Chapter 11.4, Problem 33E

(a)

To determine

To calculate:

Complete the table and limit for the given function.

Answer to Problem 33E

Table is as below:

$x$	${10}^0$	${10}^1$	${10}^2$	${10}^3$	${10}^4$	${10}^5$	${10}^6$
$f\left(x\right)$	Error	$0.77$	$0.0700$	$0.007000$	$0.00070000$	$7\times {10}^{-5}$	$7\times {10}^{-6}$

Explanation of Solution

Given information:

  $f\left(x\right)=\frac{7x}{1-x^2}$

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

Calculation:

For completing the table of the sequence,

Consider the given functions,

  $f\left(x\right)=\frac{7x}{1-x^2}$

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

  $\begin{array}{l}f\left(x\right)=\frac{7x}{1-x^2}\\ =\frac{7.1}{1-{\left(1\right)}^2}\\ =\frac{7}{0}\end{array}$

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

The value of the function $f\left(x\right)=\frac{7x}{1-x^2}$ for $x=10$ ,

  $\begin{array}{l}f\left(x\right)=\frac{7x}{1-x^2}\\ =\frac{7.10}{1-{\left(10\right)}^2}\\ =\frac{70}{99}\\ =0.77\end{array}$

The value of the function $f\left(x\right)=\frac{7x}{1-x^2}$ for $x={10}^2$ ,

  $\begin{array}{l}f\left(x\right)=\frac{7x}{1-x^2}\\ =\frac{7.100}{1-{\left(100\right)}^2}\\ =\frac{700}{9999}\\ =0.0700\end{array}$

The value of the function $f\left(x\right)=\frac{7x}{1-x^2}$ for $x={10}^3$ ,

  $\begin{array}{l}f\left(x\right)=\frac{7x}{1-x^2}\\ =\frac{7.1000}{1-{\left(1000\right)}^2}\\ =\frac{7000}{999999}\\ =0.007000\end{array}$

The value of the function $f\left(x\right)=\frac{7x}{1-x^2}$ for $x={10}^4$ ,

  $\begin{array}{l}f\left(x\right)=\frac{7x}{1-x^2}\\ =\frac{7.10000}{1-{\left(10000\right)}^2}\\ =\frac{70000}{99999999}\\ =0.00070000\end{array}$

The value of the function $f\left(x\right)=\frac{7x}{1-x^2}$ for $x={10}^5$ ,

  $\begin{array}{l}f\left(x\right)=\frac{7x}{1-x^2}\\ =\frac{7.100000}{1-{\left(100000\right)}^2}\\ =\frac{700000}{9999999999}\\ =7\times {10}^{-5}\end{array}$

The value of the function $f\left(x\right)=\frac{7x}{1-x^2}$ for $x={10}^6$ ,

  $\begin{array}{l}f\left(x\right)=\frac{7x}{1-x^2}\\ =\frac{7.1000000}{1-{\left(1000000\right)}^2}\\ =\frac{7000000}{999999999999}\\ =7\times {10}^{-6}\end{array}$

For making a table above data is used,

$x$	${10}^0$	${10}^1$	${10}^2$	${10}^3$	${10}^4$	${10}^5$	${10}^6$
$f\left(x\right)$	Error	$0.77$	$0.0700$	$0.007000$	$0.00070000$	$7\times {10}^{-5}$	$7\times {10}^{-6}$

(b)

To determine

To graph:

Estimate the limit by using graph.

Explanation of Solution

Given information:

  $f\left(x\right)=\frac{7x}{1-x^2}$

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 $0$ .

Hence, the required graph of the function and limit showing graphically is $0$ .

(c)

To determine

To calculate:

Limit of the given function algebraically..

Answer to Problem 33E

Limit of the function algebraically is $\underset{x\to \infty }{\lim }f\left(x\right)=0$

Explanation of Solution

Given information:

  $f\left(x\right)=\frac{7x}{1-x^2}$

Calculation:

For finding out the limit of the given function algebraically,

  $f\left(x\right)=\frac{7x}{1-x^2}$

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

  $\begin{array}{l}\underset{x\to \infty }{\lim }f\left(x\right)=\underset{x\to \infty }{\lim }\frac{7x}{1-x^2}\\ =\underset{x\to \infty }{\lim }\frac{7x}{x^2\left(\frac{1}{x^2}-1\right)}\\ =\underset{x\to \infty }{\lim }\frac{7}{x\left(\frac{1}{x^2}-1\right)}\\ =\frac{7}{\left(\infty \right)\left(\frac{1}{{\left(\infty \right)}^2}-1\right)}\\ =\frac{7}{\infty \left(0-1\right)}\\ =0\end{array}$

Hence, the required limit of the function algebraically is $\underset{x\to \infty }{\lim }f\left(x\right)=0$