Question

Determine o limite abaixo

Answer 1

$\lim_{t \to -4} \dfrac{t^3+3t^2-4t}{(t+4)(t-2)} = \lim_{t \to -4}\dfrac{t(t^2+3t-4)}{(t+4)(t-2)} = \\\\ \lim_{t \to -4}\dfrac{t(t+4)(t-1)}{(t+4)(t-2)} = \lim_{t \to -4}\dfrac{t(t-1)}{(t-2)} = \dfrac{-4(-4-1)}{(-4-2)} = \dfrac{-4 \cdot (-5)}{-6} = -\dfrac{20}{6} = - \dfrac{10}{3}\\\\\boxed{\boxed{ \lim_{t \to -4}\dfrac{t^3+3t^2-4t}{(t+4)(t-2)} = -\dfrac{10}{3}}}$