Question

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

• A. $\frac{2}{3\sqrt{3}}$
• B. $\frac{4}{3\sqrt{3}}$
• C. $\frac{5}{3\sqrt{3}}$
• D. $Noneofthese$

Answer 1

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

${lim}_{x\to a}\frac{\sqrt{a+2x}-\sqrt{3}x}{\sqrt{3a+x}-2\sqrt{x}}\left(\frac{0}{0}\right)$

${lim}_{x\to 0}\frac{\frac{1\times 2}{2\sqrt{a+2}x}-\frac{1\times 3}{2\sqrt{3x}}}{\frac{1}{2\sqrt{3a+x}-\frac{2}{2\sqrt{x}}}}$

$=\frac{\frac{1}{\sqrt{a+2}x}-\frac{3}{2\sqrt{3x}}}{\frac{1}{2\sqrt{3a+x}-\frac{1}{\sqrt{x}}}}$

$=\frac{\frac{-1}{2\sqrt{3a}}}{\frac{-3}{4\sqrt{a}}}$

$=\frac{2}{3\sqrt{3}}$