\text{Cancel the common factor }}t - 4 \cr
& = \mathop {\lim }\limits_{t \to 4} \left( {t + 2} \right) \cr
& \cr
& {\text{Evaluate the limit}} \cr
& \mathop {\lim }\limits_{t \to 4} \left( {t + 2} \right) = 4 + 2 = 6 \cr
& {\text{Then}} \cr
& \mathop {\lim }\limits_{t \to 4} \frac{{{t^2} - 2t - 8}}{{t - 4}} = 6 \cr} $$
