Chapter 9.1, Problem 41E

To determine

To Show: That the given sequence converges to a limit, using Sandwich Theorem.

Answer to Problem 41E

The sequence converges to $0$ .

Explanation of Solution

Given Information: The $n^{th}$ term of the sequence is given by $a_n=\frac{\sin n}{n}$

Concept Used:Sandwich Theorem

If $f_1\left(x\right)\le f\left(x\right)\le f_2\left(x\right)$ in an interval $I$ and if $\underset{x\to x_0}{\lim }{f}_1\left(x\right)=\underset{x\to x_0}{\lim }{f}_2\left(x\right)=L$ , where $x_0\in I$ , then $\underset{x\to x_0}{\lim }f\left(x\right)=L$

The magnitude of $\sin \theta$ is always less than $1$ for any value of $\theta$

  $\Rightarrow -1\le \sin n\le 1$

Divide the above inequality by $n$ to write

  $-\frac{1}{n}\le \frac{\sin n}{n}\le \frac{1}{n}$

  $\Rightarrow \underset{n\to \infty }{\lim }\left(-\frac{1}{n}\right)\le \underset{\text{n}\to \infty }{\lim }\frac{\sin n}{n}\le \underset{n\to \infty }{\lim }\frac{1}{n}$

  $\Rightarrow 0\le \underset{n\to \infty }{\lim }\frac{\sin n}{n}\le 0$ (Since $\underset{n\to \infty }{\lim }\frac{-1}{n}=\underset{n\to \infty }{\lim }\frac{1}{n}=0$ )

  $\Rightarrow \underset{n\to \infty }{\lim }\frac{\sin n}{n}=0$

Therefore, the given sequence converges to the limit $0$ .