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: 23

Answer

$$\lim\limits_{x\to\infty}\frac{\sqrt{1+4x^6}}{2-x^3}=-2$$

Work Step by Step

$$A=\lim\limits_{x\to\infty}\frac{\sqrt{1+4x^6}}{2-x^3}$$$$A=\lim\limits_{x\to\infty}\frac{X}{Y}$$ Divide both numerator and denominator by $x^3$, we have - In the numerator: $X=\frac{\sqrt{1+4x^6}}{x^3}=\frac{\sqrt{1+4x^6}}{\sqrt{x^6}}=\sqrt{\frac{1+4x^6}{x^6}}=\sqrt{\frac{1}{x^6}+4}$ - In the denominator: $Y=\frac{2-x^3}{x^3}=\frac{2}{x^3}-1$ Therefore, $$A=\lim\limits_{x\to\infty}\frac{\sqrt{\frac{1}{x^6}+4}}{\frac{2}{x^3}-1}$$$$A=\frac{\lim\limits_{x\to\infty}(\sqrt{\frac{1}{x^6}+4})}{\lim\limits_{x\to\infty}(\frac{2}{x^3})-1}$$$$A=\frac{\sqrt{\lim\limits_{x\to\infty}(\frac{1}{x^6})+4}}{2\times\lim\limits_{x\to\infty}(\frac{1}{x^3})-1}$$$$A=\frac{\sqrt{0+4}}{2\times0-1}$$$$A=-2$$
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