## Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning

# Chapter 2 - Section 2.6 - Limits at Infinity; Horizontal Asymptotes - 2.6 Exercises: 16

#### Answer

$\lim\limits_{x\to\infty}\frac{1-x^2}{x^3-x+1}=0$

#### Work Step by Step

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

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