Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 2 - Limits - Chapter Review Exercises - Page 94: 31



Work Step by Step

By substituting, we get the intermediate form $\frac{0}{0}$. So by factorizing the numerator, we have $$ \lim _{x \rightarrow b} \frac{ x^3b^3 }{x-b }=\lim _{x \rightarrow b} \frac{ (x -b)(x^2+bx+b^2)}{x-b}\\ =\lim _{x \rightarrow b} x^2+bx+b^2 = b^2+b^2+b^2=3b^2. $$
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