Chapter 11.4, Problem 41E

(a)

To determine

To calculate:

Complete the table and limit for the given function.

Answer to Problem 41E

Limit of the function is $y=\frac{-3}{4}$

$x$	${10}^0$	${10}^1$	${10}^2$	${10}^3$	${10}^4$	${10}^5$	${10}^6$
$f\left(x\right)$	$-0.7082$	$-0.7454$	$-0.7495$	$-0.74995$	$-0.749995$	$-0.7499995$	$-0.75$

Explanation of Solution

Given information:

  $f\left(x\right)=3\left(2x-\sqrt{4x^2+x}\right)$

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

Calculation:

For completing the table of the sequence,

Consider the given functions,

  $f\left(x\right)=3\left(2x-\sqrt{4x^2+x}\right)$

The value of the function $f\left(x\right)=3\left(2x-\sqrt{4x^2+x}\right)$ for $x={10}^0$ ,

  $\begin{array}{l}f\left(x\right)=3\left(2x-\sqrt{4x^2+x}\right)\\ =-0.7082\end{array}$

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

  $\begin{array}{l}f\left(x\right)=3\left(2x-\sqrt{4x^2+x}\right)\\ =-0.7454\end{array}$

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

  $\begin{array}{l}f\left(x\right)=3\left(2x-\sqrt{4x^2+x}\right)\\ =-0.7495\end{array}$

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

  $\begin{array}{l}f\left(x\right)=3\left(2x-\sqrt{4x^2+x}\right)\\ =-0.74995\end{array}$

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

  $\begin{array}{l}f\left(x\right)=3\left(2x-\sqrt{4x^2+x}\right)\\ =-0.74995\end{array}$

The value of the function $f\left(x\right)=3\left(2x-\sqrt{4x^2+x}\right)$ for $x={10}^5$ ,

  $\begin{array}{l}f\left(x\right)=3\left(2x-\sqrt{4x^2+x}\right)\\ =-0.7499995\end{array}$

The value of the function $f\left(x\right)=3\left(2x-\sqrt{4x^2+x}\right)$ for $x={10}^6$ ,

  $\begin{array}{l}f\left(x\right)=3\left(2x-\sqrt{4x^2+x}\right)\\ =-0.75\end{array}$

For making a table above data is used,

$x$	${10}^0$	${10}^1$	${10}^2$	${10}^3$	${10}^4$	${10}^5$	${10}^6$
$f\left(x\right)$	$-0.7082$	$-0.7454$	$-0.7495$	$-0.74995$	$-0.749995$	$-0.7499995$	$-0.75$

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

So first rationalize it:

  $\begin{array}{l}f\left(x\right)=3\left(2x-\sqrt{4x^2+x}\right)\times \frac{2x+\sqrt{4x^2+x}}{2x+\sqrt{4x^2+x}}\\ =3\frac{4x-4x^2-x}{2x+\sqrt{4x^2+x}}\\ =\frac{-3x}{2x+\sqrt{4x^2+x}}\end{array}$

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

  $\begin{array}{l}y=\underset{x\to \infty }{\lim }f\left(x\right)\\ =\underset{x\to \infty }{\lim }\frac{-3x}{2x+\sqrt{4x^2+x}}\\ =\underset{x\to \infty }{\lim }\frac{-3}{2+\sqrt{4+\frac{2}{x}}}\\ =\frac{-3}{2+2}\\ =\frac{-3}{4}\end{array}$

Thus, $y=\frac{-3}{4}$ .

(b)

To determine

To graph:

Estimate the limit by using graph.

Explanation of Solution

Given information:

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

Graph: