## Thomas' Calculus 13th Edition

$\dfrac{1}{32}$
Re-arrange the given equation as follows: $\lim\limits_{(x,y) \to (2,2)}\dfrac{(x-y)}{(x^4-y^4)}=\lim\limits_{(x,y) \to (2,2)}\dfrac{(x-y)}{(x^2)^2-(y^2)^2}$ This implies that $\lim\limits_{(x,y) \to (2,2)}\dfrac{(x-y)}{(x^2)^2-(y^2)^2}=\lim\limits_{(x,y) \to (2,2)}\dfrac{(x-y)}{(x-y)(x+y)(x^2+y^2)}=\lim\limits_{(x,y) \to (2,2)}\dfrac{1}{(x+y)(x^2+y^2)}$ Thus, $\lim\limits_{(x,y) \to (2,2)}\dfrac{1}{(x+y)(x^2+y^2)}=\dfrac{1}{(4)(8)}=\dfrac{1}{32}$