Chapter 11, Problem 4RE

1

To determine

  $\lim (6x-1)$

Answer to Problem 4RE

  $\lim (6x-1)=17$

Explanation of Solution

Given:

  $\lim (6x-1)$

$x$	2.9	2.99	2.999	3	3.001	3.01	3.1
$f(x)$

Calculation:

Here, $f(x)=6x-1$

Therefore, $f(2.9)=6\times 2.9-1=16.4$

  $f(2.99)=6\times 2.99-1=16.94$

  $f(2.999)=6\times 2.999-1=16.994$

  $f(3)=6\times 3-1=17$

  $f(3.001)=6\times 3.001-1=17.006$

  $f(3.01)=6\times 3.01-1=17.06$

  $f(3.1)=6\times 3.1-1=17.6$

The table is given below:

$x$	2.9	2.99	2.999	3	3.001	3.01	3.1
$f(x)$	16.4	16.94	16.994	17	17.006	17.06	17.6

Numerically, from the table it is seen, $\lim (6x-1)=17$ .

From the table it is clear, the value of $f(x)$ gradually increases to 17 from the left side of $x=3$ and gradually decreases to 17 from the right side of $x=3$ . That is, the value of $f(x)$ gradually approaches to 17 as $x$ tends to 3.

Since, $f(x)$ is continuous at $x=3$ , then limit of $f(x)$ as $x\to 3$ is the value of $f(3)$ .

  $\begin{array}{l}{\lim }_{x\to 3}(6x-1)\\ ={\lim }_{x\to 3}6x-{\lim }_{x\to 3}1\\ =6\cdot {\lim }_{x\to 3}x-1\\ =6\cdot 3-1\\ =18-1\\ =17\end{array}$

Conclusion:

  $\lim (6x-1)=17$

2

To determine

  ${\lim }_{x\to 2}(x^2-3x+1)$ .

Answer to Problem 4RE

  ${\lim }_{x\to 2}(x^2-3x+1)=-1$

Explanation of Solution

Given:

  ${\lim }_{x\to 2}(x^2-3x+1)$

$x$	1.9	1.99	1.999	2	2.001	2.01	2.1
$f(x)$

Calculation:

Here, $f(x)=x^2-3x+1$

Therefore, $f(1.9)={1.9}^2-3\times 1.9+1=3.61-5.7+1=-1.09$

  $f(1.99)={1.99}^2-3\times 1.99+1=3.9601-5.97+1=-1.0099$

  $f(1.999)={1.999}^2-3\times 1.999+1=3.996001-5.997+1=-1.000999$

  $f(2)=2^2-3\times 2+1=-1$

  $f(2.001)={2.001}^2-3\times 2.001+1=4.004001-6.003+1=-0.998999$

  $f(2.01)={2.01}^2-3\times 2.01+1=4.0401-6.03+1=-0.9899$

  $f(2.1)={2.1}^2-3\times 2.1+1=4.41-6.3+1=-0.89$

The table is given below:

$x$	1.9	1.99	1.999	2	2.001	2.01	2.1
$f(x)$	-1.09	-1.0099	-1.000999	-1	-0.998999	-0.9899	-0.89

Numerically, from the table it is seen, ${\lim }_{x\to 2}(x^2-3x+1)=-1$ .

From the table it is clear, the value of $f(x)$ gradually increases to -1 from the left side of $x=2$ and gradually decreases to -1 from the right side of $x=2$ . That is, the value of $f(x)$ gradually approaches to -1 as $x$ tends to 2.

Since, $f(x)$ is continuous at $x=2$ , then limit of $f(x)$ as $x\to 2$ is the value of $f(2)$ .

  $\begin{array}{l}{\lim }_{x\to 2}(x^2-3x+1)\\ ={\lim }_{x\to 2}{x}^2-{\lim }_{x\to 2}3x+{\lim }_{x\to 2}1\\ ={\lim }_{x\to 2}{x}^2-3{\lim }_{x\to 2}x+1\\ =4-3\cdot 2+1\\ =4-6+1\\ =-1\end{array}$

Conclusion:

  ${\lim }_{x\to 2}(x^2-3x+1)=-1$

3

To determine

  ${\lim }_{x\to 3}\frac{(x-3)}{(x^2-x-6)}$ .

Answer to Problem 4RE

  ${\lim }_{x\to 3}\frac{(x-3)}{(x^2-x-6)}=0.2$

Explanation of Solution

Given:

  ${\lim }_{x\to 3}\frac{(x-3)}{(x^2-x-6)}$

$x$	2.9	2.99	2.999	3	3.001	3.01	3.1
$f(x)$

Calculation:

Here, $f(x)=\frac{(x-3)}{(x^2-x-6)}$

Therefore, $f(2.9)=\frac{(2.9-3)}{({2.9}^2-2.9-6)}=0.20408$

  $f(2.99)=\frac{(2.99-3)}{({2.99}^2-2.99-6)}=0.20040$

  $f(2.999)=\frac{(2.999-3)}{({2.999}^2-2.999-6)}=0.20004$

  $f(3)=\frac{(3-3)}{(3^2-3-6)}=\frac{0}{0}=$ does not exist

  $f(3.001)=\frac{(3.001-3)}{({3.001}^2-3.001-6)}=0.19996$

  $f(3.01)=\frac{(3.01-3)}{({3.01}^2-3.01-6)}=0.19960$

  $f(3.1)=\frac{(3.1-3)}{({3.1}^2-3.1-6)}=0.19608$

The table is given below:

$x$	2.9	2.99	2.999	3	3.001	3.01	3.1
$f(x)$	0.20408	0.20040	0.20004	Does not exist	0.19996	0.19960	0.19608

Numerically, from the table it is seen, ${\lim }_{x\to 3}\frac{(x-3)}{(x^2-x-6)}=0.2$ .

