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: 15


$-16 $

Work Step by Step

Consider: $\lim\limits_{x \to 0}f(x)=\lim\limits_{x \to 0}\dfrac{8x^2}{\cos x-1}$ Need to check that the limit has an indeterminate form. Thus, $f(0)=\dfrac{8(0)}{\cos 0 -1}=\dfrac{0}{0}$ Now, apply L-Hospital's rule such as: $\lim\limits_{a \to b}f(x)=\lim\limits_{a \to b}\dfrac{g'(x)}{h'(x)}$ This implies: $\lim\limits_{x \to 0}\dfrac{16x}{-\sin x}=\dfrac{0}{0} $ Again apply L-Hospital's rule . we get $\lim\limits_{x \to 0}\dfrac{16}{-\cos x}=\dfrac{16}{-\cos 0}=-16 $
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