Question

The value of $\underset{x\to 0}{lim}\frac{\sqrt{3+x}-\sqrt{3}}{x}$ is

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

Answer 1

The correct option is B $\frac{1}{2\sqrt{3}}$

$\underset{x\to 0}{lim}\frac{\sqrt{3+x}-\sqrt{3}}{x}$
Rationalising both side
$\underset{x\to 0}{lim}\frac{\sqrt{3+x}-\sqrt{3}}{x}\times \frac{\sqrt{3+x}+\sqrt{3}}{\sqrt{3+x}+\sqrt{3}}$
$\underset{x\to 0}{lim}\frac{3+x-3}{x(\sqrt{3+x}+\sqrt{3})}$
$\underset{x\to 0}{lim}\frac{1}{\sqrt{3+x}+\sqrt{3}}$
$=\frac{1}{2\sqrt{3}}$

Alternative Solution

$\underset{x\to 0}{lim}\frac{\sqrt{3+x}-\sqrt{3}}{x}$
Apply L' Hospital's Rule
$\underset{x\to 0}{lim}\frac{1}{2\sqrt{3+x}}=\frac{1}{2\sqrt{3}}$