Chapter D, Problem 13E

(a).

To determine

To calculate:

Value of $\delta$ .

Answer to Problem 13E

Value of $\delta$ is $0<\delta <0.025$ .

Explanation of Solution

Given information:

If $\left|x-2\right|<\delta$ then $\left|4x-8\right|<\epsilon$

  $\epsilon =0.1$

Calculation:

Let $f$ be a function defined on some open interval that contains the number $a$ , except possibly $a$ itself then it is said that limit of $f\left(x\right)$ as $x$ approaches $a$ is $L$ .

It is write as,

  $\underset{x\to a}{\lim }f\left(x\right)=L$

If for every number $\epsilon >0$ there is a corresponding number $\delta >0$ such that

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

Consider $f\left(x\right)=4x$ .

So have to find $\delta$ such that,

  $\left|4x-8\right|<0.1$

Or $-0.1<4x-8<0.1$

Or $7.9<4x<8.1$

Or $1.975<x<2.025$

At $y=7.9,8.1$ the value of $x$ is $1.975$ and $2.025$ .

Since the interval $\left(1.975,2.025\right)$ symmetric about the line $x=2$ thus both the points $\left(1.975,7.9\right)$ and $\left(2.025,8.1\right)$ are equidistant from the line.

So the shortest distance is,

  $\left|2-1.975\right|=0.025$

Thus, $\left|x-2\right|<0.025$ then $\left|4x-8\right|<0.1$

So value of $\delta$ is $0<\delta <0.025$ .

(b).

To determine

To calculate:

Value of $\delta$ when $\epsilon =0.01$

Answer to Problem 13E

Value of $\delta$ is $0<\delta <0.0025$ .

Explanation of Solution

Given information:

If $\left|x-2\right|<\delta$ then $\left|4x-8\right|<\epsilon$

  $\epsilon =0.01$

Calculation:

Let $f$ be a function defined on some open interval that contains the number $a$ , except possibly $a$ itself then it is said that limit of $f\left(x\right)$ as $x$ approaches $a$ is $L$ .

It is write as,

  $\underset{x\to a}{\lim }f\left(x\right)=L$

If for every number $\epsilon >0$ there is a corresponding number $\delta >0$ such that

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

Consider $f\left(x\right)=4x$ .

So have to find $\delta$ such that,

If $0<\left|x-2\right|<\delta$ then $\left|f\left(x\right)-8\right|<0.01$ .

  $\left|4x-8\right|<0.01$

Or $-0.01<4x-8<0.01$

Or $7.99<4x<8.01$

Or $1.9975<x<2.0025$

At $y=7.99,8.01$ the value of $x$ is $1.9975$ and $2.0025$ .

At $y=7.9,8.01$ the value of v is $1.9975$ and $2.0025$ .

Since the interval $\left(1.9975,2.0025\right)$ symmetric about the line $x=2$ thus both the points $\left(1.9975,7.99\right)$ and $\left(2.0025,8.01\right)$ are equidistant from the line.

So the shortest distance is,

  $\left|2-1.9975\right|=0.025$

Thus, $\left|x-2\right|<0.0025$ then $\left|4x-8\right|<0.01$

So value of $\delta$ is $0<\delta <0.0025$ .