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

$\lim\limits_{y \to -3}(5-y)^{\frac{4}{3}} = 16$
$\lim\limits_{y \to -3}(5-y) = \lim\limits_{y \to -3}5 - \lim\limits_{y \to -3}y$ = 5+3 = 8 Thus: $\lim\limits_{y \to -3}(5-y)^{\frac{4}{3}} = (8)^{\frac{4}{3}} = 16$