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 - Page 138: 17

Answer

$\lim\limits_{x\to-\infty}\frac{x-2}{x^2+1}=0$

Work Step by Step

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