Calculus: Early Transcendentals 8th Edition

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

Chapter 2 - Review - Exercises: 8

Answer

$$\lim\limits_{t\to2}\frac{t^2-4}{t^3-8}=\frac{1}{3}$$

Work Step by Step

$$A=\lim\limits_{t\to2}\frac{t^2-4}{t^3-8}$$ - Consider the numerator: $t^2-4=(t-2)(t+2)$ - Consider the denominator: $t^3-8=(t-2)(t^2+2t+4)$ Therefore, $$A=\lim\limits_{t\to2}\frac{(t-2)(t+2)}{(t-2)(t^2+2t+4)}$$$$A=\lim\limits_{t\to2}\frac{t+2}{t^2+2t+4}$$$$A=\frac{2+2}{2^2+2\times2+4}$$$$A=\frac{1}{3}$$
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