Chapter 11.4, Problem 32E

(a)

To determine

To calculate:

Complete the table and limit for the given function.

Answer to Problem 32E

Table is as below:

$x$	${10}^0$	${10}^1$	${10}^2$	${10}^3$	${10}^4$	${10}^5$	${10}^6$
$f\left(x\right)$	$0.2$	$0.9615$	$0.9996$	$0.99996$	$0.99999996$	$0.9999999996$	$0.999999999996$

Explanation of Solution

Given information:

  $f\left(x\right)=\frac{x^2}{x^2+4}$

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

Calculation:

For completing the table of the sequence,

Consider the given functions,

  $f\left(x\right)=\frac{x^2}{x^2+4}$

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

  $\begin{array}{l}f\left(x\right)=\frac{x^2}{x^2+4}\\ =\frac{{\left(1\right)}^2}{{\left(1\right)}^2+4}\\ =\frac{1}{5}\\ =0.2\end{array}$

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

  $\begin{array}{l}f\left(x\right)=\frac{x^2}{x^2+4}\\ =\frac{{\left(10\right)}^2}{{\left(10\right)}^2+4}\\ =\frac{100}{104}\\ =0.9615\end{array}$

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

  $\begin{array}{l}f\left(x\right)=\frac{x^2}{x^2+4}\\ =\frac{{\left(100\right)}^2}{{\left(100\right)}^2+4}\\ =\frac{10000}{10004}\\ =0.9996\end{array}$

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

  $\begin{array}{l}f\left(x\right)=\frac{x^2}{x^2+4}\\ =\frac{{\left(1000\right)}^2}{{\left(1000\right)}^2+4}\\ =\frac{1000000}{1000004}\\ =0.99996\end{array}$

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

  $\begin{array}{l}f\left(x\right)=\frac{x^2}{x^2+4}\\ =\frac{{\left(10000\right)}^2}{{\left(10000\right)}^2+4}\\ =\frac{100000000}{100000004}\\ =0.99999996\end{array}$

The value of the function $f\left(x\right)=\frac{x^2}{x^2+4}$ for $x={10}^5$ ,

  $\begin{array}{l}f\left(x\right)=\frac{x^2}{x^2+4}\\ =\frac{{\left(100000\right)}^2}{{\left(100000\right)}^2+4}\\ =\frac{10000000000}{10000000004}\\ =0.9999999996\end{array}$

The value of the function $f\left(x\right)=\frac{x^2}{x^2+4}$ for $x={10}^6$ ,

  $\begin{array}{l}f\left(x\right)=\frac{x^2}{x^2+4}\\ =\frac{{\left(1000000\right)}^2}{{\left(1000000\right)}^2+4}\\ =\frac{1000000000000}{1000000000004}\\ =0.999999999996\end{array}$

For making a table above data is used,

$x$	${10}^0$	${10}^1$	${10}^2$	${10}^3$	${10}^4$	${10}^5$	${10}^6$
$f\left(x\right)$	$0.2$	$0.9615$	$0.9996$	$0.99996$	$0.99999996$	$0.9999999996$	$0.999999999996$

(b)

To determine

To graph:

limit by using graph.

Explanation of Solution

Given information:

  $f\left(x\right)=\frac{x^2}{x^2+4}$

Graph:

  

Interpretation:

Thus, from the tablein part (a) it is observed that $x$ approaches at infinity the function $f\left(x\right)$ get closure to $1$ .

Hence, the required graph of the function and limit showing graphically is $1$ .

(c)

To determine

To calculate:

Limit of the given function algebraically..

Answer to Problem 32E

Limit of the function algebraically is $\underset{x\to \infty }{\lim }f\left(x\right)=1$

Explanation of Solution

Given information:

  $f\left(x\right)=\frac{x^2}{x^2+4}$

Calculation:

For finding out the limit of the given function algebraically,

  $f\left(x\right)=\frac{x^2}{x^2+4}$

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

  $\begin{array}{l}\underset{x\to \infty }{\lim }f\left(x\right)=\underset{x\to \infty }{\lim }\frac{x^2}{x^2+4}\\ =\underset{x\to \infty }{\lim }\frac{x^2}{x^2\left(1-\frac{4}{x^2}\right)}\\ =\underset{x\to \infty }{\lim }\frac{x^2}{x^2\left(1-\frac{4}{x^2}\right)}\\ =\underset{x\to \infty }{\lim }\frac{1}{\left(1-\frac{4}{x^2}\right)}\\ =\frac{1}{\left(1-\frac{4}{{\left(\infty \right)}^2}\right)}\\ =\frac{1}{\left(1+0\right)}\\ =1\end{array}$

Hence, the required limit of the function algebraically is $\underset{x\to \infty }{\lim }f\left(x\right)=1$