## Calculus, 10th Edition (Anton)

t³ + 8 = (t + 2)(t² -2t + 4), then: $\frac{t³+8}{t+2}$ = t² - 2t + 4. Now we have: $\lim\limits_{t \to -2}\frac{t³+8}{t+2}$ = $\lim\limits_{t \to -2}t²-2t+4$ = (-2)² - 2$\times$(-2) + 4 = 4 + 4 + 4 = 12