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: 40



Work Step by Step

Here, we have $\lim\limits_{x \to 0} f(0)=\dfrac{0}{0}$ This shows an indeterminate form of limit, thus we will apply L-Hospital's rule such as: $\lim\limits_{x \to \infty} f(x)=\lim\limits_{x \to \infty} \dfrac{p'(x)}{q'(x)}$ $\lim\limits_{x \to 0} \dfrac{3 \sin x+(3x+1)(cos x)-1}{\sin x+x\cos x}=\dfrac{0}{0}$ Now, again use L-Hospital's rule. $\lim\limits_{x \to 0} \dfrac{6 \cos x-(3x+1)(\sin x)}{2\cos x-x \sin x}=\dfrac{6-0}{2-0}$ or, $\dfrac{6}{2}=3$
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