## Calculus, 10th Edition (Anton)

Published by Wiley

# Chapter 1 - Limits and Continuity - 1.2 Computing Limits - Exercises Set 1.2 - Page 69: 8

12

#### Work Step by Step

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

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