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: 31

Answer

$$\lim\limits_{x\to\infty}\frac{x^4-3x^2+x}{x^3-x+2}=\infty$$

Work Step by Step

$$A=\lim\limits_{x\to\infty}\frac{x^4-3x^2+x}{x^3-x+2}$$$$A=\lim\limits_{x\to\infty}\frac{X}{Y}$$ Divide both numerator and denominator by $x^3$ (the highest-power variable of the denominator), we have - The numerator: $X=\frac{x^4-3x^2+x}{x^3}=x-\frac{3}{x}+\frac{1}{x^2}$ - The denominator: $Y=\frac{x^3-x+2}{x^3}=1-\frac{1}{x^2}+\frac{2}{x^3}$ Therefore, $$A=\lim\limits_{x\to\infty}\frac{x-\frac{3}{x}+\frac{1}{x^2}}{1-\frac{1}{x^2}+\frac{2}{x^3}}$$ As $x\to\infty$, we have $x-\frac{3}{x}+\frac{1}{x^2}\to(\infty-3\times0+0)=\infty$ while $1-\frac{1}{x^2}+\frac{2}{x^3}\to(1-0+2\times0)=1$ Therefore, $$A=\infty$$
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