Chapter 11.4, Problem 36E

(a)

To determine

To calculate:

Complete the table and limit for the given function.

Answer to Problem 36E

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:

  $f\left(x\right)=\frac{2x+1}{x^2-1}$

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

Calculation:

For completing the table of the sequence,

Consider the given functions,

  $f\left(x\right)=\frac{2x+1}{x^2-1}$

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

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

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

  $\begin{array}{l}f\left(x\right)=\frac{2x+1}{x^2-1}\\ =\frac{2\left(10\right)+1}{{\left(10\right)}^2-1}\\ =\frac{21}{99}\\ =0.212121\end{array}$

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

  $\begin{array}{l}f\left(x\right)=\frac{2x+1}{x^2-1}\\ =\frac{2\left(100\right)+1}{{\left(100\right)}^2-1}\\ =\frac{201}{9999}\\ =0.02102\end{array}$

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

  $\begin{array}{l}f\left(x\right)=\frac{2x+1}{x^2-1}\\ =\frac{2\left(1000\right)+1}{{\left(1000\right)}^2-1}\\ =\frac{2001}{999999}\\ =0.0021001\end{array}$

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

  $\begin{array}{l}f\left(x\right)=\frac{2x+1}{x^2-1}\\ =\frac{2\left(10000\right)+1}{{\left(10000\right)}^2-1}\\ =\frac{20001}{999999999}\\ =0.000210001\end{array}$

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

  $\begin{array}{l}f\left(x\right)=\frac{2x+1}{x^2-1}\\ =\frac{2\left(100000\right)+1}{{\left(100000\right)}^2-1}\\ =\frac{200001}{999999999}\\ =0.00002\end{array}$

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

  $\begin{array}{l}f\left(x\right)=\frac{2x+1}{x^2-1}\\ =\frac{2\left(1000000\right)+1}{{\left(1000000\right)}^2-1}\\ =\frac{20000001}{99999999999}\\ =0.000002\end{array}$

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	$0.212121$	$0.02102$	$0.002001$	$0.00020001$	$0.00002$	$0.000002$

(b)

To determine

To graph:

Estimate the limit by using graph.

Explanation of Solution

Given information:

  $f\left(x\right)=\frac{2x+1}{x^2-1}$

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 36E

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

Explanation of Solution

Given information:

  $f\left(x\right)=\frac{2x+1}{x^2-1}$

Calculation:

For finding out the limit of the given function algebraically,

  $f\left(x\right)=\frac{2x+1}{x^2-1}$

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

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

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