Chapter 2.2, Problem 4E

(a)

To determine

To find: The value of $\underset{x\to \infty }{\lim }f\left(x\right)$ for $f\left(x\right)=\frac{3x^3-x+1}{x+3}$ with the help of graph and table.

Answer to Problem 4E

The value of $\underset{x\to \infty }{\lim }f\left(x\right)$ for $f\left(x\right)=\frac{3x^3-x+1}{x+3}$ is $\infty$ .

Explanation of Solution

Given information:The function is $f\left(x\right)=\frac{3x^3-x+1}{x+3}$ .

Calculation:

Find the limit graphically.To graph a function $y=\frac{3x^3-x+1}{x+3}$ , 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 $y=\frac{3x^3-x+1}{x+3}$ after the symbol ${\text{Y}}_1$ . Enter the keystrokes $\boxed{(}\boxed{3}\boxed{\text{X}}\boxed{^}\boxed{3}\boxed{-}\boxed{\text{X}}\boxed{+}\boxed{1}\boxed{)}\boxed{÷}\boxed{(}\boxed{\text{X}}\boxed{+}\boxed{3}\boxed{)}$ .

The display will show the equation,

  ${\text{Y}}_{\text{1}}=\left(3\text{X}^3-\text{X}+1\right)/\left(\text{X}+3\right)$

Press the window key and adjust the window to $\left[-10,10\right]$ by $\left[-100,300\right]$ .Now, press the $\boxed{\text{GRAPH}}$ key and $\boxed{\text{TRACE}}$ key to produce the graph of given function in standard window as shown below.

  

Figure (1)

As observed from graph, the function $f\left(x\right)=\frac{3x^3-x+1}{x+3}$ tends to $\infty$ as $x$ tends to $\infty$ .

Find the limit by table.To make the table of function $y=\frac{3x^3-x+1}{x+3}$ , 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 $y=\frac{3x^3-x+1}{x+3}$ after the symbol ${\text{Y}}_1$ . Enter the keystrokes $\boxed{(}\boxed{3}\boxed{\text{X}}\boxed{^}\boxed{3}\boxed{-}\boxed{\text{X}}\boxed{+}\boxed{1}\boxed{)}\boxed{÷}\boxed{(}\boxed{\text{X}}\boxed{+}\boxed{3}\boxed{)}$ .

First set the Table setup, enter the keystrokes $\boxed{2\text{nd}}\boxed{\text{WINDOW}}$ , then set to independent variable to ask.Now draw the table, enter the keystrokes $\boxed{2\text{nd}}\boxed{\text{GRAPH}}$ , and then enter the values of $\text{X}$ to $100$ , $200$ , $300$ , $400$ , $-100$ , $-200$ and $-300$ . The table is shown in window.

  

Figure (2)

As observed from table the value of $\text{Y}$ approaches to $0$ as $x$ approaches to infinity.

Therefore, the value of $\underset{x\to \infty }{\lim }f\left(x\right)$ for $f\left(x\right)=\frac{3x^3-x+1}{x+3}$ is $\infty$ .

(b)

To determine

To find:The value of $\underset{x\to -\infty }{\lim }f\left(x\right)$ for $f\left(x\right)=\frac{3x^3-x+1}{x+3}$ with the help of graph and table.

Answer to Problem 4E

The value of $\underset{x\to -\infty }{\lim }f\left(x\right)$ for $f\left(x\right)=\frac{3x^3-x+1}{x+3}$ is $\infty$ .

Explanation of Solution

Given information:The function is $f\left(x\right)=\frac{3x^3-x+1}{x+3}$ .

Calculation:

To find the limit of the function $f\left(x\right)=\frac{3x^3-x+1}{x+3}$ with the help of the graph as $x$ tends to $-\infty$ , consider Figure (1) drawn in part (a). From the graph it can be observed that the function tends to $\infty$ as $x$ tends to $-\infty$ .

To find the limit of the function $f\left(x\right)=\cos \left(\frac{1}{x}\right)$ with the help of the table as $x$ tends to $-\infty$ , consider Figure (2) drawn in part (a). From the table it can be observed that the function approaches to $\infty$ as $x$ approaches to a big value on the left side of number line or to $-\infty$ .

Therefore, the value of $\underset{x\to -\infty }{\lim }f\left(x\right)$ for $f\left(x\right)=\frac{3x^3-x+1}{x+3}$ is $\infty$ .

(c)

To determine

To find:The horizontal asymptote to the function $f\left(x\right)=\frac{3x^3-x+1}{x+3}$ .

Answer to Problem 4E

There is no horizontal asymptote to the function $f\left(x\right)=\frac{3x^3-x+1}{x+3}$ .

Explanation of Solution

Given information:The function is $f\left(x\right)=\frac{3x^3-x+1}{x+3}$ .

Calculation:

The horizontal asymptote of the graph of a function $y=f\left(x\right)$ is the line $y=a$ if $\underset{x\to \infty }{\lim }f\left(x\right)=a$ or $\underset{x\to -\infty }{\lim }f\left(x\right)=a$ .

From part (a) and (b), it can be observed that the value of both left hand and right hand limit of the function is not any real number. So, horizontal asymptote does not exist.

Therefore, there is no horizontal asymptote to the function $f\left(x\right)=\frac{3x^3-x+1}{x+3}$ .