Question

$\underset{x\to 3}{lim}\frac{\sqrt{3x}-3}{\sqrt{2x-4}-\sqrt{2}}$ is equal to :

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

Answer 1

The correct option is A $\sqrt{2}$

$\underset{x\to 3}{lim}\frac{\sqrt{3x}-3}{\sqrt{2x-4}-\sqrt{2}}$

On L'Hospital rule: $\left(\frac{0}{0}\right)$ form

$\underset{x\to 3}{lim}\frac{\frac{\sqrt{3}}{2\sqrt{x}}}{\frac{1\times 2}{2\sqrt{2x-4}}}$

$\Rightarrow \frac{\sqrt{3}\times 2\sqrt{2\times -4}}{4\sqrt{x}}$

$\therefore \underset{x\to 3}{lim}\frac{\sqrt{3}\times 2\sqrt{2\times -4}}{4\sqrt{x}}=\frac{\sqrt{3}\times 2\times \sqrt{2\times 3-4}}{4\sqrt{3}}=\frac{2\sqrt{2}}{4}=\sqrt{2}$.