## Thomas' Calculus 13th Edition

$2$
$\lim\limits_{(x,y) \to (1,1)} \dfrac{(x-y)(x+y)}{x-y}=\lim\limits_{(x,y) \to (1,1)}(x+y)$ Thus, we have $\lim\limits_{(x,y) \to (1,1)} \dfrac{(x-y)(x+y)}{x-y}=1+1=2$