Chapter 2.1, Problem 15E

(a)

To determine

To find: Complete the table given below for $f\left(x\right)=\frac{x^2+6x+2}{x+1}$ and also estimate the value of $\underset{x\to 0}{\lim }f\left(x\right)$ :

$x$	$-0.1$	$-0.01$	$-0.001$	$-0.0001$
$f\left(x\right)$	$?$	$?$	$?$	$?$

Answer to Problem 15E

The value of the $\underset{x\to 0}{\lim }f\left(x\right)$ is $2$ and the complete table is given below:

$x$	$-0.1$	$-0.01$	$-0.001$	$-0.0001$
$f\left(x\right)$	$1.5666$	$1.9596$	$1.9959$	$1.9996$

Explanation of Solution

Given information:

The function is $f\left(x\right)=\frac{x^2+6x+2}{x+1}$ .

Calculation:

Substitute $-0.1$ for $x$ in the given function.

  $\begin{array}{c}f\left(x\right)=\frac{x^2+6x+2}{x+1}\\ f\left(-0.1\right)=\frac{{\left(-0.1\right)}^2+6\times \left(-0.1\right)+2}{\left(-0.1\right)+1}\\ f\left(-0.1\right)=\frac{0.01-0.6+2}{0.9}\\ f\left(-0.1\right)=1.5666\end{array}$

So, the value of $f\left(x\right)$ at $x=-0.1$ is $1.5666$ .

Substitute $-0.01$ for $x$ in the given function.

  $\begin{array}{c}f\left(x\right)=\frac{x^2+6x+2}{x+1}\\ f\left(-0.01\right)=\frac{{\left(-0.01\right)}^2+6\times \left(-0.01\right)+2}{\left(-0.01\right)+1}\\ f\left(-0.01\right)=\frac{0.0001-0.06+2}{0.99}\\ f\left(-0.01\right)=1.9596\end{array}$

So, the value of $f\left(x\right)$ at $x=-0.01$ is $1.9596$ .

Substitute $-0.001$ for $x$ in the given function.

  $\begin{array}{c}f\left(x\right)=\frac{x^2+6x+2}{x+1}\\ f\left(-0.001\right)=\frac{{\left(-0.001\right)}^2+6\times \left(-0.001\right)+2}{\left(-0.001\right)+1}\\ f\left(-0.001\right)=\frac{0.000001-0.006+2}{0.999}\\ f\left(-0.001\right)=1.9959\end{array}$

So, the value of $f\left(x\right)$ at $x=-0.001$ is $1.9959$ .

Substitute $-0.0001$ for $x$ in the given function.

  $\begin{array}{c}f\left(x\right)=\frac{x^2+6x+2}{x+1}\\ f\left(-0.0001\right)=\frac{{\left(-0.0001\right)}^2+6\times \left(-0.0001\right)+2}{\left(-0.0001\right)+1}\\ f\left(-0.0001\right)=\frac{0.00000001-0.0006+2}{0.9999}\\ f\left(-0.0001\right)\approx 1.9996\end{array}$

So, the value of $f\left(x\right)$ at $x=-0.0001$ is $1.9996$ .

Therefore, the complete table is given by:

$x$	$-0.1$	$-0.01$	$-0.001$	$-0.0001$
$f\left(x\right)$	$1.5666$	$1.9596$	$1.9959$	$1.9996$

(b)

To determine

To find: Complete the table given below for $f\left(x\right)=\frac{x^2+6x+2}{x+1}$ and also estimate the value of $\underset{x\to 0}{\lim }f\left(x\right)$ :

$x$	$0.1$	$0.01$	$0.001$	$0.0001$
$f\left(x\right)$	$?$	$?$	$?$	$?$

Answer to Problem 15E

The value of the $\underset{x\to 0}{\lim }f\left(x\right)$ is $2$ and the complete table is given below:

$x$	$0.1$	$0.01$	$0.001$	$0.0001$
$f\left(x\right)$	$2.3727$	$2.0397$	$2.0039$	$2.0004$

Explanation of Solution

Given information:

The function is $f\left(x\right)=\frac{x^2+6x+2}{x+1}$ .

Calculation:

Substitute $0.1$ for $x$ in the given function.

  $\begin{array}{c}f\left(x\right)=\frac{x^2+6x+2}{x+1}\\ f\left(0.1\right)=\frac{{\left(0.1\right)}^2+6\times \left(0.1\right)+2}{\left(0.1\right)+1}\\ f\left(0.1\right)=\frac{0.01+0.6+2}{1.1}\\ f\left(0.1\right)=2.3727\end{array}$

So, the value of $f\left(x\right)$ at $x=0.1$ is $2.3727$ .

Substitute $0.01$ for $x$ in the given function.

  $\begin{array}{c}f\left(x\right)=\frac{x^2+6x+2}{x+1}\\ f\left(0.01\right)=\frac{{\left(0.01\right)}^2+6\times \left(0.01\right)+2}{\left(0.01\right)+1}\\ f\left(0.01\right)=\frac{0.0001+0.06+2}{1.01}\\ f\left(0.01\right)=2.0397\end{array}$

So, the value of $f\left(x\right)$ at $x=0.01$ is $2.0397$ .

Substitute $0.001$ for $x$ in the given function.

  $\begin{array}{c}f\left(x\right)=\frac{x^2+6x+2}{x+1}\\ f\left(0.001\right)=\frac{{\left(0.001\right)}^2+6\times \left(0.001\right)+2}{\left(0.001\right)+1}\\ f\left(0.001\right)=\frac{0.000001+0.006+2}{1.001}\\ f\left(0.001\right)=2.0039\end{array}$

So, the value of $f\left(x\right)$ at $x=0.001$ is $2.0039$ .

Substitute $0.0001$ for $x$ in the given function.

  $\begin{array}{c}f\left(x\right)=\frac{x^2+6x+2}{x+1}\\ f\left(0.0001\right)=\frac{{\left(0.0001\right)}^2+6\times \left(0.0001\right)+2}{\left(0.0001\right)+1}\\ f\left(0.0001\right)=\frac{0.00000001+0.0006+2}{1.0001}\\ f\left(0.0001\right)\approx 2.0004\end{array}$

So, the value of $f\left(x\right)$ at $x=0.0001$ is $2.0004$ .

Therefore, the complete table is given by:

$x$	$0.1$	$0.01$	$0.001$	$0.0001$
$f\left(x\right)$	$2.3727$	$2.0397$	$2.0039$	$2.0004$