Question

Evaluate the limit:
$\underset{x\to 0}{lim}\frac{\sqrt{1+3x}-\sqrt{1-3x}}{x}$

Answer 1

At $x\to 0$,

$\frac{\sqrt{1+3x}-\sqrt{1-3x}}{x}$ becomes $\frac{0}{0}$ form

$\underset{x\to 0}{lim}\frac{\sqrt{1+3x}-\sqrt{1-3x}}{x}$

On rationalising numerator, we get

$=\underset{x\to 0}{lim}\frac{(\sqrt{1+3x}-\sqrt{1-3x})(\sqrt{1+3x}+\sqrt{1-3x})}{x(\sqrt{1+3x}+\sqrt{1-3x})}$

$=\underset{x\to 0}{lim}\frac{\left(\right(\sqrt{1+3x}{)}^2-\left(\sqrt{1-3x}{)}^2\right)}{x(\sqrt{1+3x}+\sqrt{1-3x})}$

$=\underset{x\to 0}{lim}\frac{(1+3x-1+3x)}{x(\sqrt{1+3x}+\sqrt{1-3x})}$

$=\underset{x\to 0}{lim}\frac{6x}{x(\sqrt{1+3x}+\sqrt{1-3x})}$

$=\underset{x\to 0}{lim}\frac{6}{(\sqrt{1+3x}+\sqrt{1-3x})}[\because x\ne 0]$

$=\frac{6x}{(\sqrt{1+3\left(0\right)}+\sqrt{1-3\left(0\right)})}$

$=\frac{6}{2}=3$

$\therefore \underset{x\to 0}{lim}\frac{\sqrt{1+3x}-\sqrt{1-3x}}{x}=3$