Chapter 11.4, Problem 31E

(a)

To determine

To calculate:

Complete the table and limit for the given function.

Answer to Problem 31E

Table is as below:

$x$	${10}^0$	${10}^1$	${10}^2$	${10}^3$	${10}^4$	${10}^5$	${10}^6$
$f\left(x\right)$	$1$	$-2.8571$	$-2.06185$	$-0.2006018$	$-0.200060018$	$-0.2000060002$	$-0.2000006$

Explanation of Solution

Given information:

  $f\left(x\right)=\frac{2x}{3-x}$

$x$	${10}^0$	${10}^1$	${10}^2$	${10}^3$	${10}^4$	${10}^5$	${10}^6$
$f\left(x\right)$

Calculation:

For completing the table of the sequence,

Consider the given functions,

  $f\left(x\right)=\frac{2x}{3-x}$

The value of the function $f\left(x\right)$ for $x={10}^0=1$ ,

  $\begin{array}{l}f\left(x\right)=\frac{2x}{3-x}\\ =\frac{2.1}{3-1}\\ =\frac{2}{2}\\ =1\end{array}$

Thus, function $f\left(x\right)$ is not defined at $x={10}^0$

The value of the function $f\left(x\right)=\frac{2x}{3-x}$ for $x=10$ ,

  $\begin{array}{l}f\left(x\right)=\frac{2x}{3-x}\\ =\frac{2.10}{3-10}\\ =\frac{20}{-7}\\ =-2.8571\end{array}$

The value of the function $f\left(x\right)=\frac{2x}{3-x}$ for $x={10}^2$ ,

  $\begin{array}{l}f\left(x\right)=\frac{2x}{3-x}\\ =\frac{2.100}{3-100}\\ =\frac{200}{-97}\\ =-2.06185\end{array}$

The value of the function $f\left(x\right)=\frac{2x}{3-x}$ for $x={10}^3$ ,

  $\begin{array}{l}f\left(x\right)=\frac{2x}{3-x}\\ =\frac{2.1000}{3-1000}\\ =\frac{2000}{-997}\\ =-0.2006018\end{array}$

The value of the function $f\left(x\right)=\frac{2x}{3-x}$ for $x={10}^4$ ,

  $\begin{array}{l}f\left(x\right)=\frac{2x}{3-x}\\ =\frac{{2.10}^4}{3-{10}^4}\\ =\frac{20000}{-9997}\\ =-0.200060018\end{array}$

The value of the function $f\left(x\right)=\frac{2x}{3-x}$ for $x={10}^5$ ,

  $\begin{array}{l}f\left(x\right)=\frac{2x}{3-x}\\ =\frac{{2.10}^5}{3-{10}^5}\\ =\frac{200000}{-99997}\\ =-0.2000060002\end{array}$

The value of the function $f\left(x\right)=\frac{2x}{3-x}$ for $x={10}^6$ ,

  $\begin{array}{l}f\left(x\right)=\frac{2x}{3-x}\\ =\frac{{2.10}^6}{3-{10}^6}\\ =\frac{2000000}{-999997}\\ =-0.2000006\end{array}$

For making a table above data is used,

$x$	${10}^0$	${10}^1$	${10}^2$	${10}^3$	${10}^4$	${10}^5$	${10}^6$
$f\left(x\right)$	$1$	$-2.8571$	$-2.06185$	$-0.2006018$	$-0.200060018$	$-0.2000060002$	$-0.2000006$

(b)

To determine

To graph:

limit by using graph.

Explanation of Solution

Given information:

  $f\left(x\right)=\frac{2x}{3-x}$

Graph:

  

Interpretation:

Thus, from the table in part (a) it is observed that $x$ approaches at infinity the function $f\left(x\right)$ get closure to $-0.66$ .

Hence, the required graph of the function and limit showing graphically is $-0.66$ .

(c)

To determine

To calculate:

Limit of the given function algebraically..

Answer to Problem 31E

Limit of the function algebraically is $\underset{x\to \infty }{\lim }f\left(x\right)=-0.66$

Explanation of Solution

Given information:

  $f\left(x\right)=\frac{2x}{3-x}$

Calculation:

For finding out the limit of the given function algebraically,

  $f\left(x\right)=\frac{2x}{3-x}$

Limit of the function $f\left(x\right)$ as $x$ approaches at infinity is

  $\begin{array}{l}\underset{x\to \infty }{\lim }f\left(x\right)=\underset{x\to \infty }{\lim }\frac{2x}{3-x}\\ =\underset{x\to \infty }{\lim }\frac{2x}{x\left(\frac{1}{x}-3\right)}\\ =\underset{x\to \infty }{\lim }\frac{2x}{x\left(\frac{1}{x}-3\right)}\\ =\underset{x\to \infty }{\lim }\frac{2}{\left(\frac{1}{x}-3\right)}\\ =\frac{2}{\left(\frac{1}{\infty }-3\right)}\\ =\frac{2}{\left(0-3\right)}\\ =-0.66\end{array}$

Hence, the required limit of the function algebraically is $\underset{x\to \infty }{\lim }f\left(x\right)=-0.66$