Answer
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
Calculus, 10th Edition (Anton)
by Anton, Howard
Published by
Wiley
ISBN 10:
0-47064-772-8
ISBN 13:
978-0-47064-772-1
Chapter 1 - Limits and Continuity - 1.2 Computing Limits - Exercises Set 1.2 - Page 69: 8
Update this answer!
You can help us out by revising, improving and updating this answer.
Update this answerAfter you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
