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: 33

Answer

$\lim _{x\rightarrow \infty }\dfrac {40x^{5}+x^{2}}{16x^{4}-2x}=\infty ;\lim _{x\rightarrow -\infty }\dfrac {4x^{5}+x^{2}}{16x^{4}-2x}=-\infty $ Doesn’t have horizontal asymptote

Work Step by Step

$\lim _{x\rightarrow \infty }\dfrac {40x^{5}+x^{2}}{16x^{4}-2x}=\dfrac {40x+\dfrac {1}{x^{2}}}{16-\dfrac {2}{x^{3}}}=\dfrac {40x+0}{16-0}=\dfrac {40x}{16}=\dfrac {5}{2}x=\infty $ $\lim _{x\rightarrow -\infty }\dfrac {40x^{5}+x^{2}}{16x^{4}-2x}=\dfrac {40x+\dfrac {1}{x^{2}}}{16-\dfrac {2}{x^{3}}}=\dfrac {40x+0}{16-0}=\dfrac {40x}{16}=\dfrac {5}{2}x=-\infty $ Since both limits are infinite this equation doesn’t have horizontal asymptote
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