Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 1 - Limits and Their Properties - 1.3 Exercises: 72

Answer

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

Work Step by Step

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