Question

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

Answer 1

Given: $\underset{x\to \sqrt{3}}{lim}\frac{x^2-3}{x^2+3\sqrt{3}x-12}becomes\frac{0}{0}form$

Using Factorization method

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

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

$\Rightarrow \underset{x\to \sqrt{3}}{lim}\frac{(x-\sqrt{3})(x+\sqrt{3})}{x^2+4\sqrt{3}x-\sqrt{3}x-12}$

$[\because (a^2-b^2)=(a+b\left)\right(a-b\left)\right]$

$\Rightarrow \underset{x\to \sqrt{3}}{lim}\frac{(x-\sqrt{3})(x+\sqrt{3})}{x(x+4\sqrt{3})-\sqrt{3}(x+4\sqrt{3})}$

$\Rightarrow \underset{x\to \sqrt{3}}{lim}\frac{(x-\sqrt{3})(x+\sqrt{3})}{(x-\sqrt{3})(x+4\sqrt{3})}$

$\Rightarrow \underset{x\to \sqrt{3}}{lim}\frac{(x+\sqrt{3})}{(x+4\sqrt{3})}=\frac{(\sqrt{3}+\sqrt{3})}{(\sqrt{3}+4\sqrt{3})}=\frac{2\sqrt{3}}{5\sqrt{3}}$

$=\frac{2}{5}$

$\therefore \underset{x\to \sqrt{3}}{lim}\frac{x^2-3}{x^2+3\sqrt{3}x-12}=\frac{2}{5}$