University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson

Chapter 4 - Section 4.5 - Indeterminate Forms and L'Hôpital's Rule - Exercises - Page 249: 79

Answer

$\dfrac{27}{10}$

Work Step by Step

L'Hospital's rule states that $\lim\limits_{x \to \infty} f(x)=\lim\limits_{x \to \infty} \dfrac{A'(x)}{B'(x)}$ Here, $\lim\limits_{x \to 0} f(x)=f(0)$ 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 the limit, so we need to use L'Hospital's rule. $\lim\limits_{x \to 0} f(x)=\lim\limits_{x \to 0} \dfrac{9-9 \cos 3x}{15x^2}=\dfrac{0}{0}$ This shows an indeterminate form of the limit, so we need to use L'Hospital's rule. $\lim\limits_{x \to 0} f(x)=\lim\limits_{x \to 0} \dfrac{27 \sin 3x}{30 x}=\dfrac{0}{0}$ This shows an indeterminate form of the limit, so we need to use L'Hospital's rule. $\lim\limits_{x \to 0} f(x)=\lim\limits_{x \to 0} \dfrac{81 \cos 3x}{30 }=\dfrac{27}{10}$

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