Question

$\underset{x\to \sqrt{3}}{\lim }\frac{x^4-9}{x^2+4\sqrt{3}x-15}$

Answer 1

$$\begin{gathered} \underset{x\to \sqrt{3}}{\lim }\left[\frac{x^4-9}{x^2+4\sqrt{3}x-15}\right] \\ \mathrm{It}\mathrm{is}\mathrm{of}\mathrm{the}\mathrm{form}\frac{0}{0}. \\ \underset{x\to \sqrt{3}}{\lim }\left[\frac{{\left(x^2\right)}^2-{\left(3\right)}^2}{x^2+5\sqrt{3}x-\sqrt{3}x-15}\right] \\ =\underset{x\to \sqrt{3}}{\lim }\left[\frac{\left(x^2-3\right)\left(x^2+3\right)}{x\left(x+5\sqrt{3}\right)-\sqrt{3}\left(x+5\sqrt{3}\right)}\right] \\ =\underset{x\to \sqrt{3}}{\lim }\left[\frac{\left\{x^2-{\left(\sqrt{3}\right)}^2\right\}\left(x^2+3\right)}{\left(x-\sqrt{3}\right)\left(x+5\sqrt{3}\right)}\right] \\ =\underset{x\to \sqrt{3}}{\lim }\left[\frac{\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)\left(x^2+3\right)}{\left(x-\sqrt{3}\right)\left(x+5\sqrt{3}\right)}\right] \\ =\frac{\left(\sqrt{3}+\sqrt{3}\right)\left(3+3\right)}{\sqrt{3}+5\sqrt{3}} \\ =\frac{2\sqrt{3}\times 6}{6\sqrt{3}} \\ =2 \end{gathered}$$