Question

$\displaystyle\lim_{x\to3} \frac{\sqrt{3x} - 3}{\sqrt{2x - 4} - \sqrt{2}}$ is equal to :

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

Answer 1

Answer: (B)

$\displaystyle\lim_{x \to 3}$ $\frac{\sqrt{3x}-3}{\sqrt{2x-4}-\sqrt{2}}$
Rationalize
$\displaystyle\lim_{x \to 3}$ $\frac{\left(3x-9\right)\times\left(\sqrt{2x-4}+4\sqrt{2}\right)}{\left\{\left(2x-4\right)-2\right\}\times\left(\sqrt{3x+3}\right)}$
$= \displaystyle\lim_{x \to 3}$ $\frac{3\left(x-3\right)}{2\left(x-3\right)}\times\frac{\sqrt{2x-4}+\sqrt{2}}{\left(\sqrt{3x}+3\right)}$
$= \frac{3}{2}\times\frac{2\sqrt{2}}{6} = \frac{1}{\sqrt{2}}$