Answer
$\frac{9}{2}$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{t \to 3} \frac{{{t^3} - 27}}{{{t^2} - 9}} \cr
& {\text{Evaluate the limit by using the Direct Substitution Property}} \cr
& \mathop {\lim }\limits_{t \to a} f\left( t \right) = f\left( a \right) \cr
& \cr
& {\text{Therefore}}{\text{,}} \cr
& \mathop {\lim }\limits_{t \to 3} \frac{{{t^3} - 27}}{{{t^2} - 9}} = \frac{{{{\left( 3 \right)}^3} - 27}}{{{{\left( 3 \right)}^2} - 9}} \cr
& \cr
& {\text{Simplifying}} \cr
& \mathop {\lim }\limits_{t \to 3} \frac{{{t^3} - 27}}{{{t^2} - 9}} = \frac{{{{\left( 3 \right)}^3} - 27}}{{{{\left( 3 \right)}^2} - 9}} = \frac{0}{0}{\text{ Indeterminate}} \cr
& \cr
& {\text{Factoring the numerator and denominator}} \cr
& \mathop {\lim }\limits_{t \to 3} \frac{{{t^3} - 27}}{{{t^2} - 9}} = \mathop {\lim }\limits_{t \to 3} \frac{{\left( {t - 3} \right)\left( {{t^2} + 3t + 9} \right)}}{{\left( {t + 3} \right)\left( {t - 3} \right)}} \cr
& {\text{Cancel the common factor }}t - 3 \cr
& = \mathop {\lim }\limits_{t \to 3} \frac{{{t^2} + 3t + 9}}{{t + 3}} \cr
& \cr
& {\text{Evaluate the limit}} \cr
& \mathop {\lim }\limits_{t \to 3} \frac{{{t^2} + 3t + 9}}{{t + 3}} = \frac{{{{\left( 3 \right)}^2} + 3\left( 3 \right) + 9}}{{\left( 3 \right) + 3}} = \frac{{27}}{6} = \frac{9}{2} \cr
& {\text{Then}} \cr
& \mathop {\lim }\limits_{t \to 3} \frac{{{t^2} + 3t + 9}}{{t + 3}} = \frac{9}{2} \cr} $$
