Chapter 3.3, Problem 46E

a.

To determine

Find the value of thegiven limit .

Answer to Problem 46E

The $\underset{x\to \infty }{\lim }x\sin \frac{1}{x}=1$ .

Explanation of Solution

Given:

The given limitis $\underset{x\to \infty }{\lim }x\sin \frac{1}{x}$ .

Calculation:

  $\underset{x\to \infty }{\lim }x\sin \frac{1}{x}$

Let $x=\frac{1}{t}$ .

If $\frac{1}{t}\to \infty$ , then $t\to 0$ .

  $\begin{array}{l}\underset{x\to \infty }{\lim }x\sin \frac{1}{x}\\ =\underset{t\to 0}{\lim }\frac{1}{t}\sin t\end{array}$

Use limit formula $\underset{x\to 0}{\lim }\frac{\sin x}{x}=1$ .

  $=1$

Hence the value of the $\underset{x\to \infty }{\lim }x\sin \frac{1}{x}=1$ .

b.

To determine

Find the value of the given limit .

Answer to Problem 46E

The $\underset{x\to 0}{\lim }x\sin \frac{1}{x}=0$ .

Explanation of Solution

Given:

The given limitis $\underset{x\to 0}{\lim }x\sin \frac{1}{x}$ .

Calculation:

  $\underset{x\to 0}{\lim }x\sin \frac{1}{x}$

Let $x=\frac{1}{t}$ .

If $\frac{1}{t}\to 0$ , then $t\to \infty$ .

  $\begin{array}{l}\underset{x\to 0}{\lim }x\sin \frac{1}{x}\\ =\underset{t\to \infty }{\lim }\frac{1}{t}\sin t\end{array}$

From Squeeze theorem $\underset{x\to \infty }{\lim }\frac{\sin x}{x}=0$ .

  $=0$

Hence the value of the $\underset{x\to 0}{\lim }x\sin \frac{1}{x}=0$ .

c.

To determine

Find the value of the given limit in part (a) and (b) using graph.

Explanation of Solution

Given:

The given limitsare $\underset{x\to \infty }{\lim }x\sin \frac{1}{x}$ and $\underset{x\to 0}{\lim }x\sin \frac{1}{x}$ .

Calculation:

Graph of the function $y=x\sin \left(\frac{1}{x}\right)$ is given below.

  

From the graph,

When $x\to \infty$ , then the value of the $\underset{x\to \infty }{\lim }x\sin \frac{1}{x}=1$ .

When $x\to 0$ , then the value of the $\underset{x\to 0}{\lim }x\sin \frac{1}{x}=0$ .