Calculus: Early Transcendentals 8th Edition

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

Chapter 2 - Review - Exercises: 11

Answer

$$\lim\limits_{u\to1}\frac{u^4-1}{u^3+5u^2-6u}=\frac{4}{7}$$

Work Step by Step

$$A=\lim\limits_{u\to1}\frac{u^4-1}{u^3+5u^2-6u}$$ - Consider the numerator: $u^4-1=(u^2-1)(u^2+1)=(u-1)(u+1)(u^2+1)$ - Consider the denominator: $u^3+5u^2-6u=u(u^2+5u-6)=u(u-1)(u+6)$ Therefore, $$A=\lim\limits_{u\to1}\frac{(u-1)(u+1)(u^2+1)}{u(u-1)(u+6)}$$$$A=\lim\limits_{u\to1}\frac{(u+1)(u^2+1)}{u(u+6)}$$$$A=\frac{(1+1)(1^2+1)}{1\times(1+6)}$$$$A=\frac{4}{7}$$
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