University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 2 - Section 2.4 - One-Sided Limits - Exercises - Page 85: 41

Answer

$$\lim_{\theta\to0}\frac{\tan\theta}{\theta^2\cot3\theta}=3$$

Work Step by Step

$$A=\lim_{\theta\to0}\frac{\tan\theta}{\theta^2\cot3\theta}=\lim_{\theta\to0}\frac{\frac{\sin\theta}{\cos\theta}}{\theta^2\frac{\cos3\theta}{\sin3\theta}}=\lim_{\theta\to0}\frac{\sin\theta\sin3\theta}{\theta^2\cos\theta\cos3\theta}$$ $$A=\lim_{\theta\to0}\frac{\sin\theta}{\theta}\times\lim_{\theta\to0}\frac{\sin3\theta}{\theta}\times\lim_{\theta\to0}\frac{1}{\cos\theta\cos3\theta}$$ $$A=X\times Y\times Z$$ 1) Solve $X$: $$X=\lim_{\theta\to0}\frac{\sin\theta}{\theta}$$ Appy Theorem 7: $$X=1$$ 2) Solve $Y$: $$Y=\lim_{\theta\to0}\frac{\sin3\theta}{\theta}$$ Multiply both numerator and denominator by $3$ $$Y=3\lim_{\theta\to0}\frac{\sin3\theta}{3\theta}$$ Apply Theorem 7 with $3\theta$: $$Y=3\times1=3$$ 3) Solve $Z$: $$Z=\lim_{\theta\to0}\frac{1}{\cos\theta\cos3\theta}=\frac{1}{\cos0\cos0}=\frac{1}{1\times1}=1$$ Therefore, $$A=X\times Y\times Z=1\times3\times1=3$$
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