Question

$\underset{x\to a}{lim}\frac{\sqrt{a+2x}-\sqrt{3x}}{\sqrt{3a+x}-2\sqrt{x}};(a\ne 0)$ is equal to

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

Answer 1

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

$L=\underset{x\to a}{lim}\frac{\sqrt{a+2x}-\sqrt{3x}}{\sqrt{3a+x}-2\sqrt{x}}$
This is $\frac{0}{0}$ form
$\Rightarrow L=\underset{x\to a}{lim}\frac{\sqrt{a+2x}-\sqrt{3x}}{\sqrt{3a+x}-\sqrt{4x}}\times \frac{\sqrt{a+2x}+\sqrt{3x}}{\sqrt{a+2x}+\sqrt{3x}}\times \frac{\sqrt{3a+x}+\sqrt{4x}}{\sqrt{3a+x}+\sqrt{4x}}\Rightarrow L=\underset{x\to a}{lim}\frac{(a+2x-3x)(\sqrt{3a+x}+\sqrt{4x})}{(3a+x-4x)(\sqrt{a+2x}+\sqrt{3x})}\Rightarrow L=\underset{x\to a}{lim}\frac{(a-x)(\sqrt{3a+x}+\sqrt{4x})}{3(a-x)(\sqrt{a+2x}+\sqrt{3x})}\Rightarrow L=\underset{x\to a}{lim}\frac{(\sqrt{3a+x}+\sqrt{4x})}{3(\sqrt{a+2x}+\sqrt{3x})}\Rightarrow L=\frac{\sqrt{4a}+\sqrt{4a}}{3(\sqrt{3a}+\sqrt{3a})}\Rightarrow L=\frac{4}{3\left(2\sqrt{3}\right)}=\frac{2}{3\sqrt{3}}$