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: 13

Answer

$\lim\limits_{n \to \infty} \frac{3^{n+1}+3}{3^n} =3$

Work Step by Step

$\lim\limits_{n \to \infty} \frac{3^{n+1}+3}{3^n} = \lim\limits_{n \to \infty} \frac{\frac{3^{n+1}}{3^n}+\frac{3}{3^n}}{\frac{3^n}{3^n}} =\lim\limits_{n \to \infty} \frac{3+\frac{3}{3^n}}{1} = \frac{3+0}{1} = 3$
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