Calculus: Early Transcendentals 8th Edition

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

Chapter 2 - Review - Exercises: 7



Work Step by Step

$$A=\lim\limits_{h\to0}\frac{(h-1)^3+1}{h}$$ - Consider the numerator: $(h-1)^3+1=(h^3-3h^2+3h-1)+1=h^3-3h^2+3h=h(h^2-3h+3)$ Therefore, $$A=\lim\limits_{h\to0}\frac{h(h^2-3h+3)}{h}$$$$A=\lim\limits_{h\to0}(h^2-3h+3)$$$$A=0^2-3\times0+3$$$$A=3$$
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