Question

${lim}_{x\to a}\frac{\sqrt{a+2x}-\sqrt{3x}}{\sqrt{3a+x}-2\sqrt{x}}$

Answer 1

$\underset{x\to a}{lt}\frac{\sqrt{a+2x}-\sqrt{3x}}{\sqrt{3a+x}=2\sqrt{x}}$

Now rationalise.

$\underset{x\to a}{lt}$ $\sqrt{a+2x}-\sqrt{3x}\times$ $\frac{(\sqrt{a+2x}+\sqrt{3x})}{(\sqrt{a+2x}+\sqrt{3x})}$ $\times \frac{1}{\sqrt{3a+x}-2\sqrt{x}}$ $\times \frac{(\sqrt{3a+x}+2\sqrt{x})}{(\sqrt{3a+x}+2\sqrt{x})}$

$\underset{x\to a}{lt}\frac{a+2x-3x)}{(3a+x-9x)}\times \frac{\sqrt{3a+x}+2\sqrt{x}}{\sqrt{a+2x}+\sqrt{3x}}$

$\underset{x\to a}{lt}$ $\frac{\text{/}a-x}{3\left(\text{/}a-x\right)}\times \frac{\sqrt{3a+x}+2\sqrt{x}}{\sqrt{a+2x}+\sqrt{3x}}$

$\Rightarrow \frac{1}{3}\times \frac{\sqrt{4a}+2\sqrt{a}}{\sqrt{3a}+\sqrt{3a}}$

$\Rightarrow \frac{1}{3}\text{/}\sqrt{a}$ $(\frac{(2+2)}{2\sqrt{3\text{/}a}}=\frac{2}{3\sqrt{3}}or\frac{2\sqrt{3}}{9})$.