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$\lim _{x \rightarrow a} \frac{\sqrt{a+2 x}-\sqrt{3 x}}{\sqrt{3 a+x}-2 \sqrt{x}}=$
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The correct answer is:
$\frac{2}{3 \sqrt{3}}$
$\begin{aligned} & \lim _{x \rightarrow a} \frac{\sqrt{a+2 x}-\sqrt{3 x}}{\sqrt{3 a+x}-2 \sqrt{x}} \\ & =\lim _{x \rightarrow a} \frac{(\sqrt{a+2 x}-\sqrt{3 x})(\sqrt{a+2 x}+\sqrt{3 x})(\sqrt{3 a+x}+2 \sqrt{x})}{(\sqrt{3 a+x}-2 \sqrt{x})(\sqrt{3 a+x}+2 \sqrt{x})(\sqrt{a+2 x}+\sqrt{3 x})} \\ & =\lim _{x \rightarrow a} \frac{(a+2 x-3 x)(\sqrt{3 a+x}+2 \sqrt{x})}{(3 a+x-4 x)(\sqrt{a+2 x}+\sqrt{3 x})} \\ & =\frac{(\sqrt{3 a+a}+2 \sqrt{a})}{3(\sqrt{a+2 a}+\sqrt{3 a})} \\ & =\frac{1}{3} \cdot \frac{(2 \sqrt{a}+2 \sqrt{a})}{(\sqrt{3 a}+\sqrt{3 a})} \\ & =\frac{4 \sqrt{a}}{6 \sqrt{3 a}} \\ & =\frac{2}{3 \sqrt{3}}\end{aligned}$
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