Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 13 - Section 13.4 - Limits at Infinity; Limits of Sequences - 13.4 Exercises: 10

Answer

$\lim_{x\to-\infty}\dfrac{x^{2}+2}{x^{3}+x+1}=0$

Work Step by Step

$\lim_{x\to-\infty}\dfrac{x^{2}+2}{x^{3}+x+1}$ Since $\lim_{x\to-\infty}f(x)=\lim_{x\to\infty}f(-x)$, substitute $x$ by $-x$ in the function and simplify: $f(-x)=\dfrac{(-x)^{2}+2}{(-x)^{3}-x+1}=\dfrac{x^{2}+2}{-x^{3}-x+1}$ Evaluate $\lim_{x\to\infty}f(-x)$: $\lim_{x\to\infty}\dfrac{x^{2}+2}{-x^{3}-x+1}$ Divide the numerator and the denominator by $x^{3}$: $\lim_{x\to\infty}\dfrac{x^{2}+2}{-x^{3}-x+1}=\lim_{x\to\infty}\dfrac{\dfrac{x^{2}+2}{x^{3}}}{\dfrac{-x^{3}-x+1}{x^{3}}}=...$ Simplify: $...=\lim_{x\to\infty}\dfrac{\dfrac{x^{2}}{x^{3}}+\dfrac{2}{x^{3}}}{\dfrac{-x^{3}}{x^{3}}-\dfrac{x}{x^{3}}+\dfrac{1}{x^{3}}}=\lim_{x\to\infty}\dfrac{\dfrac{1}{x}+\dfrac{2}{x^{3}}}{-1-\dfrac{1}{x^{2}}+\dfrac{1}{x^{3}}}=...$ Apply direct substitution: $...=\dfrac{\dfrac{1}{\infty}+\dfrac{2}{\infty^{3}}}{-1-\dfrac{1}{\infty^{2}}+\dfrac{1}{\infty^{3}}}=\dfrac{0+0}{-1-0+0}=\dfrac{0}{-1}=0$
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