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 73: 37



Work Step by Step

Transforming algebraically and canceling, we have $\frac{x^{3}-8}{x-2}=\frac{(x-2)(x^{2}+2x+2^{2})}{x-2}=x^{2}+2x+4$ Evaluating using continuity, we get $\lim\limits_{x \to 2}\frac{x^{3}-8}{x-2}=\lim\limits_{x \to 2}(x^{2}+2x+4)=2^{2}+2(2)+4=12$
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