Question

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

Answer 1

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

$=\underset{x\to 0}{lim}\frac{(\sqrt{1+3x}-\sqrt{1-3x})}{x}\times \frac{(\sqrt{1+3x}+\sqrt{1-3x})}{(\sqrt{1+3x}+\sqrt{1-3x})}$ [on rationalising the numerator]

$=\underset{x\to 0}{lim}\frac{(1+3x)-(1-3x)}{x(\sqrt{1+3x}+\sqrt{1-3x})}$ [$\because (a+b)(a-b)=a^2-b^2$]

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

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

$=\frac{6}{(\sqrt{1+3\times 0}+\sqrt{1-3\times 0})}$

$=\frac{6}{(\sqrt{1}+\sqrt{1})}=\frac{6}{2}=3$