Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 11 - Section 11.1 - Sequences - 11.1 Exercises - Page 704: 28

Answer

The sequence converges to 3.

Work Step by Step

Since the highest degree of $n$ in the numerator is 1/2 and the highest degree in the denominator is 1/2, the limit as $n$ approaches infinity will be that of $\frac{3\sqrt{n}}{\sqrt{n}} = 3$. Hence the sequence converges to 3.
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