Calculus (3rd Edition)

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

Chapter 2 - Limits - 2.6 Trigonometric Limits - Exercises - Page 77: 30

Answer

1

Work Step by Step

We set $\theta=4h$. As $h\rightarrow0$, $\theta$ which is the multiple of $h$ also tends to $0$. Therefore, we can write $\lim\limits_{h \to 0}\frac{\sin 4h}{4h}=\lim\limits_{\theta \to 0}\frac{\sin \theta}{\theta}=1$ (from theorem 2)
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