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: 27



Work Step by Step

Using Theorem one, we have $$\lim\limits_{n \to \infty}a_n=\lim\limits_{n \to \infty}{4+\frac{1}{n}}=4.$$ Now, since $\sqrt{x}$ is continuous for all $x\gt 0$, then by Theorem 4, we have $$\lim_{n\to \infty}\sqrt{4+\frac{1}{n}}=\sqrt{ \lim_{n\to \infty}(4+\frac{1}{n})}=\sqrt{4}=2.$$
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