Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 2 - Derivatives - 2.4 Derivatives of Trigonometric Functions - 2.4 Exercises - Page 151: 45


$$\lim _{\theta \rightarrow 0} \frac{\sin \theta}{\theta+\tan \theta}=\frac{1}{2} $$

Work Step by Step

Given $$\lim _{\theta \rightarrow 0} \frac{\sin \theta}{\theta+\tan \theta}$$ by dividing numerator and denominator by $\theta$, we get \begin{align} \lim _{\theta \rightarrow 0} \frac{\sin \theta}{\theta+\tan \theta}&=\lim _{\theta \rightarrow 0} \frac{\frac{\sin \theta}{\theta}}{1+\frac{\sin \theta}{\theta} \cdot \frac{1}{\cos \theta}}\\ &=\frac{\lim _{\theta \rightarrow 0} \frac{\sin \theta}{\theta}}{1+\lim _{\theta \rightarrow 0} \frac{\sin \theta}{\theta} \lim _{\theta \rightarrow 0} \frac{1}{\cos \theta}}\\ &=\frac{1}{1+1 \cdot 1}=\frac{1}{2}, \ \ \ (since \lim _{\theta \rightarrow 0} \frac{\sin \theta}{\theta}=1 ) \end{align}
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