Chapter D, Problem 10E

i.

To determine

To explain:

Formulation of the $\epsilon$ .

Answer to Problem 10E

  $\left|f\left(x\right)-L\right|<\epsilon$ is formulated as below.

Explanation of Solution

Given information:

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

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

ii.

To determine

To prove:

  $\underset{x\to 0^{+}}{\lim }\sqrt{x}=0$ using definition $I$

Answer to Problem 10E

  $\underset{x\to 0^{+}}{\lim }\sqrt{x}=0$ is proved.

Explanation of Solution

Given information:

  $\underset{x\to 0^{+}}{\lim }\sqrt{x}=0$

Calculation:

Here given function is,

  $f\left(x\right)=\sqrt{x}$

Need to prove that $\underset{x\to 0^{+}}{\lim }\sqrt{x}=0$

Consider $\epsilon$ be a positive number.

So need to find out $\delta$ such that,

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

i.e. if $0<x-0<\delta$ then $\left|\sqrt{x}-0\right|<\epsilon$ .

  $0<x<\delta$ then $\left|\sqrt{x}\right|<\epsilon$ .

We select, $\delta ={\epsilon }^2$ then

  $\begin{array}{l}0<x<\delta \\ 0<x<{\epsilon }^2\\ 0<\sqrt{x<\epsilon }\end{array}$

If $0<x-0<\delta$ then $\left|f\left(x\right)-0\right|<\epsilon$ where $\delta ={\epsilon }^2$

Thus, $\underset{x\to 0^{+}}{\lim }\sqrt{x}=0$