Answer
$\dfrac{1}{32}$
Work Step by Step
$\lim\limits_{(x,y) \to (2,2)}\dfrac{(x-y)}{(x^4-y^4)}=\lim\limits_{(x,y) \to (2,2)}\dfrac{(x-y)}{(x^2)^2-(y^2)^2}$
Thus, we get $\lim\limits_{(x,y) \to (2,2)}\dfrac{(x-y)}{(x-y)(x+y)(x^2+y^2)}=\lim\limits_{(x,y) \to (2,2)}\dfrac{1}{(x+y)(x^2+y^2)}$
Hence, $\dfrac{1}{(2+2)(4+4)}=\dfrac{1}{32}$
