Chapter 11.2, Problem 8E

a.

To determine

To find: The value of $\underset{x\to 0}{\lim }f\left(x\right)$ with the help of the graph. Also identify another function that agrees with the given function at all one point.

Answer to Problem 8E

The value of $\underset{x\to 0}{\lim }f\left(x\right)=0$ . Another function is $g\left(x\right)=x\left(x-2\right)$

Explanation of Solution

Given information:

Given function is $f\left(x\right)=\frac{x^3-4x}{x+2}$

And graph is given as,

  

Consider the graph as,

  

This is observed from the graph that the function cuts the x − axis at $x=0$ and $x=2$ , then $\underset{x\to 0}{\lim }f\left(x\right)=0$

Now consider the function as $f\left(x\right)=\frac{x^3-4x}{x+2}$

Direct substitute $x=0$ in the function,

  $f\left(0\right)=\frac{0^3-4\left(0\right)}{0+2}=0$

Then

  $\begin{array}{c}\underset{x\to 0}{\lim }f\left(x\right)=\underset{x\to 0}{\lim }\frac{x^3-4x}{x+2}\\ =f\left(0\right)\\ =0\end{array}$

Now, the function $f\left(x\right)$ can be written as,

  $\begin{array}{c}f\left(x\right)=\frac{x^3-4x}{x+2}\\ =\frac{x\left(x^2-4\right)}{\left(x+2\right)}\\ =\frac{x\left(x-2\right)\left(x+2\right)}{\left(x+2\right)}\\ =x\left(x-2\right)\end{array}$

Hence, the new function can be written as $g\left(x\right)=x\left(x-2\right)$

Therefore, the value of $\underset{x\to 0}{\lim }f\left(x\right)=0$ and another function is $g\left(x\right)=x\left(x-2\right)$

b.

To determine

To find: The value of $\underset{x\to -2}{\lim }f\left(x\right)$ with the help of the graph. Also identify another function that agrees with the given function at all one point.

Answer to Problem 8E

The value of $\underset{x\to 0}{\lim }f\left(x\right)=8$ and another function is $g\left(x\right)=x\left(x-2\right)$

Explanation of Solution

Given information:

Given function is $f\left(x\right)=\frac{x^3-4x}{x+2}$

And graph is given as,

  

Consider the graph as,

  

Now consider the function as $f\left(x\right)=\frac{x^3-4x}{x+2}$

Direct substitute $x=-2$ in the function,

  $\begin{array}{c}f\left(-2\right)=\frac{{\left(-2\right)}^3-4\left(-2\right)}{-2+2}\\ =\frac{-8+8}{-2+2}\\ =\frac{0}{0}\end{array}$

This means the function is not defined at $x=-2$

Now, the function $f\left(x\right)$ can be written as,

  $\begin{array}{c}f\left(x\right)=\frac{x^3-4x}{x+2}\\ =\frac{x\left(x^2-4\right)}{\left(x+2\right)}\\ =\frac{x\left(x-2\right)\left(x+2\right)}{\left(x+2\right)}\\ =x\left(x-2\right)\end{array}$

Then, limiting value of the function is,

  $\begin{array}{c}\underset{x\to -2}{\lim }f\left(x\right)=\underset{x\to -2}{\lim }\left(x\left(x-2\right)\right)\\ =-2\left(-2-2\right)\\ =-2\left(-4\right)\\ =8\end{array}$

Hence, the new function can be written as $g\left(x\right)=x\left(x-2\right)$

Therefore, the value of $\underset{x\to -2}{\lim }f\left(x\right)=8$ and another function is $g\left(x\right)=x\left(x-2\right)$

c.

To determine

To find: The value of $\underset{x\to 1}{\lim }f\left(x\right)$ with the help of the graph. Also identify another function that agrees with the given function at all one point.

Answer to Problem 8E

The value of $\underset{x\to 1}{\lim }f\left(x\right)=-1$ and another function is $g\left(x\right)=x\left(x-2\right)$

Explanation of Solution

Given information:

Given function is $f\left(x\right)=\frac{x^3-4x}{x+2}$

And graph is given as,

  

Consider the graph as,

  

This is observed from the graph that $\underset{x\to 1}{\lim }f\left(x\right)=-1$

Now consider the function as $f\left(x\right)=\frac{x^3-4x}{x+2}$

Direct substitute $x=1$ in the function,

  $\begin{array}{c}f\left(1\right)=\frac{{\left(1\right)}^3-4\left(1\right)}{1+2}\\ =\frac{1-4}{3}\\ =\frac{-3}{3}\\ =-1\end{array}$

Now, the function $f\left(x\right)$ can be written as,

  $\begin{array}{c}f\left(x\right)=\frac{x^3-4x}{x+2}\\ =\frac{x\left(x^2-4\right)}{\left(x+2\right)}\\ =\frac{x\left(x-2\right)\left(x+2\right)}{\left(x+2\right)}\\ =x\left(x-2\right)\end{array}$

Hence, the new function can be written as $g\left(x\right)=x\left(x-2\right)$

Therefore, the value of $\underset{x\to 1}{\lim }f\left(x\right)=-1$ and another function is $g\left(x\right)=x\left(x-2\right)$

d.

To determine

To find: The value of $\underset{x\to 2}{\lim }f\left(x\right)$ with the help of the graph. Also identify another function that agrees with the given function at all one point.

Answer to Problem 8E

The value of $\underset{x\to 2}{\lim }f\left(x\right)=0$ and another function is $g\left(x\right)=x\left(x-2\right)$

Explanation of Solution

Given information:

Given function is $f\left(x\right)=\frac{x^3-4x}{x+2}$

And graph is given as,

  

Consider the graph as,

  

This is observed from the graph that $\underset{x\to 2}{\lim }f\left(x\right)=0$

Now consider the function as $f\left(x\right)=\frac{x^3-4x}{x+2}$

Direct substitute $x=2$ in the function,

  $\begin{array}{c}f\left(2\right)=\frac{{\left(2\right)}^3-4\left(2\right)}{2+2}\\ =\frac{8-8}{4}\\ =\frac{0}{4}\\ =0\end{array}$

Now, the function $f\left(x\right)$ can be written as,

  $\begin{array}{c}f\left(x\right)=\frac{x^3-4x}{x+2}\\ =\frac{x\left(x^2-4\right)}{\left(x+2\right)}\\ =\frac{x\left(x-2\right)\left(x+2\right)}{\left(x+2\right)}\\ =x\left(x-2\right)\end{array}$

Hence, the new function can be written as $g\left(x\right)=x\left(x-2\right)$

Therefore, the value of $\underset{x\to 2}{\lim }f\left(x\right)=0$ and another function is $g\left(x\right)=x\left(x-2\right)$