Chapter 2.2, Problem 11E

To determine

To find: The value of $\underset{x\to -\infty }{\lim }\left(\frac{\sin x}{x}\right)$ and confirm the answer with the help of Sandwich theorem.

Answer to Problem 11E

The value of $\underset{x\to -\infty }{\lim }\left(\frac{\sin x}{x}\right)$ is $0$ .

Explanation of Solution

Given information:The limit is $\underset{x\to -\infty }{\lim }\left(\frac{\sin x}{x}\right)$ .

Calculation:

To graph a function $y=\frac{\sin x}{x}$ , follow the steps using graphing calculator.

First press “ON” button on graphical calculator, press $\boxed{\text{Y}=}$ key and enter right hand side of the equation $y=\frac{\sin x}{x}$ after the symbol ${\text{Y}}_1$ . Enter the keystrokes $\boxed{(}\boxed{\text{SIN}}\boxed{\text{X}}\boxed{)}\boxed{)}\boxed{÷}\boxed{\text{X}}$ .

The display will show the equation,

  ${\text{Y}}_{\text{1}}=\left(\sin \left(\text{X}\right)\right)/\left(\text{X}\right)$

Press the window key and adjust the window to $\left[-10,10\right]$ by $\left[-5,5\right]$ .Now, press the $\boxed{\text{GRAPH}}$ key and $\boxed{\text{TRACE}}$ key to produce the graph of given function in standard window as shown in below.

  

Figure (1)

As observed from graph, the function $f\left(x\right)=\frac{\sin x}{x}$ tends to $0$ as $x$ tends to $-\infty$ .So, the value of $\underset{x\to -\infty }{\lim }\left(\frac{\sin x}{x}\right)$ is $0$ .

Check the answer by sandwich theorem.

It is known that thesine function lies between $-1$ and $1$ . So,

  $-1\le \sin x\le 1$

Divide all the sides of inequality by $x$ .

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

Now calculate the limits.

  $\underset{x\to -\infty }{\lim }\left(-\frac{1}{x}\right)=0$ and $\underset{x\to -\infty }{\lim }\left(\frac{1}{x}\right)=0$

So,

  $\begin{array}{c}\underset{x\to -\infty }{\lim }0\le \underset{x\to -\infty }{\lim }\frac{\sin x}{x}\le \underset{x\to -\infty }{\lim }\frac{1}{x}\\ 0\le \underset{x\to -\infty }{\lim }\frac{\sin x}{x}\le 0\end{array}$

Therefore,the value of $\underset{x\to -\infty }{\lim }\left(\frac{\sin x}{x}\right)$ is $0$ .