The value of $\lim _{x \rightarrow a} \frac{\sqrt{a+2 x}-\sqrt{3 x}}{\sqrt{3 a+x}-2 \sqrt{x}}$ is

The value of $\lim _{x \rightarrow a} \frac{\sqrt{a+2 x}-\sqrt{3 x}}{\sqrt{3 a+x}-2 \sqrt{x}}$ is
  1. $\frac{1}{3 \sqrt{3}}$
  2. $\frac{2}{\sqrt{3}}$
  3. $\frac{2}{3 \sqrt{3}}$
  4. $\frac{4}{3 \sqrt{3}}$

Solution

$\begin{aligned} & \lim _{x \rightarrow a} \frac{\sqrt{a+2 x}-\sqrt{3 x}}{\sqrt{3 a+x}-2 \sqrt{x}} \\ & =\lim _{x \rightarrow a} \frac{(\sqrt{a+2 x}-\sqrt{3 x})(\sqrt{a+2 x}+\sqrt{3 x})(\sqrt{3 a+x}+2 \sqrt{x})}{(\sqrt{3 a+x}-2 \sqrt{x})(\sqrt{3 a+x}+2 \sqrt{x})(\sqrt{a+2 x}+\sqrt{3 x})} \\ & =\lim _{x \rightarrow a} \frac{(a+2 x-3 x)(\sqrt{3 a+x}+2 \sqrt{x})}{(3 a+x-4 x)(\sqrt{a+2 x}+\sqrt{3 x})} \\ & =\lim _{x \rightarrow a} \frac{(a-x)(\sqrt{3 a+x}+2 \sqrt{x})}{3(a-x)(\sqrt{a+2 x}+\sqrt{3 x})}\end{aligned}$ $\begin{aligned} & =\frac{1}{3} \cdot \frac{(\sqrt{3 a+a}+2 \sqrt{a})}{(\sqrt{a+2 a}+\sqrt{3 a})} \\ & =\frac{4 \sqrt{a}}{3 \times 2 \times \sqrt{3} \cdot \sqrt{a}} \\ & =\frac{2}{3 \sqrt{3}} \end{aligned}$

Asked in: MHT CET 2023 (10 May Shift 1)

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