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 410: 79



Work Step by Step

$\lim\limits_{x \to 0} f(x)=f(0)$ The function will be defined when $c=f(0)$ Thus, $\lim\limits_{x \to 0} f(x)=\lim\limits_{x \to 0} \dfrac{9x-3 \sin 3x}{5x^3}=\dfrac{0}{0}$ This shows an indeterminate form of limit, thus we will apply L-Hospital's rule such as: $\lim\limits_{x \to \infty} f(x)=\lim\limits_{x \to \infty} \dfrac{p'(x)}{q'(x)}$ $\lim\limits_{x \to 0} \dfrac{9-9 \cos 3x}{15x^2}=\dfrac{0}{0}$ Again apply L-Hospital's rule. $\lim\limits_{x \to 0} \dfrac{27 \sin 3x}{30 x}=\dfrac{0}{0}$ Again apply L-Hospital's rule. $\lim\limits_{x \to 0} \dfrac{81 \cos 3x}{30 }=\dfrac{27}{10}$ Thus, $\lim\limits_{x \to 0} \dfrac{81 \cos (0)}{30 }=\dfrac{27}{10}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.