Thomas' Calculus 13th Edition

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

Chapter 7: Transcendental Functions - Section 7.6 - Inverse Trigonometric Functions - Exercises 7.6 - Page 421: 20

Answer

$0$

Work Step by Step

$y=\csc^{-1}x$ is the number in $[-\pi/2,0) \cup(0, \pi/2]$ for which $\csc y=x$ $($In terms of sine, $\displaystyle \sin y=\frac{1}{x}$.) We want an angle for which sine approaches 0 from the left, so the cosecant $\rightarrow\infty$. This would be $0$, as we observe $[-\pi/2,0) \cup(0, \pi/2]$. Alternatively, we can reach the same conclusion by observing the graph of $y=\csc x$ (also written as $\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{c}\mathrm{s}\mathrm{c} x$) when $ x\rightarrow-\infty$. See below.
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