University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 3 - Practice Exercises - Page 204: 128



Work Step by Step

$$A=\lim_{\theta\to0^+}\frac{1-2\cot^2\theta}{5\cot^2\theta-7\cot\theta-8}$$ Divide both numerator and denominator by $\cot^2\theta$: $$A=\lim_{\theta\to0^+}\frac{\frac{1}{\cot^2\theta}-2}{5-\frac{7}{\cot\theta}-\frac{8}{\cot^2\theta}}$$ As $\theta\to0^+$, we have $\sin\theta\to0$ while $\cos\theta\to1$, so $\cot\theta=\frac{\cos\theta}{\sin\theta}\to\infty$. Therefore, as $\theta\to0^+$, $\frac{1}{\cot^2\theta}\to0$ and $\frac{7}{\cot\theta}\to0$ and $\frac{8}{\cot^2\theta}\to0$. $$A=\frac{0-2}{5-0-0}=-\frac{2}{5}$$
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