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 83: 41



Work Step by Step

Since $\lim\limits_{x \to \infty}\frac{x^{2}+9}{9-x}=\lim\limits_{x \to \infty}-\frac{x^{2}}{x}=\lim\limits_{x \to \infty}-x=-\infty$ And $\lim\limits_{t \to -\infty}tan^{-1}(t)=-\frac{\pi}{2}$ Then $\lim\limits_{x \to \infty} tan^{-1}(\frac{x^{2}+9}{9-x} )=-\frac{\pi}{2}$
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