Question

Evaluate $\displaystyle \lim_{x \to a}$$\frac{\sqrt{a+2x}-\sqrt{3x}}{\sqrt{3a+x}-2\sqrt{x}}$

• A. $\frac{2}{3\sqrt{3}}$
• B. $3\sqrt{3}$
• C. $\frac{1}{3\sqrt{3}}$
• D. $\frac{2\sqrt{3}}{8}$

Answer 1

Answer: (A)

We have, $\displaystyle \lim_{x \to a}$$\frac{\sqrt{a+2x}-\sqrt{3x}}{\sqrt{3a+x}-2\sqrt{x}}$
$=\displaystyle \lim_{x \to a}$$\frac{\sqrt{a+2x}-\sqrt{3x}}{\sqrt{3a+x}-2\sqrt{x}}\times\frac{\sqrt{a+2x}+\sqrt{3x}}{\sqrt{a+2x}+\sqrt{3x}}$
$=\displaystyle \lim_{x \to a}$$\frac{a+2x-3x}{\left(\sqrt{3a+x}-2\sqrt{x}\right)\left(\sqrt{a+2x}+\sqrt{3x}\right)}$
$=\displaystyle \lim_{x \to a}$$\frac{\left(a-x\right)\left[\sqrt{3a+x}+2\sqrt{x}\right]}{\left(\sqrt{a+2x}+\sqrt{3x}\right)\left(3a+x-4x\right)}$
$=\frac{4\sqrt{a}}{3 \times 2\sqrt{3a}}=\frac{2}{3\sqrt{3}}$