Question

$\underset{x\to \sqrt{3}}{\lim }\frac{x^2-3}{x^2+3\sqrt{3}x-12}$

Answer 1

$$\begin{gathered} \underset{x\to \sqrt{3}}{\lim }\left[\frac{x^2-3}{x^2+3\sqrt{3}x-12}\right] \\ \mathrm{It}\mathrm{is}\mathrm{of}\mathrm{the}\mathrm{form}\frac{0}{0}. \\ \underset{x\to \sqrt{3}}{\lim }\left[\frac{x^2-{\left(\sqrt{3}\right)}^2}{x^2+4\sqrt{3}x-\sqrt{3}x-12}\right] \\ =\underset{x\to \sqrt{3}}{\lim }\left[\frac{\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)}{x\left(x+4\sqrt{3}\right)-\sqrt{3}\left(x+4\sqrt{3}\right)}\right] \\ =\underset{x\to \sqrt{3}}{\lim }\left[\frac{\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)}{\left(x-\sqrt{3}\right)\left(x+4\sqrt{3}\right)}\right] \\ =\frac{\sqrt{3}+\sqrt{3}}{\sqrt{3}+4\sqrt{3}} \\ =\frac{2}{5} \end{gathered}$$