Chapter D, Problem 8E

To determine

To calculate:

Value of $\delta$ .

Answer to Problem 8E

The value of number $\delta$ is $0.7$ and $0.11$

Explanation of Solution

Given information:

Condition given: $\epsilon =0.5,\epsilon =0.1$ .

Calculation:

Let, $f$ be a function defined on some open interval that contains the number $a$ , except possibly $a$ itself.

So, limit of $f\left(x\right)$ as $x$ approaches $a$ is $L$ , so it is write as $\underset{x\to a}{\lim }f\left(x\right)=L$ .

In case every number $\epsilon >0$ there is a corresponding number $\delta >0$ such that

If $0<\left|x-a\right|<\delta$ and $\left|f\left(x\right)-L\right|<\epsilon$

The given function is,

  $f\left(x\right)=\frac{e^x-1}{x}$

So for first case $\epsilon =0.5$

Then we have to find $\delta$ such that,

if $\left|x-0\right|<\delta$ then $\left|f\left(x\right)-1\right|<0.5$

now,

  $\left|\frac{e^x-1}{x}-1\right|<0.5$

Or $-0.5<\frac{e^x-1}{x}-1<0.5$

Or $0.5<\frac{e^x-1}{x}-1<1.5$

Here need to determine values for $x$ for which given curve lies between horizontal lines $y=0.5$ and $y=1.5$ .

Graph and curves as shows below:

  

From the graph it is observed that curve intersects the lines $y=0.5$ and $y=1.5$ when $x\approx 0.77,-1.6$ .

Nearest point from $x=0$ is $0.77$ .

Value of $\delta$ is $0<\delta <0.77$

When choose $\delta =0.7$ then $\left|x-0\right|<0.7$ implies $\left|f\left(x\right)-1\right|<0.5$ .

For second case, $\epsilon =0.1$

Then we have to find $\delta$ such that,

if $\left|x-0\right|<\delta$ then $\left|f\left(x\right)-1\right|<0.1$

now,

  $\left|\frac{e^x-1}{x}-1\right|<0.1$

Or $-0.1<\frac{e^x-1}{x}-1<0.1$

Or $0.9<\frac{e^x-1}{x}-1<1.1$

Here need to determine values for $x$ for which the c $y=\frac{e^x-1}{x}$ urve lies between horizontal lines $y=0.9$ and $y=1.1$ .

Graph and curves as shows below:

  

From the graph it is observed that curve intersects the lines $y=0.9$ and $y=1.1$ when $x\approx -0.2152,0.187$ .

Nearest point from $x=1$ is $0.187$

Value of $\delta$ is $0<\delta <0.187$

When choose $\delta =0.11$ then $\left|x-1\right|<0.11$ implies $\left|f\left(x\right)-1\right|<0.1$ .

Hence by definition result is $\underset{x\to a}{\lim }\left(\frac{e^x-1}{x}\right)=1$