Calculus 10th Edition

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

Chapter 1 - Limits and Their Properties - 1.3 Exercises - Page 69: 116

Answer

False.

Work Step by Step

The questions specifies $x$ tending to $\pi$ not $0.$ The limit can be evaluated by plugging in $x=\pi$ since we don't get an indeterminate form. $\lim\limits_{x\to\pi}\dfrac{\sin{x}}{x}=\dfrac{\sin{\pi}}{\pi}=0.$
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