Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 13 - Review - Exercises - Page 941: 7

Answer

(a) See explanations. (b) $0$

Work Step by Step

(a) $\lim_{n\to\infty}a_n=L$ means that the sequence will get arbitrarily close to $L$ with sufficiently larger $n$, and quantity $L$ is the limit of this convergent sequence. In other words, a convergent sequence will have a limit when the number $n$ approaches infinity. (b) As $\lim_{n\to\infty}\frac{1}{n}=0$, and the term $(-1)^n$ will only change the sign of the result ($\pm0$), thus we have $\lim_{n\to\infty}\frac{(-1)^n}{n}=0$
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