Chapter 2.1, Problem 32E

To determine

To find: The value of $\underset{x\to 0}{\lim }\left(\frac{x+\sin x}{x}\right)$ with the help of graph and check the results algebraically.

Answer to Problem 32E

The value of $\underset{x\to 0}{\lim }\left(\frac{x+\sin x}{x}\right)$ is $2$ .

Explanation of Solution

Given information: The limit is $\underset{x\to 0}{\lim }\left(\frac{x+\sin x}{x}\right)$ .

Calculation:

Let $y=\frac{x+\sin x}{x}$ to find the value of $\underset{x\to 0}{\lim }\left(\frac{x+\sin x}{x}\right)$ graphically.

To graph a function $y=\frac{x+\sin x}{x}$ , 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 $y=\frac{x+\sin x}{x}$ after the symbol ${\text{Y}}_1$ . Enter the keystrokes $\boxed{(}\boxed{\text{X}}\boxed{+}\boxed{\text{SIN}}\boxed{\text{X}}\boxed{)}\boxed{÷}\boxed{\text{X}}$ .

The display will show the equation,

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

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)

The function $\frac{x+\sin x}{x}$ tends to $2$ as $x$ tends to $0$ from both the sides. So, the value of $\underset{x\to 0}{\lim }\left(\frac{x+\sin x}{x}\right)$ graphically is $2$ .

Check the limit algebraically. Use the property of limit.

  $\underset{x\to 0}{\lim }\left(\frac{\sin \left(ax\right)}{ax}\right)=1$

Simplify the limit.

  $\begin{array}{c}\underset{x\to 0}{\lim }\left(\frac{x+\sin x}{x}\right)=\underset{x\to 0}{\lim }\left(\frac{x}{x}+\frac{\sin x}{x}\right)\\ =\underset{x\to 0}{\lim }1+\underset{x\to 0}{\lim }\left(\frac{\sin x}{x}\right)\\ =1+1\\ =2\end{array}$

So, the value of $\underset{x\to 0}{\lim }\left(\frac{x+\sin x}{x}\right)$ is equal to $2$ .

Therefore, the value of $\underset{x\to 0}{\lim }\left(\frac{x+\sin x}{x}\right)$ is $2$ .