Question

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

Answer 1

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

$x\to \sqrt{3}$ $\frac{4\times (\sqrt{3}{)}^3}{2\times \sqrt{3}+4\sqrt{3}}$

$\Rightarrow \frac{12\sqrt{3}}{6\sqrt{3}}$

using L' Hospital rule

$\Rightarrow lim$ $\frac{4x^3}{2x+4\sqrt{3}}$

$x\to \frac{12\sqrt{3}}{6\sqrt{3}}=2$