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: 32

Answer

$\lim _{x\rightarrow \infty }\dfrac {12x^{8}-3}{3x^{8}-2x^{7}}=4;\lim _{x\rightarrow -\infty }\dfrac {12x^{8}-3}{3x^{8}-2x^{7}}=4;y=4$

Work Step by Step

$\lim _{x\rightarrow \infty }\dfrac {12x^{8}-3}{3x^{8}-2x^{7}}=\dfrac {12-\dfrac {3}{x^{8}}}{3-\dfrac {2}{x}}=\dfrac {12-0}{3-0}=\dfrac {12}{3}=4$ $\lim _{x\rightarrow -\infty }\dfrac {12x^{8}-3}{3x^{8}-2x^{7}}=\dfrac {12-\dfrac {3}{x^{8}}}{3-\dfrac {2}{x}}=\dfrac {12-0}{3-0}=\dfrac {12}{3}=4$ So the horizontal asymptote will be $y=4$
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