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: 22

Answer

$\lim\limits_{x \to -\infty} \frac{4x^2+5}{x^2+3} = 4$

Work Step by Step

We only look at the terms in the numerator and denominator with highest degree, as they are the only terms with significance as x approaches infinity. $\lim\limits_{x \to -\infty}\frac{4x^2+5}{x^2+3} = \lim\limits_{x \to -\infty}\frac{4x^2}{x^2} = \lim\limits_{x \to -\infty}4 = 4$
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