Chapter 11.4, Problem 42E

(a)

To determine

To calculate:

Complete the table and limit for the given function.

Answer to Problem 42E

Limit of the function is $y=\frac{1}{2}$

$x$	${10}^0$	${10}^1$	${10}^2$	${10}^3$	${10}^4$	${10}^5$	${10}^6$
$f\left(x\right)$	$0.508$	$0.5008$	$0.50008$	$0.500008$	$0.5000008$	$0.50000008$	$0.5$

Explanation of Solution

Given information:

  $f\left(x\right)=4\left(4x-\sqrt{16x^2-x}\right)$

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

Calculation:

For completing the table of the sequence,

Consider the given functions,

  $f\left(x\right)=4\left(4x-\sqrt{16x^2-x}\right)$

The value of the function $f\left(x\right)=4\left(4x-\sqrt{16x^2-x}\right)$ for $x={10}^0$ ,

  $\begin{array}{l}f\left(x\right)=4\left(4\times {10}^0-\sqrt{16\times {10}^0-{10}^0}\right)\\ =0.508\end{array}$

The value of the function $f\left(x\right)=4\left(4x-\sqrt{16x^2-x}\right)$ for $x={10}^1$ ,

  $\begin{array}{l}f\left(x\right)=4\left(4\times {10}^1-\sqrt{16\times {10}^1-{10}^1}\right)\\ =0.005008\end{array}$

The value of the function $f\left(x\right)=4\left(4x-\sqrt{16x^2-x}\right)$ for $x={10}^2$ ,

  $\begin{array}{l}f\left(x\right)=4\left(4\times {10}^2-\sqrt{16\times {10}^2-{10}^2}\right)\\ =0.50008\end{array}$

The value of the function $f\left(x\right)=4\left(4x-\sqrt{16x^2-x}\right)$ for $x={10}^3$ ,

  $\begin{array}{l}f\left(x\right)=4\left(4\times {10}^3-\sqrt{16\times {10}^3-{10}^3}\right)\\ =0.500008\end{array}$

The value of the function $f\left(x\right)=4\left(4x-\sqrt{16x^2-x}\right)$ for $x={10}^4$ ,

  $\begin{array}{l}f\left(x\right)=4\left(4\times {10}^4-\sqrt{16\times {10}^4-{10}^4}\right)\\ =0.5000008\end{array}$

The value of the function $f\left(x\right)=4\left(4x-\sqrt{16x^2-x}\right)$ for $x={10}^5$ ,

  $\begin{array}{l}f\left(x\right)=4\left(4\times {10}^5-\sqrt{16\times {10}^5-{10}^5}\right)\\ =0.50000008\end{array}$

The value of the function $f\left(x\right)=4\left(4x-\sqrt{16x^2-x}\right)$ for $x={10}^6$ ,

  $\begin{array}{l}f\left(x\right)=4\left(4\times {10}^6-\sqrt{16\times {10}^6-{10}^6}\right)\\ =0.5\end{array}$

For making a table above data is used,

$x$	${10}^0$	${10}^1$	${10}^2$	${10}^3$	${10}^4$	${10}^5$	${10}^6$
$f\left(x\right)$	$0.508$	$0.5008$	$0.50008$	$0.500008$	$0.5000008$	$0.50000008$	$0.5$

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

So first rationalize it:

  $\begin{array}{l}f\left(x\right)=4\left(4x-\sqrt{16x^2-x}\right)\times \frac{4x+\sqrt{16x^2-x}}{4x+\sqrt{16x^2-x}}\\ =4\frac{16x-16x^2-x}{4x+\sqrt{16x^2-x}}\\ =\frac{4x}{4x+\sqrt{16x^2-x}}\end{array}$

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

  $\begin{array}{l}y=\underset{x\to \infty }{\lim }f\left(x\right)\\ =\underset{x\to \infty }{\lim }\frac{4x}{4x+\sqrt{16x^2-x}}\\ =\underset{x\to \infty }{\lim }\frac{4x}{4+\sqrt{16-\frac{1}{x}}}\\ =\frac{4}{4+4}\end{array}$

Thus, $y=\frac{1}{2}$ .

(b)

To determine

To graph:

Estimate the limit by using graph.

Explanation of Solution

Given information:

  $f\left(x\right)=4\left(4x-\sqrt{16x^2-x}\right)$

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

Graph: