University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 3 - Section 3.9 - Inverse Trigonometric Functions - Exercises - Page 181: 13

Answer

$\displaystyle \frac{\pi}{2}$

Work Step by Step

$y=\sin^{-1}x$ is the number in $[-\pi/2, \pi/2]$ for which $\sin y=x.$ As the value of sine approaches 1 (from the left, because it can't approach it from the right - sine is never greater than 1), the angle we approach (in radians) is $\displaystyle \frac{\pi}{2}$. Alternatively (if in doubt), we can reach the same conclusion by observing the graph of $y=\sin^{-1}x$ (also written as $\arcsin x$) in the vicinity of x=1. See below.
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