Limits.

The above requires one iteration of L'Hoptial's Rule, which yields:

$\frac{1/(\sqrt{a+2x}) - 3/(2\sqrt{3x})}{1/(2\sqrt{3a+x}) - 1/\sqrt{x}}$

and as $x \rightarrow a$ , we obtain $\frac{1/(\sqrt{3a}) - 3/(2\sqrt{3a})}{1/(4\sqrt{a}) - 1/\sqrt{a}} = \frac{1/\sqrt{3} - \sqrt{3}/2}{-3/4} = \frac{2-3}{2\sqrt{3}} \cdot -\frac{4}{3} = \boxed{\frac{2}{3\sqrt{3}}}.$