Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Section 3.3 - Derivatives of Trigonometric Functions - 3.3 Exercises: 49

Answer

$$\lim\limits_{x\to\pi/4}\frac{1-\tan x}{\sin x-\cos x}=-\sqrt 2$$

Work Step by Step

$$A=\lim\limits_{x\to\pi/4}\frac{1-\tan x}{\sin x-\cos x}$$ $$A=\lim\limits_{x\to\pi/4}\frac{1-\frac{\sin x}{\cos x}}{\sin x-\cos x}$$ $$A=\lim\limits_{x\to\pi/4}\frac{\frac{\cos x-\sin x}{\cos x}}{\sin x-\cos x}$$ $$A=\lim\limits_{x\to\pi/4}\frac{\cos x-\sin x}{\cos x(\sin x-\cos x)}$$ $$A=\lim\limits_{x\to\pi/4}\frac{-(\sin x-\cos x)}{\cos x(\sin x-\cos x)}$$ $$A=\lim\limits_{x\to\pi/4}\frac{-1}{\cos x}$$ $$A=\frac{-1}{\cos(\pi/4)}$$ $$A=\frac{-1}{\frac{\sqrt 2}{2}}$$ $$A=\frac{-2}{\sqrt 2}=-\sqrt 2$$
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