Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 2 - Section 2.6 - Limits at Infinity; Horizontal Asymptotes - 2.6 Exercises: 34



Work Step by Step

$$A=\lim\limits_{x\to-\infty}\frac{1+x^6}{x^4+1}$$$$=\lim\limits_{x\to-\infty}\frac{X}{Y}$$ Divide both numerator and denominator by $x^4$ (the highest-power variable in the denominator), we have - In the numerator: $X=\frac{1+x^6}{x^4}=\frac{1}{x^4}+x^2$ - In the denominator: $Y=\frac{x^4+1}{x^4}=1+\frac{1}{x^4}$ Therefore, $$A=\lim\limits_{x\to-\infty}\frac{\frac{1}{x^4}+x^2}{1+\frac{1}{x^4}}$$ As $x\to-\infty$, $\frac{1}{x^4}+x^2$ approaches $0+\infty=\infty$ while $1+\frac{1}{x^4}$ approaches $1+0=1$ Therefore, $$A=\infty$$
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