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

$-\dfrac{23}{7}$
Consider: $\lim\limits_{t \to -3}f(t)=\lim\limits_{t \to -3}\dfrac{t^3-4t +15}{t^2-t-12}$ Now, $f(-3)=\dfrac{(-3)^3-4(-3)+15}{(-3)^2-(-3)-12}=\dfrac{-27+12+15}{9+3-12}=\dfrac{0}{0}$ The limit shows an indeterminate form. Thus, apply L-Hospital's rule: $\lim\limits_{a \to b}f(x)=\lim\limits_{a \to b}\dfrac{g'(x)}{h'(x)}$ Thus: $\lim\limits_{t \to -3}\dfrac{3t^2-4}{2t-1}=\dfrac{3(-3)^2 -4}{2(-3)-1}=-\dfrac{23}{7}$