Question

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

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

Answer 1

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

Given,

$\underset{x\to 3}{lim}\left(\frac{\sqrt{3x}-3}{\sqrt{2x-4}-\sqrt{2}}\right)$

apply L-Hospital's rule,

$=\underset{x\to 3}{lim}\left(\frac{\frac{\sqrt{3}}{2\sqrt{x}}}{\frac{1}{\sqrt{2x-4}}}\right)$

$=\underset{x\to 3}{lim}\left(\frac{\sqrt{3}\sqrt{2x-4}}{2\sqrt{x}}\right)$

$=\frac{\sqrt{3}\sqrt{2\cdot 3-4}}{2\sqrt{3}}$

$=\frac{1}{\sqrt{2}}$