Chapter 11.4, Problem 39E

(a)

To determine

To calculate:

Complete the table and limit for the given function.

Answer to Problem 39E

Limit of the function is $y=0$

$x$	${10}^0$	${10}^1$	${10}^2$	${10}^3$	${10}^4$	${10}^5$	${10}^6$
$f\left(x\right)$	$-0.7321$	$-0.0995$	$-0.01$	$-0.001$	$-1.0\times {10}^{-4}$	$-1.0\times {10}^{-5}$	$-1.0\times {10}^{-6}$

Explanation of Solution

Given information:

  $f\left(x\right)=x-\sqrt{x^2+2}$

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

Calculation:

For completing the table of the sequence,

Consider the given functions,

  $f\left(x\right)=x-\sqrt{x^2+2}$

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

  $\begin{array}{l}f\left(x\right)=x-\sqrt{x^2+2}\\ ={10}^0-\sqrt{{10}^0+2}\\ =1-\sqrt{3}\\ =1-1.7321\\ =-0.7321\end{array}$

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

  $\begin{array}{l}f\left(x\right)=x-\sqrt{x^2+2}\\ ={10}^1-\sqrt{{10}^1+2}\\ =10-\sqrt{102}\\ =10-10.0995\\ =-0.0995\end{array}$

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

  $\begin{array}{l}f\left(x\right)=x-\sqrt{x^2+2}\\ ={10}^2-\sqrt{{10}^2+2}\\ =100-\sqrt{10002}\\ =10-10.01\\ =-0.01\end{array}$

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

  $\begin{array}{l}f\left(x\right)=x-\sqrt{x^2+2}\\ ={10}^3-\sqrt{{10}^6+2}\\ =-0.0010\end{array}$

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

  $\begin{array}{l}f\left(x\right)=x-\sqrt{x^2+2}\\ ={10}^4-\sqrt{{10}^8+2}\\ =-1.0\times {10}^{-4}\end{array}$

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

  $\begin{array}{l}f\left(x\right)=x-\sqrt{x^2+2}\\ ={10}^5-\sqrt{{10}^{10}+2}\\ =-1.0\times {10}^{-5}\end{array}$

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

  $\begin{array}{l}f\left(x\right)=x-\sqrt{x^2+2}\\ ={10}^6-\sqrt{{10}^{12}+2}\\ =-1.0\times {10}^{-6}\end{array}$

For making a table above data is used,

$x$	${10}^0$	${10}^1$	${10}^2$	${10}^3$	${10}^4$	${10}^5$	${10}^6$
$f\left(x\right)$	$-0.7321$	$-0.0995$	$-0.01$	$-0.001$	$-1.0\times {10}^{-4}$	$-1.0\times {10}^{-5}$	$-1.0\times {10}^{-6}$

For numerical estimate the limit as $x$ approaches infinity:

So first rationalize it:

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

Now apply the limit and then divide numerator and denominator by $x$ , highest power of denominator.

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

Thus, $y=0$ .

(b)

To determine

To graph:

Estimate the limit by using graph.

Explanation of Solution

Given information:

  $f\left(x\right)=x-\sqrt{x^2+2}$

Graph: