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: 21

Answer

$$\lim\limits_{x\to\infty}\frac{(2x^2+1)^2}{(x-1)^2(x^2+x)}=4$$

Work Step by Step

$$A=\lim\limits_{x\to\infty}\frac{(2x^2+1)^2}{(x-1)^2(x^2+x)}$$$$A=\lim\limits_{x\to\infty}\frac{4x^4+4x^2+1}{(x^2-2x+1)(x^2+x)}$$$$A=\lim\limits_{x\to\infty}\frac{4x^4+4x^2+1}{x^4-x^3-x^2+x}$$ Divide both numerator and denominator by $x^4$$$A=\lim\limits_{x\to\infty}\frac{\frac{4x^4+4x^2+1}{x^4}}{\frac{x^4-x^3-x^2+x}{x^4}}$$$$A=\lim\limits_{x\to\infty}\frac{4+\frac{4}{x^2}+\frac{1}{x^4}}{1-\frac{1}{x}-\frac{1}{x^2}+\frac{1}{x^3}}$$$$A=\frac{\lim\limits_{x\to\infty}(4+4\times\frac{1}{x^2}+\frac{1}{x^4})}{\lim\limits_{x\to\infty}(1-\frac{1}{x}-\frac{1}{x^2}+\frac{1}{x^3})}$$$$A=\frac{4+4\times0+0}{1-0-0+0}$$$$A=4$$
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