Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 11 - Infinite Series - Chapter Review Exercises - Page 591: 9



Work Step by Step

Since $$a_{n}=\sqrt{n+5}-\sqrt{n+2}$$ Then \begin{align*} \lim _{n \rightarrow \infty} a_{n}&=\lim _{n \rightarrow \infty} \sqrt{n+5}-\sqrt{n+2} \\ &= \lim _{n \rightarrow \infty} (\sqrt{n+5}-\sqrt{n+2})\frac{(\sqrt{n+5}+\sqrt{n+2})}{\sqrt{n+5}+\sqrt{n+2}}\\ &=\lim _{n \rightarrow \infty}\frac{(n+5)-(n+2)}{\sqrt{n+5}+\sqrt{n+2}}=\frac{3}{\sqrt{n+5}+\sqrt{n+2}} \\ &=\lim _{n \rightarrow \infty} \frac{3}{\sqrt{n+5}+\sqrt{n+2}}\\ &=0 \end{align*}
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