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: 13

Answer

$\lim_{x\to\infty}\dfrac{x^{4}}{1-x^{2}+x^{3}}$ does not exist

Work Step by Step

$\lim_{x\to\infty}\dfrac{x^{4}}{1-x^{2}+x^{3}}$ Try to evaluate the limit applying direct substitution: $\lim_{x\to\infty}\dfrac{x^{4}}{1-x^{2}+x^{3}}=\dfrac{\infty^{4}}{1-\infty^{2}+\infty^{3}}=\dfrac{\infty}{\infty}$ Indeterminate form Divide the numerator and the denominator of the function by $x^{4}$: $\lim_{x\to\infty}\dfrac{x^{4}}{1-x^{2}+x^{3}}=\lim_{x\to\infty}\dfrac{\dfrac{x^{4}}{x^{4}}}{\dfrac{1-x^{2}+x^{3}}{x^{4}}}=...$ Simplify: $...=\lim_{x\to\infty}\dfrac{1}{\dfrac{1}{x^{4}}-\dfrac{1}{x^{2}}+\dfrac{1}{x}}=...$ Apply direct substitution again: $...=\dfrac{1}{\dfrac{1}{\infty^{4}}-\dfrac{1}{\infty^{2}}+\dfrac{1}{\infty}}=\dfrac{1}{0-0+0}=\dfrac{1}{0}=\infty$
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