Chapter 2.1, Problem 16E

(a)

To determine

To find: Complete the table given below for $f\left(x\right)=\frac{x^2-x}{x}$ and also estimate the value of $\underset{x\to 0}{\lim }f\left(x\right)$ :

$x$	$-0.1$	$-0.01$	$-0.001$	$-0.0001$
$f\left(x\right)$	$?$	$?$	$?$	$?$

Answer to Problem 16E

The value of the $\underset{x\to 0}{\lim }f\left(x\right)$ is $-1$ and the complete table is given below:

$x$	$-0.1$	$-0.01$	$-0.001$	$-0.0001$
$f\left(x\right)$	$-1.1$	$-1.01$	$-1.001$	$-1.0001$

Explanation of Solution

Given information:

The function is $f\left(x\right)=\frac{x^2-x}{x}$ .

Calculation:

Simplify the given function.

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

Substitute $-0.1$ for $x$ in the given function.

  $\begin{array}{c}f\left(x\right)=x-1\\ f\left(-0.1\right)=-0.1-1\\ f\left(-0.1\right)=-1.1\end{array}$

So, the value of $f\left(x\right)$ at $x=-0.1$ is $-1.1$ .

Substitute $-0.01$ for $x$ in the given function.

  $\begin{array}{c}f\left(x\right)=x-1\\ f\left(-0.01\right)=-0.01-1\\ f\left(-0.01\right)=-1.01\end{array}$

So, the value of $f\left(x\right)$ at $x=-0.01$ is $-1.01$ .

Substitute $-0.001$ for $x$ in the given function.

  $\begin{array}{c}f\left(x\right)=x-1\\ f\left(-0.001\right)=-0.001-1\\ f\left(-0.001\right)=-1.001\end{array}$

So, the value of $f\left(x\right)$ at $x=-0.001$ is $-1.001$ .

Substitute $-0.0001$ for $x$ in the given function.

  $\begin{array}{c}f\left(x\right)=x-1\\ f\left(-0.0001\right)=-0.0001-1\\ f\left(-0.0001\right)=-1.0001\end{array}$

So, the value of $f\left(x\right)$ at $x=-0.0001$ is $-1.0001$ .

Therefore, the complete table is given by:

$x$	$-0.1$	$-0.01$	$-0.001$	$-0.0001$
$f\left(x\right)$	$-1.1$	$-1.01$	$-1.001$	$-1.0001$

The value of the function tends to $-1$ as $x$ tends to $0$ .

Therefore, The value of the $\underset{x\to 0}{\lim }f\left(x\right)$ is $-1$ .

(b)

To determine

To find: Complete the table given below for $f\left(x\right)=\frac{x^2-x}{x}$ and also estimate the value of $\underset{x\to 0}{\lim }f\left(x\right)$ :

$x$	$0.1$	$0.01$	$0.001$	$0.0001$
$f\left(x\right)$	$?$	$?$	$?$	$?$

Answer to Problem 16E

The value of the $\underset{x\to 0}{\lim }f\left(x\right)$ is $2$ and the complete table is given below:

$x$	$0.1$	$0.01$	$0.001$	$0.0001$
$f\left(x\right)$	$-0.9$	$-0.99$	$-0.999$	$-0.9999$

Explanation of Solution

Given information:

The function is $f\left(x\right)=\frac{x^2-x}{x}$ .

Calculation:

Simplify the given function.

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

Substitute $0.1$ for $x$ in the given function.

  $\begin{array}{c}f\left(x\right)=x-1\\ f\left(0.1\right)=0.1-1\\ f\left(0.1\right)=-0.9\end{array}$

So, the value of $f\left(x\right)$ at $x=0.1$ is $-0.9$ .

Substitute $0.01$ for $x$ in the given function.

  $\begin{array}{c}f\left(x\right)=x-1\\ f\left(0.01\right)=0.01-1\\ f\left(0.01\right)=-0.99\end{array}$

So, the value of $f\left(x\right)$ at $x=0.01$ is $-0.99$ .

Substitute $0.001$ for $x$ in the given function.

  $\begin{array}{c}f\left(x\right)=x-1\\ f\left(0.001\right)=0.001-1\\ f\left(0.001\right)=-0.999\end{array}$

So, the value of $f\left(x\right)$ at $x=0.001$ is $-0.999$ .

Substitute $0.0001$ for $x$ in the given function.

  $\begin{array}{c}f\left(x\right)=x-1\\ f\left(0.0001\right)=0.0001-1\\ f\left(0.0001\right)=-0.9999\end{array}$

So, the value of $f\left(x\right)$ at $x=0.0001$ is $-0.9999$ .

Therefore, the complete table is given by:

$x$	$0.1$	$0.01$	$0.001$	$0.0001$
$f\left(x\right)$	$-0.9$	$-0.99$	$-0.999$	$-0.9999$

The value of the function tends to $-1$ as $x$ tends to $0$ .

Therefore, The value of the $\underset{x\to 0}{\lim }f\left(x\right)$ is $-1$ .