University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 9 - Section 9.10 - The Binomial Series and Applications of Taylor Series - Practice Exercises - Page 553: 79


$ r=-3, s=\frac{9}{2}$

Work Step by Step

The Taylor series for $\sin x $ can be defined as: $\sin x= x-\dfrac{x^3}{3!}+\dfrac{ x^5}{5!}-....$ Now, $\lim\limits_{x \to 0} (\dfrac{\sin 3x}{x^3}+\dfrac{r}{x^2}+s)=0$ So, $\lim\limits_{x \to 0} [\dfrac{x(3-\dfrac{(3x)^3}{3!}+....)}{x^3}+\dfrac{r}{x^2}+s]=0$ or, $=\lim\limits_{x \to 0} [\dfrac{3}{x^2}-\dfrac{9}{2}+\dfrac{81 x^2}{40}+......\dfrac{r}{x^2}+s]=0$ Now, $\dfrac{3}{x^2}+\dfrac{r}{x^2}=0 \implies r=-3$ and $ s-\dfrac{9}{2}=0 \implies s=\frac{9}{2}$ So, $ r=-3, s=\frac{9}{2}$
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