Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 9 - Infinite Series - 9.1 Sequences - Exercises Set 9.1 - Page 606: 35

Answer

The limit of the given sequence is $0$.

Work Step by Step

Step 1: Intuitively speaking, the given function is a limit of sine function and a polynomial function of degree $1$. We know that sine function remains bounded throughout the domain while the polynomial function would increase for increasing values of $n$. Thus, the denominator will dominate and the limit must be zero. Step 2: Let the given sequence be represented as ${c_n}=\frac{\sin^2 n}{n}$. Consider two sequences as follows: ${a_n}=0$ and ${b_n}=\frac{1}{n}$. Note that for each $n$, $0 \leq \frac{\sin^2 n}{n} \leq \frac{1}{n}$ holds. Conditions for Squeezing Theorem are satisfied. The limit of both sequences, ${a_n}$ and ${b_n}$ is $0$. Thus, the limit of ${c_n}$ must also be $0$. Result: The limit of the given sequence is $0$.
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