Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 2: Limits and Continuity - Section 2.6 - Limits Involving Infinity; Asymptotes of Graphs - Exercises 2.6 - Page 98: 50

Answer

$+\infty$

Work Step by Step

$\displaystyle \sec x=\frac{1}{\cos x}$ You can sketch a unit circle. 0 radians is at (1,0)$;$ the positive direction is counterclockwise; the negative direction is clockwise$;$ $-\displaystyle \frac{\pi}{2}$ radians is positioned at $(0,-1).$ When $x\displaystyle \rightarrow(-\frac{\pi}{2})^{+}$, x is a radian angle measure in the fourth quadrant, (slightly more than $-\displaystyle \frac{\pi}{2}$ means on the counterclockwise side of $-\displaystyle \frac{\pi}{2}$) where cosine is positive. The numerator, $\rightarrow 1$, (positive) and the denominator, $\cos x\rightarrow 0$, (positive). Thus, $\displaystyle \sec x=\frac{1}{\cos x}\rightarrow+\infty$
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