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

a) 1 b) 0 c) $\frac{16}{3}$
a) $\lim\limits_{x \to -2}(p(x)+r(x)+s(x)) = 4+0-3 = 1$ b) $\lim\limits_{x \to -2}p(x).r(x).s(x) = 4(0)(-3) = 0$ c) $\lim\limits_{x \to -2}\frac{-4p(x)+5r(x)}{s(x)} = \frac{-4(4)+5(0)}{-3}=\frac{16}{3}$