Answer
The limit does not exist
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to 2} \frac{{{x^2} + 5x + 4}}{{x - 2}} \cr
& {\text{Evaluate the limit by using the Direct Substitution Property}} \cr
& \mathop {\lim }\limits_{t \to a} f\left( x \right) = f\left( a \right) \cr
& \cr
& {\text{Therefore}}{\text{,}} \cr
& \mathop {\lim }\limits_{x \to 2} \frac{{{x^2} + 5x + 4}}{{x - 2}} = \frac{{{{\left( 2 \right)}^2} + 5\left( 2 \right) + 4}}{{\left( 2 \right) - 2}} \cr
& \cr
& {\text{Simplifying}} \cr
& \mathop {\lim }\limits_{x \to 2} \frac{{{x^2} + 5x + 4}}{{x - 2}} = \frac{{4 + 10 + 4}}{{4 - 2}} = \frac{{18}}{0}{\text{ }} \cr
& \cr
& {\text{Evaluate the one sided limits}} \cr
& \mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right){\text{ and }}\mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right) \cr
& \mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ - }} \frac{{{x^2} + 5x + 4}}{{x - 2}} \cr
& = \frac{{{{\left( {1.999} \right)}^2} + 5\left( {1.999} \right) + 4}}{{\left( {1.999} \right) - 2}} = - \infty \cr
& and \cr
& \mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ + }} \frac{{{x^2} + 5x + 4}}{{x - 2}} \cr
& = \frac{{{{\left( {2.001} \right)}^2} + 5\left( {2.001} \right) + 4}}{{\left( {2.001} \right) - 2}} = + \infty \cr
& {\text{Because}}\mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right) \ne \mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right),{\text{ we can conclude that}} \cr
& {\text{the limit does not exist.}} \cr} $$
