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What is the value of $\lim _{x \rightarrow a} \frac{\sqrt{\alpha+2 x}-\sqrt{3 x}}{\sqrt{3 \alpha+x}-2 \sqrt{x}}$ ?
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$\frac{1}{\sqrt{3}}$
$\begin{aligned} & \lim _{x \rightarrow \infty} \frac{\sqrt{\alpha+2 x}-\sqrt{3 x}}{\sqrt{3 \alpha+x}-2 \sqrt{x}} \\ &=\lim _{x \rightarrow \alpha} \frac{(\sqrt{\alpha+2 x})^{2}-(\sqrt{3 x})^{2}}{\sqrt{\alpha+2 x}+\sqrt{3 x}} \times \frac{\sqrt{3 \alpha+x}+2 \sqrt{x}}{(\sqrt{3 \alpha+x})^{2}-(2 \sqrt{x})^{2}} \\ &=\lim _{x \rightarrow \alpha} \frac{\alpha+2 x-3 x}{\sqrt{\alpha+2 x}+\sqrt{3 x}} \times \frac{\sqrt{3 \alpha+x}+2 \sqrt{x}}{3 \alpha+x-4 x} \\ &=\lim _{x \rightarrow \alpha} \frac{\sqrt{3 \alpha+x}+2 \sqrt{x}}{\sqrt{\alpha+2 x}+\sqrt{3 x}} \times \frac{(\alpha-x)}{3(\alpha-x)} \\ &=\frac{1}{3} \lim _{x \rightarrow \alpha} \frac{\sqrt{3 \alpha+x}+2 \sqrt{x}}{\sqrt{\alpha+2 x}+\sqrt{3 x}} \\ &=\frac{1}{3} \frac{\sqrt{4 \alpha}+2 \sqrt{\alpha}}{\sqrt{3 \alpha}+\sqrt{3 \alpha}}=\frac{1}{2}\left(\frac{4 \sqrt{\alpha}}{2 \sqrt{3} \sqrt{\alpha}}\right)=\frac{1}{\sqrt{3}} \end{aligned}$
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