University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 2 - Practice Exercises - Page 111: 38

Answer

$g(\theta)$ has a continuous extension at $\theta=\pi/2$ and can be extended to include the value $g(\pi/2)=-1.25$
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Work Step by Step

$$g(\theta)=\frac{5\cos\theta}{4\theta-2\pi}$$ The graph is enclosed below. Condition to have a continuous extension for $g(\theta)$ at $\theta=c$: $\lim_{\theta\to c}g(\theta)$ must exist. Here, $\lim_{\theta\to\pi/2}g(\theta)$ does seem to exist, because as $\theta\to\pi/2$ from the left and the right, $g(\theta)$ both approaches a value of $-1.25$, meaing that $\lim_{\theta\to\pi/2}g(\theta)=-1.25$ So function $g(\theta)$ can be extended to include the value $g(\pi/2)=-1.25$ so that $\lim_{\theta\to\pi/2}g(\theta)=g(\pi/2)$ and $g$ would then be continuous at $\theta=\pi/2$.
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