Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 2 - Section 2.2 - The Limit of a Function - 2.2 Exercises: 39

Answer

$\lim\limits_{x \to 2\pi^-}x\csc x=-\infty$

Work Step by Step

$\lim\limits_{x \to 2\pi^-}x\csc x=\lim\limits_{x \to 2\pi^-}\frac{x}{\sin x}=\frac{\lim\limits_{x \to 2\pi^-} x}{\lim\limits_{x \to 2\pi^-}\sin x}=\frac{2\pi^-}{0^-}=-\infty$
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