Chapter 8, Problem 1RWDT

To determine

To find: the limit $L=\underset{n\to \infty }{\lim }{a}_n$ and the value of $M$ such that $n>M$ guarantees that $L-0.01<a_n<L+0.01$ .

Answer to Problem 1RWDT

The value of the limit is $L=\underset{n\to \infty }{\lim }{a}_n=0$ and $M=100$ .

Explanation of Solution

Given information:

The sequence $a_n=\frac{\sin n}{n}$ .

Calculation:

Since $-1\le \sin n\le 1$ , then $-\frac{1}{n}\le \frac{\sin n}{n}\le \frac{1}{n}$ .

Now, since $\underset{n\to \infty }{\lim }\left(-\frac{1}{n}\right)=0$ and $\underset{n\to \infty }{\lim }\left(\frac{1}{n}\right)=0$ , by squeeze theorem $\underset{n\to \infty }{\lim }\left(\frac{\sin n}{n}\right)=0$ .

Therefore, $L=\underset{n\to \infty }{\lim }{a}_n=\underset{n\to \infty }{\lim }\left(\frac{\sin n}{n}\right)=0$ .

Since $\left|\sin n\right|\le 1$ , then $\left|\frac{\sin n}{n}\right|\le \frac{1}{n}$ .

For $n=100$ , $\left|\frac{\sin n}{n}\right|\le 0.01$ . That means, $-0.01\le \frac{\sin n}{n}\le 0.01$ .

As the value of $n$ increases, $\left|\frac{\sin n}{n}\right|$ tends to 0. Thus, $M=100$ guarantees that $-0.01<\frac{\sin n}{n}<0.01$ or $-0.01<a_n<0.01$ or $L-0.01<a_n<L+0.01$ .

Conclusion:

The value of the limit is $L=\underset{n\to \infty }{\lim }{a}_n=0$ and $M=100$ .