Answer
$\frac{1}{2}$
Work Step by Step
$\lim_{r\to0}\frac{sin(r)}{tan2r}=\lim_{r\to0}\frac{sin(r)cos2r}{sin2r}=\lim_{r\to0}\frac{sin(r)cos2r}{2sin(r)cos(r)}=\lim_{r\to0}\frac{cos2r}{2cos(r)}=\frac{1}{2}$
Thomas' Calculus 13th Edition
by Thomas Jr., George B.
Published by
Pearson
ISBN 10:
0-32187-896-5
ISBN 13:
978-0-32187-896-0
Chapter 3: Derivatives - Practice Exercises - Page 180: 97
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