Question

Evaluate the limit:

$\underset{x\to 3}{lim}\frac{\sqrt{x+3}-\sqrt{6}}{x^2-9}$

Answer 1

At $x\to 3,\frac{\sqrt{x+3}-\sqrt{6}}{x^2-9}\text{becomes}\frac{0}{0}\text{form}$

$\underset{x\to 3}{lim}\frac{\sqrt{x+3}-\sqrt{6}}{x^2-9}$

On rationalising numerator, we get

$=\underset{x\to 3}{lim}\frac{(\sqrt{x+3}-\sqrt{6})\times (\sqrt{x+3}+\sqrt{6})}{(x^2-9)(\sqrt{x+3}+\sqrt{6})}$

$=\underset{x\to 3}{lim}\frac{({\left(\sqrt{x+3}\right)}^2-{\left(\sqrt{6}\right)}^2)}{(x^2-9)(\sqrt{x+3}+\sqrt{6})}$

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

$=\underset{x\to 3}{lim}\frac{(x+3-6)}{(x^2-9)(\sqrt{x+3}+\sqrt{6})}$

$=\underset{x\to 3}{lim}\frac{(x-3)}{(x+3)(x-3)(\sqrt{x+3}+\sqrt{6})}\left[\right(x-3)\ne 0]$

$=\underset{x\to 3}{lim}\frac{1}{(x+3)(\sqrt{x+3}+\sqrt{6})}$

$=\frac{1}{(3+3)(\sqrt{3+3}+\sqrt{6})}$

$=\frac{1}{6\left(2\sqrt{6}\right)}$

$\therefore \underset{x\to 3}{lim}\frac{\sqrt{x+3}-\sqrt{6}}{x^2-9}=\frac{1}{12\sqrt{6}}$