Q.$\displaystyle\lim_{x\to a}\left[\frac{\sqrt{a+2x}-\sqrt{3x}}{\sqrt{3a+x}-2\sqrt{x}}\right]$

Multiply numerator by $\sqrt{a+2x}+\sqrt{3x}$: it becomes $(a+2x)-3x=a-x=-(x-a)$. Multiply denominator by $\sqrt{3a+x}+2\sqrt x$: it becomes $(3a+x)-4x=3a-3x=-3(x-a)$. So the expression becomes $\dfrac{-(x-a)/(\sqrt{a+2x}+\sqrt{3x})}{-3(x-a)/(\sqrt{3a+x}+2\sqrt x)}=\dfrac{\sqrt{3a+x}+2\sqrt x}{3(\sqrt{a+2x}+\sqrt{3x})}$ after cancelling $(x-a)$. At $x=a$: $\dfrac{\sqrt{4a}+2\sqrt a}{3(\sqrt{3a}+\sqrt{3a})}=\dfrac{2\sqrt a+2\sqrt a}{6\sqrt{3a}}=\dfrac{4\sqrt a}{6\sqrt{3a}}=\dfrac{2}{3\sqrt3}=\dfrac{2\sqrt3}{9}$ — the $\sqrt a$ cancels out of the ratio, so the answer is the same constant for every $a>0$.

Rationalize numerator and denominator by their separate conjugates, cancel the shared $(x-a)$, substitute and simplify the surviving square roots.

Not noticing the final answer is independent of $a$ (the $\sqrt a$ factors cancel), and stopping the simplification too early leaving $a$ in the answer.