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: 30

Answer

$\lim\limits_{x \to 2}\sin\dfrac{\pi x}{2}=0$

Work Step by Step

$\lim\limits_{x \to 2}\sin\dfrac{\pi x}{2}$ Apply direct substitution to evaluate the limit: $\lim\limits_{x \to 2}\sin\dfrac{\pi x}{2}=\sin\dfrac{\pi(2)}{2}=\sin\pi=0$
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