University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 2 - Section 2.2 - Limit of a Function and Limit Laws - Exercises: 28

Answer

$\lim\limits_{t \to -1}\frac{(t^{2}+3t+2)}{(t^{2}-t-2)} = -\frac{1}{3}= 0.33$

Work Step by Step

Simplify: $\frac{(t^{2}+3t+2)}{(t^{2}-t-2)} =\frac{(t+1)(t+2)}{(t-2)(t+1)} = \frac{(t+2)}{(t-2)}$ Now, $\lim\limits_{t \to -1}\frac{(t^{2}+3t+2)}{(t^{2}-t-2)} = \lim\limits_{t \to -1}\frac{(t+2)}{(t-2)} = -\frac{1}{3}= 0.33$
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