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: 2

Answer

$ f(3)=\frac{9-3^{2}}{3-3}=\frac{0}{0}$ The function has the indeterminate form $\frac{0}{0}$ at h=3. Transforming algebraically and canceling, we have $\frac{9-h^{2}}{h-3}=\frac{-(h+3)(h-3)}{h-3}=-h-3$ Evaluating using continuity, we get $\lim\limits_{h \to 3}\frac{9-h^{2}}{h-3}=\lim\limits_{h \to 3}(-h-3)=-3-3=-6$

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