Chapter 1.2, Problem 68E

To determine

To explain: The given statement.

Answer to Problem 68E

This is not the case of $f(x)=\sin x$ . $f(x)$ will consistently fluctuate between -1 and 1.

And the distance between $f(x)$ and $x=0$ will always be in the interval $(0,1)$ .

There is no such thing as a limit.

Explanation of Solution

Given information: No value $L$ for that $\underset{x\to \infty }{\lim }f(x)=L$.

Graph:

Assume that when $x$ increases, the distance between $f(x)$ and $L$ will decrease.

The graph of $f(x)=\frac{\sin x}{x}$ ,there exists a limit at $y=0$ is shown below:

  

The graph oscillates, but the distance from the $x$ -axis gets less and smaller.

This is not the case of $f(x)=\sin x$ . $f(x)$ will consistently fluctuate between -1 and 1.

And the distance between $f(x)$ and $x=0$ will always be in the interval $(0,1)$ .

Hence, there is no such thing as a limit.