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 7.5 - Indeterminate Forms and L'Hopital's Rule - Exercises 7.5 - Page 409: 42



Work Step by Step

Here, we have $\lim\limits_{x \to 0} f(0)=\lim\limits_{x \to 0}\dfrac{1-\cos x-\cos x \sin x}{\sin 0}=\dfrac{0}{0}$ This shows an indeterminate form of limit, thus we will apply L-Hospital's rule such as: $\lim\limits_{x \to \infty} f(x)=\lim\limits_{x \to \infty} \dfrac{p'(x)}{q'(x)}$ $\lim\limits_{x \to 0} \dfrac{\sin x+\cos 2x}{\cos x}=\dfrac{\sin 0+\cos 0}{\cos 0}=1$ or, $\dfrac{0+1}{1}=1$
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