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: 127



Work Step by Step

$$A=\lim_{\theta\to(\pi/2)^-}\frac{4\tan^2\theta+\tan\theta+1}{\tan^2\theta+5}$$ Divide both numerator and denominator by $\tan^2\theta$: $$A=\lim_{\theta\to(\pi/2)^-}\frac{4+\frac{1}{\tan\theta}+\frac{1}{\tan^2\theta}}{1+\frac{5}{\tan^2\theta}}$$ As $\theta\to(\pi/2)^-$, we have $\sin\theta\to1$ while $\cos\theta\to0$, so $\tan\theta=\frac{\sin\theta}{\cos\theta}\to\infty$. Therefore, as $\theta\to(\pi/2)^-$, $\frac{1}{\tan\theta}\to0$ and $\frac{1}{\tan^2\theta}\to0$ and $\frac{5}{\tan^2\theta}\to0$. $$A=\frac{4+0+0}{1+0}=4$$
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