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: 31


$$\lim _{x\rightarrow \infty }\dfrac {2x+1}{3x^{4}-2}=\lim _{x\rightarrow -\infty }\dfrac {2x+1}{3x^{4}-2}=y=0$$

Work Step by Step

$$\lim _{x\rightarrow \infty }\dfrac {2x+1}{3x^{4}-2}=\dfrac {\dfrac {2}{x^{3}}+\dfrac {1}{x^{4}}}{3-\dfrac {2}{x^{4}}}=\dfrac {0+0}{3-0}=\dfrac {0}{3}=0$$ $$\lim _{x\rightarrow -\infty }\dfrac {2x+1}{3x^{4}-2}=\dfrac {\dfrac {2}{x^{3}}+\dfrac {1}{x^{4}}}{3-\dfrac {2}{x^{4}}}=\dfrac {0+0}{3-0}=\dfrac {0}{3}=0$$ So the horizontal asymptote will be $y=0$
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