Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 3 - Applications of Differentiation - 3.5 Exercises: 24

Answer

$\lim\limits_{x \to \infty}\frac{5x^3+1}{10x^3-3x^2+7} = \frac{1}{2}$

Work Step by Step

We look at the term with the highest degre in the numerator and denominator, as these will be the only values that have any significance when x approaches infinity. $\lim\limits_{x \to \infty} \frac{5x^3+1}{10x^3-3x^2+7} = \lim\limits_{x \to \infty}\frac{5x^3}{10x^3} = \lim\limits_{x \to \infty}\frac{1}{2} = \frac{1}{2}$
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