Q74 / 180 · KCET 2011 · Mathematics
$\lim _{x \rightarrow a}\left[\frac{\sqrt{a+2 x}-\sqrt{3 x}}{\sqrt{3 a+x}-2 \sqrt{x}}\right]$ is equal to
- A$\frac{2}{3}$
- B$\frac{2}{\sqrt{3}}$
- C$\frac{3 \sqrt{3}}{2}$
- Correct answerD$\frac{2}{3 \sqrt{3}}$
Solution
$\begin{aligned} \lim _{x \rightarrow a} \frac{\sqrt{a+2 x}-\sqrt{3 x}}{\sqrt{3 a+x}-2 \sqrt{x}} \\ &=\lim _{x \rightarrow a} \frac{\sqrt{3 a+x}+2 \sqrt{x}}{\sqrt{a+2 x}+\sqrt{3 x}} \times \frac{a+2 x-3 x}{3 a+x-4 x} \\ &=\lim _{x \rightarrow a} \frac{\sqrt{3 a+x}+2 \sqrt{x}}{\sqrt{a+2 x}+\sqrt{3 x}} \times \frac{a-x}{3 a-3 x} \\ &=\lim _{x \rightarrow a} \cdot \frac{1}{3}\left\{\frac{\sqrt{3 a+x}+2 \sqrt{x}}{\sqrt{a+2 x}+\sqrt{3 x}}\right\} \cdot\left(\frac{a-x}{a-x}\right) \\ &=\frac{1}{3}\left\{\frac{\sqrt{4 a}+2 \sqrt{a}}{\sqrt{a+2 a}+\sqrt{3 a}}\right\}=\frac{1}{3}\left(\frac{2 \sqrt{a}+2 \sqrt{a}}{\sqrt{3 a}+\sqrt{3 a}}\right) \\ &=\frac{1}{3} \cdot \frac{4 \sqrt{a}}{2 \sqrt{3} \cdot \sqrt{a}}=\frac{2}{3 \sqrt{3}} \end{aligned}$
Hence, the correct answer is option: 'D'.