Answer
$\dfrac{1}{4}$
Work Step by Step
Re-arrange the given equation as: $\lim\limits_{(x,y) \to (2,0)} \dfrac{\sqrt {2x-y}-2}{(2x-y)-4}=\lim\limits_{(x,y) \to (2,0)}\dfrac{\sqrt {2x-y}-2}{(\sqrt {2x-y}+2)(\sqrt {2x-y}-2)}$
This implies that$\lim\limits_{(x,y) \to (2,0)}\dfrac{\sqrt {2x-y}-2}{(\sqrt {2x-y}+2)(\sqrt {2x-y}-2)}=\lim\limits_{(x,y) \to (2,0)}\dfrac{1}{(\sqrt {2x-y}+2)}$
Plug the limits, then we get $\dfrac{1}{(\sqrt {2(2)-0}+2)}=\dfrac{1}{(\sqrt {4}+2)}=\dfrac{1}{4}$
