Answer
$ - \infty $
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to {3^ - }} \frac{{{x^2} + 4x}}{{{x^2} - 2x - 3}} \cr
& {\text{Finding the one sided limits}} \cr
& \cr
& {\text{* }}x \to {3^ - }{\text{ taking }}x = 2.9 \cr
& \approx \frac{{{{\left( {2.9} \right)}^2} + 4\left( {2.9} \right)}}{{{{\left( {2.9} \right)}^2} - 2\left( {2.9} \right) - 3}} \cr
& \approx - 51.3 \cr
& \cr
& {\text{* }}x \to {3^ - }{\text{ taking }}x = 2.99 \cr
& \approx \frac{{{{\left( {2.99} \right)}^2} + 4\left( {2.99} \right)}}{{{{\left( {2.99} \right)}^2} - 2\left( {2.99} \right) - 3}} \cr
& \approx - 523.8 \cr
& \cr
& {\text{* }}x \to {3^ - }{\text{ taking }}x = 2.999 \cr
& \approx \frac{{{{\left( {2.999} \right)}^2} + 4\left( {2.999} \right)}}{{{{\left( {2.999} \right)}^2} - 2\left( {2.999} \right) - 3}} \cr
& \approx - 5248.81 \cr
& \cr
& {\text{* }}x \to {3^ - }{\text{ taking }}x = 2.9999 \cr
& \approx \frac{{{{\left( {2.9999} \right)}^2} + 4\left( {2.9999} \right)}}{{{{\left( {2.9999} \right)}^2} - 2\left( {2.9999} \right) - 3}} \cr
& \approx - 52498.8 \cr
& \cr
& {\text{When }}x \to {3^ - }{\text{ }}\mathop {\lim }\limits_{x \to {3^ - }} \frac{{{x^2} + 4x}}{{{x^2} - 2x - 3}}{\text{ tends to }} - \infty \cr
& \cr
& {\text{Therefore}}{\text{,}} \cr
& \mathop {\lim }\limits_{x \to {3^ - }} \frac{{{x^2} + 4x}}{{{x^2} - 2x - 3}} = - \infty \cr} $$
