Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 11 - Infinite Series - 11.1 Sequences - Exercises - Page 537: 29



Work Step by Step

Using Theorem 1, we have $$ \lim\limits_{n \to \infty}{\frac{n^3}{2n^3+1}}=\frac{1}{2}.$$ Now, since $\cos^{-1}{x}$ is continuous, then by Theorem 4, we have $$\lim_{n\to \infty}\cos^{-1}{\frac{n^3}{2n^3+1}}= \cos^{-1}{ \lim_{n\to \infty}\frac{n^3}{2n^3+1}}=\cos^{-1}\frac{1}{2}=\frac{\pi}{3}.$$
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