From the table it is clear, the value of $f(x)$ gradually increases to 0.2 from the left side of $x=3$ and gradually decreases to 0.2 from the right side of $x=3$ . That is, the value of $f(x)$ gradually approaches to 0.2 as $x$ tends to 3.

  $\begin{array}{l}{\lim }_{x\to 3}\frac{(x-3)}{(x^2-x-6)}\\ ={\lim }_{x\to 3}\frac{(x-3)}{(x^2-3x+2x-6)}\\ ={\lim }_{x\to 3}\frac{(x-3)}{\{x(x-3)+2(x-3)\}}\end{array}$

  $\begin{array}{l}={\lim }_{x\to 3}\frac{(x-3)}{(x-3)(x+2)}\\ ={\lim }_{x\to 3}\frac{1}{(x+2)}[\because x\ne 3,(x-3)\ne 0]\\ =\frac{{\lim }_{x\to 3}1}{{\lim }_{x\to 3}(x+2)}\end{array}$

  $\begin{array}{l}=\frac{1}{{\lim }_{x\to 3}x+{\lim }_{x\to 3}2}\\ =\frac{1}{{\lim }_{x\to 3}x+2}\\ =\frac{1}{3+2}\end{array}$

  $\begin{array}{l}=\frac{1}{5}\\ =0.2\end{array}$

Conclusion:

  ${\lim }_{x\to 3}\frac{(x-3)}{(x^2-x-6)}=0.2$

4

To determine

  ${\lim }_{x\to 0}\frac{\ln (1-x)}{x}$ .

Answer to Problem 4RE

  ${\lim }_{x\to 0}\frac{\ln (1-x)}{x}=-1$

Explanation of Solution

Given:

  ${\lim }_{x\to 0}\frac{\ln (1-x)}{x}$

$x$	-0.1	-0.01	-0.001	0	0.001	0.01	0.1
$f(x)$

Calculation:

Here, $f(x)=\frac{\ln (1-x)}{x}$

Therefore, $f(-0.1)=\frac{\ln (1+0.1)}{-0.1}=-0.9531$

  $f(-0.01)=\frac{\ln (1+0.01)}{-0.01}=-0.9950$

  $f(-0.001)=\frac{\ln (1+0.001)}{-0.001}==-0.9995$

  $f(0)=\frac{\ln (1-0)}{0}=\frac{0}{0}=$ does not exist

  $f(0.001)=\frac{\ln (1-0.001)}{0.001}=-1.0005$

  $f(0.01)=\frac{\ln (1-0.01)}{0.01}=-1.0050$

  $f(0.1)=\frac{\ln (1-0.1)}{0.1}=-1.0536$

The table is given below:

$x$	-0.1	-0.01	-0.001	0	0.001	0.01	0.1
$f(x)$	-0.9531	-0.9950	-0.9995	Does not exist	-1.0005	-1.0050	-1.0536

Numerically, from the table it is seen, ${\lim }_{x\to 0}\frac{\ln (1-x)}{x}=-1$ .

From the table it is clear, the value of $f(x)$ gradually decreases to -1 from the left side of $x=0$ and gradually increases to -1 from the right side of $x=0$ . That is, the value of $f(x)$ gradually approaches to -1 as $x$ tends to 0.

  ${\lim }_{x\to 0}\frac{\ln (1-x)}{x}$

Since, $\frac{0}{0}$ form, applying L’Hospital rule,

  $\begin{array}{l}{\lim }_{x\to 0}\frac{\ln (1-x)}{x}\\ ={\lim }_{x\to 0}\frac{\frac{d}{dx}\ln (1-x)}{\frac{d}{dx}x}\\ ={\lim }_{x\to 0}\frac{\frac{-1}{(1-x)}}{1}\end{array}$

  $\begin{array}{l}={\lim }_{x\to 0}\frac{-1}{(1-x)}\\ =\frac{{\lim }_{x\to 0}-1}{{\lim }_{x\to 0}(1-x)}\\ =\frac{-1}{{\lim }_{x\to 0}1-{\lim }_{x\to 0}x}\end{array}$

  $\begin{array}{l}=\frac{-1}{1-{\lim }_{x\to 0}x}\\ =\frac{-1}{1-0}\\ =\frac{-1}{1}\\ =-1\end{array}$

Conclusion:

  ${\lim }_{x\to 0}\frac{\ln (1-x)}{x}=-1$