Chapter 2.1, Problem 29E

To determine

To find: The value of $\underset{x\to 0}{\lim }\left(\frac{{\left(2+x\right)}^3-8}{x}\right)$ with the help of graph and check the results algebraically.

Answer to Problem 29E

The value of $\underset{x\to 0}{\lim }\left(\frac{{\left(2+x\right)}^3-8}{x}\right)$ is $12$ .

Explanation of Solution

Given information: The limit is $\underset{x\to 0}{\lim }\left(\frac{{\left(2+x\right)}^3-8}{x}\right)$ .

Calculation:

Let $y=\frac{{\left(2+x\right)}^3-8}{x}$ to find the value of $\underset{x\to 0}{\lim }\left(\frac{{\left(2+x\right)}^3-8}{x}\right)$ graphically.

To graph a function $y=\frac{{\left(2+x\right)}^3-8}{x}$ , 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 $y=\frac{{\left(2+x\right)}^3-8}{x}$ after the symbol ${\text{Y}}_1$ . Enter the keystrokes $\boxed{(}\boxed{(}\boxed{2}\boxed{+}\boxed{\text{X}}\boxed{)}\boxed{^}\boxed{3}\boxed{-}\boxed{8}\boxed{)}\boxed{÷}\boxed{\text{X}}$ .

The display will show the equation,

  ${\text{Y}}_{\text{1}}=\left(\left(2+\text{X}\right)^3-8\right)/\text{X}$

Now, press the $\boxed{\text{GRAPH}}$ key and $\boxed{\text{TRACE}}$ key to produce the graph of given function in standard window as shown below.

  

Figure (1)

The function $\frac{{\left(2+x\right)}^3-8}{x}$ tends to $12$ as $x$ tends to $0$ from both the sides. So, the value of $\underset{x\to 0}{\lim }\left(\frac{{\left(2+x\right)}^3-8}{x}\right)$ graphically is $12$ .

Check the limit algebraically.

  $\begin{array}{c}\underset{x\to 0}{\lim }\left(\frac{{\left(2+x\right)}^3-8}{x}\right)=\underset{x\to 0}{\lim }\left(\frac{8+x^3+6x^2+12x-8}{x}\right)\\ =\underset{x\to 0}{\lim }\left(\frac{x\left(x^2+6x+12\right)}{x}\right)\\ =\underset{x\to 0}{\lim }\left(x^2+6x+12\right)\end{array}$

Further simplify.

  $\begin{array}{c}\underset{x\to 0}{\lim }\left(\frac{{\left(2+x\right)}^3-8}{x}\right)=0+6\times 0+12\\ =12\end{array}$

So, the value of $\underset{x\to 0}{\lim }\left(\frac{{\left(2+x\right)}^3-8}{x}\right)$ is equal to $12$ .

Therefore, the value of $\underset{x\to 0}{\lim }\left(\frac{{\left(2+x\right)}^3-8}{x}\right)$ is $12$ .