Rationalize numerator and denominator:

\[ \frac{\sqrt{3x}-3}{\sqrt{2x-4}-\sqrt{2}} =\frac{\dfrac{3(x-3)}{\sqrt{3x}+3}}{\dfrac{2(x-3)}{\sqrt{2x-4}+\sqrt{2}}} =\frac{3}{\sqrt{3x}+3}\cdot\frac{\sqrt{2x-4}+\sqrt{2}}{2}. \]

Now let \(x\to 3\): \(\sqrt{3x}\to 3\) and \(\sqrt{2x-4}\to \sqrt{2}\). Hence

\[ \lim_{x\to 3}=\frac{3}{3+3}\cdot\frac{\sqrt{2}+\sqrt{2}}{2} =\frac{1}{2}\cdot\sqrt{2}=\boxed{\frac{1}{2\sqrt{2}}}. \]