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

$\dfrac{1}{4}$
$\lim\limits_{(x,y) \to (4,3)} \dfrac{\sqrt x-\sqrt {y+1}}{(x-y)-1}=\lim\limits_{(x,y) \to (4,3)} \dfrac{\sqrt x-\sqrt {y+1}}{(\sqrt x+\sqrt {y+1})(\sqrt x-\sqrt {y+1})}$ This implies that$\lim\limits_{(x,y) \to (4,3)} \dfrac{\sqrt x-\sqrt {y+1}}{(\sqrt x+\sqrt {y+1})(\sqrt x-\sqrt {y+1})}=\lim\limits_{(x,y) \to (4,3)} \dfrac{1}{(\sqrt x-\sqrt {y+1})}$ Thus, we get $\dfrac{1}{(\sqrt 4-\sqrt {3+1})}=\dfrac{1}{4}$