Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 4 - Section 4.4 - Indeterminate Forms and l''Hospital''s Rule - 4.4 Exercises - Page 314: 88

Answer

$a = \frac{4}{3}$ $b = -2$

Work Step by Step

$\lim\limits_{x \to 0}(\frac{sin~2x}{x^3}+a+\frac{b}{x^2}) = \lim\limits_{x \to 0}(\frac{sin~2x+ax^3+bx}{x^3}) = \frac{0}{0}$ We can apply L'Hospital's Rule. $\lim\limits_{x \to 0}(\frac{2~cos~2x+3ax^2+b}{3x^2})$ If this limit is $\frac{0}{0}$, then we can apply L'Hospital's Rule again. $\lim\limits_{x \to 0}(2~cos~2x+3ax^2+b) = 0$ $2+0+b = 0$ $b = -2$ If $ b = -2$, then the limit is $\frac{0}{0}$, and we can apply L'Hospital's Rule again. $\lim\limits_{x \to 0}(\frac{-4~sin~2x+6ax}{6x}) = \frac{0}{0}$ We can apply L'Hospital's Rule. $\lim\limits_{x \to 0}(\frac{-8~cos~2x+6a}{6}) = \frac{-8+6a}{6}$ If $(-8+6a) = 0,$ then this limit is $0$ $-8+6a = 0$ $a = \frac{4}{3}$ If $~~a = \frac{4}{3}~~$ and $~~b = -2,~~$ then $~~\lim\limits_{x \to 0}(\frac{sin~2x}{x^3}+a+\frac{b}{x^2}) = 0$
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