## University Calculus: Early Transcendentals (3rd Edition)

$3$
$\lim\limits_{(x,y) \to (1,-1)}\dfrac{(x^3-y^3)}{(x+y)}=\lim\limits_{(x,y) \to (1,-1)}\dfrac{(x+y)(x^2-xy+y^2)}{(x+y)}$ Thus, we get $\lim\limits_{(x,y) \to (1,-1)}\dfrac{(x+y)(x^2-xy+y^2)}{(x+y)}=\lim\limits_{(x,y) \to (1,-1)}(x^2-xy+y^2)$ Hence, $1-(1)(-1)+(-1)^2=3$