Calculus (3rd Edition)

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

Chapter 2 - Limits - 2.2 Limits: A Numerical and Graphical Approach - Exercises - Page 54: 21



Work Step by Step

$\lim\limits_{x \to 0}\frac{sin2x}{x}$= $\lim\limits_{x \to 0}[\frac{sin2x}{2x}.2]$ = 2.$\lim\limits_{2x \to 0}[\frac{sin2x}{2x}]$ (as x tends to 0, 2x also tends to 0) =2•1 (As $ \lim\limits_{x \to 0}\frac{sinx}{x}=1$) = 2
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