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: 27

Answer

$\lim\limits_{t \to 1}\frac{(t^{2}+t-2)}{(t^{2}-1)} = \frac{3}{2} = 1.5$

Work Step by Step

Simplify: $\frac{(t^{2}+t-2)}{(t^{2}-1)} = \frac{(t+2)(t-1)}{(t+1)(t-1)} = \frac{(t+2)}{(t+1)}$ $\lim\limits_{t \to 1}\frac{(t^{2}+t-2)}{(t^{2}-1)} = \lim\limits_{t \to 1}\frac{(t+2)}{(t+1)} = \frac{3}{2} = 1.5$
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