Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 8 - Sequences and Infinite Series - 8.2 Sequences - 8.2 Exercises - Page 616: 28

Answer

$-\ln (3)$

Work Step by Step

$\lim\limits_{n \to \infty} [\ln (n^3+1)-\ln (3n^3+10n)]=\lim\limits_{n \to \infty} \ln \dfrac{n^3+1}{3n^3+10n}$ or, $=\lim\limits_{n \to \infty} \ln \dfrac{1+1/n^3}{3+10/n^2}$ or, $=\ln \lim\limits_{n \to \infty} \dfrac{1+1/n^3}{3+10/n^2}$ or, $=\ln (\dfrac{1}{3})$ or, $=\ln (1)-\ln (3)$ or, $=0-\ln (3)$ or, $=-\ln (3)$
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