Chapter A3, Problem 15E

(a)

To determine

To find: the interval.

Answer to Problem 15E

Here, the interval is $I=\left(5,5+{\epsilon }^2\right)$ .

Explanation of Solution

Given:

  $\epsilon >0$ and $I=\left(5,5+\delta \right),\delta >0$

Such that if x lies in I then $\sqrt{x-5<\epsilon }$ .

Calculation:

The interval is calculated as,

  $\begin{array}{l}\sqrt{\left(5+\delta \right)-5}=\delta \\ \Rightarrow \sqrt{\delta }=\epsilon \\ \Rightarrow \delta ={\epsilon }^2\\ \Rightarrow I=\left(5,5+{\epsilon }^2\right)\end{array}$

Conclusion:

Hence, the interval is $I=\left(5,5+{\epsilon }^2\right)$ .

(b)

To determine

To find: the limit that is verified.

Answer to Problem 15E

Here, the limit is $\underset{x\to 5^{+}}{\lim }\sqrt{x-5}=0$ .

Explanation of Solution

Calculation:

  $\begin{array}{l}\underset{x\to 5^{+}}{\lim }\sqrt{x-5}=\sqrt{5-5}\\ \text{ = 0}\end{array}$

Therefore, $\underset{x\to 5^{+}}{\lim }\sqrt{x-5}=0$ .

Conclusion:

Hence, the limit is $\underset{x\to 5^{+}}{\lim }\sqrt{x-5}=0$ .