Answer
$2$
Work Step by Step
$$\lim_{x\to 0}\frac{\sin{2x}}{x}\\=\lim_{x\to 0}\frac{2\sin{x}\cos{x}}{x}\\=\lim_{x\to 0}\frac{\sin{x}}{x}2\cos{x}\\=\lim_{x\to 0}1\cdot2\cos{x}\\=2\cos{0}\\=2(1)=2.$$
Precalculus (10th Edition)
by Sullivan, Michael
Published by
Pearson
ISBN 10:
0-32197-907-9
ISBN 13:
978-0-32197-907-0
Chapter 14 - A Preview of Calculus: The Limit, Derivative, and Integral of a Function - 14.2 Algebra Techniques for Finding Limits - 14.2 Assess Your Understanding - Page 884: 54
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