Calculus (3rd Edition)

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

Chapter 2 - Limits - 2.5 Evaluating Limits Algebraically - Exercises - Page 72: 8



Work Step by Step

Numerator at $x=8:\,\,\,x^{3}-64x=(8)^{3}-64(8)=0$ Denominator at $x=8:\,\,\,x-8=8-8=0$ The function has the indeterminate form $\frac{0}{0}$ at $x=7$. Transforming algebraically and canceling the common factor, we have $\frac{x^{3}-64x}{x-8}=\frac{(x-8)(x^{2}+8x)}{x-8}=x^{2}+8x$ Therefore, $\lim\limits_{x \to 8}\frac{x^{3}-64x}{x-8}=\lim\limits_{x \to 8}x^{2}+8x=(8)^{2}+8(8)=128$
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