Question

The value of $\underset{x\to \sqrt{3}}{lim}\frac{x^2-3}{x^2+3\sqrt{3}x-12}$ is equal to?

• A. $\frac{1}{5}$
• B. $\frac{2}{5}$
• C. $\frac{\sqrt{3}}{5}$
• D. $\frac{3}{5}$

Answer 1

The correct option is B $\frac{2}{5}$

We have,
$\underset{x\to \sqrt{3}}{lim}\frac{x^2-3}{x^2+3\sqrt{3}x-12}$
$=\underset{x\to \sqrt{3}}{lim}\frac{(x-\sqrt{3})(x+\sqrt{3})}{(x+4\sqrt{3})(x-\sqrt{3})}$

$=\underset{x\to \sqrt{3}}{lim}\frac{x+\sqrt{3}}{x+4\sqrt{3}}=\frac{2}{5}$
Hence, option 'B' is correct.