Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 2: Limits and Continuity - Section 2.2 - Limit of a Function and Limit Laws - Exercises 2.2 - Page 57: 54

Answer

a) 0 b) 0 c) 9 d) 3

Work Step by Step

a.$ \lim\limits_{x \to 4}(g(x)+3) = (-3+3) =-3+3= 0$ b .$ \lim\limits_{x \to 4}[xf(x)] = (4)(0) = 0$ c. $ \lim\limits_{x \to 4}[g(x)]^{2}= (-3)^{2} = 9$ d . $ \lim\limits_{x \to 4}\dfrac{g(x)}{f(x)-1}= \dfrac{(-3)}{(0-1)} = 3$
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