Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 2 - Limits - 2.7 Limits at Infinity - Exercises - Page 82: 7



Work Step by Step

Dividing the numerator and denominator by x, we have $\lim\limits_{x \to \infty}\frac{x}{x+9}=\lim\limits_{x \to \infty}\frac{1}{1+\frac{9}{x}}=\frac{\lim\limits_{x \to \infty}1}{\lim\limits_{x \to \infty}1+\lim\limits_{x \to \infty}\frac{9}{x}}=\frac{1}{1+0}=1$
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