Mind your limits

As the limit is in the indeterminate form $0/0$ we could use L'Hopital's rule for a quick approach, but avoiding it, we first note that, for $x \ne 3$ ,

$\dfrac{\sqrt{3x} - 3}{\sqrt{2x - 4} - \sqrt{2}} = \dfrac{\sqrt{3}}{\sqrt{2}} \times \dfrac{\sqrt{x} - \sqrt{3}}{\sqrt{x - 2} - 1} = \dfrac{\sqrt{3}}{\sqrt{2}} \times \dfrac{\sqrt{x} - \sqrt{3}}{\sqrt{x - 2} - 1} \times \dfrac{\sqrt{x - 2} + 1}{\sqrt{x - 2} + 1} =$

$\dfrac{\sqrt{3}}{\sqrt{2}} \times \dfrac{(\sqrt{x} - \sqrt{3})(\sqrt{x - 2} + 1)}{x - 3} = \dfrac{\sqrt{3}}{\sqrt{2}} \times \dfrac{(\sqrt{x} - \sqrt{3})(\sqrt{x - 2} + 1)}{(\sqrt{x} - \sqrt{3})(\sqrt{x} + \sqrt{3})} = \dfrac{\sqrt{3}}{\sqrt{2}} \times \dfrac{\sqrt{x - 2} + 1}{\sqrt{x} + \sqrt{3}}$ .

The given limit is then equivalent to $\dfrac{\sqrt{3}}{\sqrt{2}} \displaystyle\lim_{x \to 3} \dfrac{\sqrt{x - 2} + 1}{\sqrt{x} + \sqrt{3}}$ , which is no longer in the indeterminate form $0/0$ , so we can just plug in $x = 3$ to find that the limit is

$y = \dfrac{\sqrt{3}}{\sqrt{2}} \times \dfrac{2}{2\sqrt{3}} = \dfrac{1}{\sqrt{2}} \approx 0.707 \Longrightarrow \lfloor y \rfloor = \boxed{0}$ .