## Thomas' Calculus 13th Edition

$\dfrac{1}{4}$
Re-arrange the equation as follows: $\lim\limits_{(x,y) \to (4,3)} \dfrac{\sqrt x-\sqrt {y+1}}{(x-y)-1}=\lim\limits_{(x,y) \to (4,3)} \dfrac{\sqrt x-\sqrt {y+1}}{(\sqrt x+\sqrt {y+1})(\sqrt x-\sqrt {y+1})}$ and $\lim\limits_{(x,y) \to (4,3)} \dfrac{\sqrt x-\sqrt {y+1}}{(\sqrt x+\sqrt {y+1})(\sqrt x-\sqrt {y+1})}=\lim\limits_{(x,y) \to (4,3)} \dfrac{1}{(\sqrt x-\sqrt {y+1})}$ Plug the limits, then we get $\dfrac{1}{(\sqrt 4+\sqrt {3+1})}=\dfrac{1}{(\sqrt 4+\sqrt {4})}=\dfrac{1}{4}$