Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 1 - Functions and Limits - Review - Exercises - Page 97: 30



Work Step by Step

$\lim\limits_{t \to 2}\dfrac{t^2-4}{t^3-8}$ Factor: $\lim\limits_{t \to 2}\dfrac{(t-2)(t+2)}{(t-2)(t^2+2t+4)}$ Simplify: $\lim\limits_{t \to 2}\dfrac{t+2}{t^2+2t+4}$ Plug in 2: $\dfrac{2+2}{2^2+2(2)+4}=\dfrac{4}{12}=\dfrac{1}{3}$ Therefore, $\lim\limits_{t \to 2}\dfrac{t^2-4}{t^3-8}=\dfrac{1}{3}$
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