Chapter 11.4, Problem 35E

(a)

To determine

To calculate:

Complete the table and limit for the given function.

Answer to Problem 35E

Limit of the function is $y=0$

$x$	${10}^0$	${10}^1$	${10}^2$	${10}^3$	${10}^4$	${10}^5$	${10}^6$
$f\left(x\right)$	$-2$	$0.97$	$-0.9997$	$-0.999997$	$0.99999997$	$0.9999999997$	$0.999999999997$

Explanation of Solution

Given information:

  $f\left(x\right)=1-\frac{3}{x^2}$

Calculation:

For completing the table of the sequence,

Consider the given functions,

  $f\left(x\right)=1-\frac{3}{x^2}$

The value of the function $f\left(x\right)=1-\frac{3}{x^2}$ for $x={10}^0=1$ ,

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

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

  $\begin{array}{l}f\left(x\right)=1-\frac{3}{x^2}\\ ={\left(10\right)}^2-\frac{3}{{\left(10\right)}^2}\\ =\frac{100-3}{100}\\ =0.97\end{array}$

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

  $\begin{array}{l}f\left(x\right)=1-\frac{3}{x^2}\\ ={\left(100\right)}^2-\frac{3}{{\left(100\right)}^2}\\ =\frac{1000-3}{1000}\\ =0.9997\end{array}$

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

  $\begin{array}{l}f\left(x\right)=1-\frac{3}{x^2}\\ ={\left(1000\right)}^2-\frac{3}{{\left(1000\right)}^2}\\ =\frac{10000-3}{10000}\\ =0.999997\end{array}$

The value of the function $f\left(x\right)=1-\frac{3}{x^2}$ for $x={10}^4$ ,

  $\begin{array}{l}f\left(x\right)=1-\frac{3}{x^2}\\ ={\left(10000\right)}^2-\frac{3}{{\left(10000\right)}^2}\\ =\frac{100000-3}{100000}\\ =0.99999997\end{array}$

The value of the function $f\left(x\right)=1-\frac{3}{x^2}$ for $x={10}^5$ ,

  $\begin{array}{l}f\left(x\right)=1-\frac{3}{x^2}\\ =\frac{x^2-3}{x^2}\\ =\frac{{\left(1000000\right)}^2-3}{{\left(1000000\right)}^2}\\ =\frac{10000000-3}{10000000}\\ =0.9999999997\end{array}$

The value of the function $f\left(x\right)=1-\frac{3}{x^2}$ for $x={10}^6$ ,

  $\begin{array}{l}f\left(x\right)=1-\frac{3}{x^2}\\ =\frac{x^2-3}{x^2}\\ =\frac{{\left(1000000\right)}^2-3}{{\left(1000000\right)}^2}\\ =\frac{10000000-3}{10000000}\\ =0.999999999997\end{array}$

For making a table above data is used,

$x$	${10}^0$	${10}^1$	${10}^2$	${10}^3$	${10}^4$	${10}^5$	${10}^6$
$f\left(x\right)$	$-2$	$0.97$	$-0.9997$	$-0.999997$	$0.99999997$	$0.9999999997$	$0.999999999997$

(b)

To determine

To graph:

Estimate the limit by using graph.

Explanation of Solution

Given information:

  $f\left(x\right)=1-\frac{3}{x^2}$

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 35E

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

Explanation of Solution

Given information:

  $f\left(x\right)=1-\frac{3}{x^2}$

Calculation:

For finding out the limit of the given function algebraically,

  $f\left(x\right)=1-\frac{3}{x^2}$

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

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

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