Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 2 - Section 2.3 - Calculating Limits Using the Limit Laws - 2.3 Exercises: 5

Answer

$\lim\limits_{t\to -2}\frac{t^4-2}{2t^2-3t+2}=\frac{7}{8}$

Work Step by Step

$\lim\limits_{t\to -2}\frac{t^4-2}{2t^2-3t+2}$ $=\frac{\lim\limits_{t\to -2}(t^4-2)}{\lim\limits_{t\to -2}(2t^2-3t+2)}$ (quotient law) $=\frac{\lim\limits_{t\to -2}t^4-\lim\limits_{t\to -2}2}{\lim\limits_{t\to -2}2t^2-\lim\limits_{t\to -2}3t+\lim\limits_{t\to -2}2}$ (difference and addition law) $=\frac{\lim\limits_{t\to -2}t^4-\lim\limits_{t\to -2}2}{2\lim\limits_{t\to -2}t^2-3\lim\limits_{t\to -2}t+\lim\limits_{t\to -2}2}$ (constant multiple law) $=\frac{(-2)^4-2}{2\times(-2)^2-3\times(-2)+2}$ $=\frac{7}{8}$
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