University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 3 - Section 3.5 - Derivatives of Trigonometric Functions - Exercises - Page 152: 48


$$\lim_{x\to-\pi/6}\sqrt{1+\cos(\pi+\csc x)}=\sqrt{2}$$

Work Step by Step

$$A=\lim_{x\to-\pi/6}\sqrt{1+\cos(\pi\csc x)}=\lim_{x\to-\pi/6}\sqrt{1+\cos\Big(\frac{\pi}{\sin x}\Big)}$$ $$A=\sqrt{1+\cos\Big(\frac{\pi}{\sin(-\pi/6)}\Big)}$$ $$A=\sqrt{1+\cos\Big(\frac{\pi}{-\frac{1}{2}}\Big)}$$ $$A=\sqrt{1+\cos(-2\pi)}$$ $$A=\sqrt{1+1}=\sqrt2$$
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