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: 33


$\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
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.