Evaluate: $\lim\limits_{x \to \sqrt{3}} \;\; \dfrac{x^2 -3}{x^2+3\sqrt{3}x - 12}$
$\begin{aligned} \lim\limits_{x \to \sqrt{3}} \;\; \dfrac{x^2 - 3}{x^2+3\sqrt{3}x-12} & = \lim\limits_{x \to \sqrt{3}} \;\; \dfrac{\left(x+\sqrt{3}\right) \left(x-\sqrt{3}\right)}{x^2+4\sqrt{3}x - \sqrt{3}x -12} \\ & \\ & = \lim\limits_{x \to \sqrt{3}} \;\; \dfrac{\left(x+\sqrt{3}\right) \left(x-\sqrt{3}\right)}{x \left(x+4\sqrt{3}\right)-\sqrt{3} \left(x+4\sqrt{3}\right)} \\ & \\ & = \lim\limits_{x \to \sqrt{3}} \;\; \dfrac{\left(x+\sqrt{3}\right) \left(x-\sqrt{3}\right)}{\left(x+4\sqrt{3}\right) \left(x-\sqrt{3}\right)} \\ & \\ & = \lim\limits_{x \to \sqrt{3}} \;\; \dfrac{x+\sqrt{3}}{x+4\sqrt{3}} \\ & \\ & = \dfrac{\sqrt{3}+ \sqrt{3}}{\sqrt{3}+ 4\sqrt{3}} \\ & \\ & = \dfrac{2\sqrt{3}}{5\sqrt{3}} \\ & \\ & = \dfrac{2}{5} \end{aligned}$