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: 28



Work Step by Step

Using Theorem one, we have $$ \lim\limits_{n \to \infty}{\frac{4n}{3n+9}}=\frac{4}{3}.$$ Now, since $e^{x}$ is continuous for all $x$, then by Theorem 4, we have $$\lim_{n\to \infty}e^{\frac{4n}{3n+9}}=e^{ \lim_{n\to \infty}\frac{4n}{3n+9}}=e^{4/3}.$$
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