Si $x \in [2,\infty)\setminus \{4\}$ entonces
$\begin{eqnarray*}\frac{\sqrt{2x+1}-3}{\sqrt{x-2}-\sqrt{2}} &=& \frac{\sqrt{2x+1}-3}{\sqrt{x-2}-\sqrt{2}} \cdot \frac{\sqrt{2x+1}+3}{\sqrt{2x+1}+3} \cdot \frac{\sqrt{x-2}+\sqrt{2}}{\sqrt{x-2}+\sqrt{2}}\\ &=& \frac{(2x-8)}{(x-4)}\cdot\frac{\sqrt{x-2}+\sqrt{2}}{\sqrt{2x+1}+3}\\ &=&\frac{2(\sqrt{x-2}+\sqrt{2})}{\sqrt{2x+1}+3}\end{eqnarray*}.$
Así, el límite buscado es igual a
$\begin{eqnarray*}\frac{4\sqrt{2}}{6} = \frac{2\sqrt{2}}{3}.\end{eqnarray*}$