Chapter 1.1, Problem 78E

(a)

To determine

To prove: $\underset{x\to 2}{\lim }f(x)=f(2)$

Explanation of Solution

Given information: The limit $\underset{x\to 2}{\lim }f(x)=f(2)$

  $\begin{array}{c}\underset{x\to 2}{\lim }f(x)=f(2)\\ =\sqrt{3(2)-2}\\ =\sqrt{4}\\ =2\end{array}$

Hence, the given statement $\underset{x\to 2}{\lim }f(x)=f(2)$ is proved.

(b)

To determine

To calculate: The possible value of $a$ and $b$

Answer to Problem 78E

The possible value of $a$ : $1.7467$

The possible value of $b<2.28$

Explanation of Solution

Given information: The function is $1.8<f(x)<2.2$

Graph:

In the same coordinate plane $f(x)$

  $\begin{array}{l}y=1.8\\ y=2.2\end{array}$

Use the viewing window $[1.4,2.8],[1.4,2.6]$

  

Locate the intersections at approximately $(1.7467,1.8)$ and exactly $(2.28,2.2)$

Must be careful to tighten the interval $(a,b)$ so that the inequality is satisfied since it is crucial to remember that we are requested to ensure that $1.8<f(x)<2.2$ is true.

Thus, it follows that as a result of $f(2.28)=2.2$ then $b$ must be less than $2.28$

Choose $b=2.279$ for instance or $b=2.27$ or, any value of $b<2.28$

On the other hand, the lower bound has an approximate value of $1.7467$ , therefore we must round it up to ensure the inequality $1.747,1.75$ and so on are potential values for $a$

Any number that is rounded up from $1.747,1.75$ is a candidate for the value of $a$ .

The possible value of $b$ : $b<2.28$

(c)

To determine

To calculate: The possible value of $a$ and $b$

Answer to Problem 78E

The possible value of $a>1.9867$

And $b=2.0134$

Explanation of Solution

Given information: The function is $1.99<f(x)<2.01$

Graph:

In the same coordinate plane $f(x)$

  $\begin{array}{l}y=1.99\\ y=2.01\end{array}$

Use the viewing window $[1.98,2.02],[1.98,2.02]$

  

Locate the intersections at approximately $(1.9867,1.99)$ and about $(2.0134,2.01)$

Must be careful to tighten the interval $(a,b)$ so that the inequality is satisfied since it is crucial to remember that we are requested to ensure that $1.99<f(x)<2.01$ is true.

Thus, it follows that as a result of $f(1.9867)=1.99$ , $a$ must be greater than $1.9867$ for instance $a=1.987$ for instance or $a=1.99$ or, the value of $a>1.9867$

On the other hand, the upper bound has an approximate value of $2.0134$ , therefore we must round it down to ensure the inequality. $b$ could have values of $2.013,2.01$ , and so on.

This exercise's goal is to demonstrate how, as we tighten the interval $(a,b)$ about 2

The interval of $f$ around of 2

Hence $\underset{x\to 2}{\lim }f(x)=2$ If $1.9999<x<2.0001$ then $1.999925<f(x)<2.000074$

Therefore, the possible value for $a:$

  $a>1.9867$

The possible value for $b=2.0134$