Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 2 - Limits - 2.5 Limits at Infinity - 2.5 Exercises - Page 96: 14



Work Step by Step

$0\leq \dfrac {\sin ^{4}x^{3}}{x^{2}}\leq \dfrac {1}{x^{2}}$ $\lim _{x\rightarrow -\infty }\dfrac {1}{x^{2}}=0\Rightarrow 0\leq \dfrac {\sin ^{4}x^{3}}{x^{2}}\leq 0\Rightarrow \dfrac {\sin ^{4}x^{3}}{x^{2}}=0$ $\Rightarrow \lim _{x\rightarrow -\infty }\left( 5+\dfrac {100}{x}+\dfrac {\sin ^{4}x^{3}}{x^{2}}\right) =5+0+0=5$
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