## Calculus: Early Transcendentals 8th Edition

$-2.5$
From the table, we observe that the function $f(x,y)$ approaches to $-2.5$ as $(x,y)$ approaches $(0,0)$. Our guess can be proved by using the substitution method as follows: $\lim\limits_{(x,y) \to(0,0)}f(x,y)=\lim\limits_{(x,y) \to(0,0)}f(x,y)\frac{x^{2}y^{3}+x^{3}y^{2}-5}{2-xy}$ $=\frac{-5}{2}=-2.5$