Answer
The limit does not exist
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to 4} \frac{{{x^2} + 3x}}{{{x^2} - x - 12}} \cr
& {\text{Evaluate the limit by using the Direct Substitution Property}} \cr
& \mathop {\lim }\limits_{t \to a} f\left( x \right) = f\left( a \right) \cr
& \cr
& {\text{Therefore}}{\text{,}} \cr
& \mathop {\lim }\limits_{x \to 4} \frac{{{x^2} + 3x}}{{{x^2} - x - 12}} = \frac{{{{\left( 4 \right)}^2} + 3\left( 4 \right)}}{{{{\left( 4 \right)}^2} - \left( 4 \right) - 12}} \cr
& \cr
& {\text{Simplifying}} \cr
& \mathop {\lim }\limits_{x \to 4} \frac{{{x^2} + 3x}}{{{x^2} - x - 12}} = \frac{{16 + 12}}{{16 - 4 - 12}} = \frac{{28}}{0} \cr
& \cr
& {\text{Evaluate the one sided limits}} \cr
& \mathop {\lim }\limits_{x \to {4^ - }} f\left( x \right){\text{ and }}\mathop {\lim }\limits_{x \to {4^ + }} f\left( x \right) \cr
& \mathop {\lim }\limits_{x \to {4^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {4^ - }} \frac{{{x^2} + 3x}}{{{x^2} - x - 12}} \cr
& = \frac{{{{\left( {3.999} \right)}^2} + 3\left( {3.999} \right)}}{{{{\left( {3.999} \right)}^2} - \left( {3.999} \right) - 12}} = - \infty \cr
& and \cr
& \mathop {\lim }\limits_{x \to {4^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {4^ + }} \frac{{{x^2} + 3x}}{{{x^2} - x - 12}} \cr
& = \frac{{{{\left( {4.001} \right)}^2} + 3\left( {4.001} \right)}}{{{{\left( {4.001} \right)}^2} - \left( {4.001} \right) - 12}} = + \infty \cr
& {\text{Because}}\mathop {\lim }\limits_{x \to {4^ - }} f\left( x \right) \ne \mathop {\lim }\limits_{x \to {4^ + }} f\left( x \right),{\text{ we can conclude that}} \cr
& {\text{the limit does not exist.}} \cr} $$
