Chapter 1.2, Problem 66E

(a)

To determine

To show:. $\frac{x-1}{x}<\frac{\mathrm{int}(x)}{x}\le 1$ and $\frac{x-1}{x}>\frac{\mathrm{int}(x)}{x}\ge 1$

Explanation of Solution

Given information:

  $\frac{x-1}{x}<\frac{\mathrm{int}(x)}{x}\le 1$

  $\frac{x-1}{x}>\frac{\mathrm{int}(x)}{x}\ge 1$

Let $x$  stand in for a genuine number. if $x$ is a positive integer

  $\mathrm{int}(x)=x$

There will be two consecutive integers k and $k+1$ such that if $x$ is not an integer, then

All components of the inequality are subtracted by 1,

  $k-1<x-1<k$

  $x-1<k<x$

By the definition of the greatest integer function, $\mathrm{int}(x)=k$ and so

  $x-1<\mathrm{int}(x)<x$

If combine the scenarios in which $x$ is both an integer and not, you will obtain the inequality.

  $x-1<\mathrm{int}(x)\le x$

Divide all members of the inequality by $x$ , if $x>0$ result in

  $\frac{x-1}{x}<\frac{\mathrm{int}(x)}{x}\le 1$

If $x<0$ ,

  $\frac{x-1}{x}>\frac{\mathrm{int}(x)}{x}\ge 1$

(b)

To determine

To find: The value of $\underset{x\to \infty }{\lim }\frac{\text{ int }x}{x}=1$

Answer to Problem 66E

The limit is 1.

Explanation of Solution

Given information: The limit is $\underset{x\to \infty }{\lim }\frac{\text{ int }x}{x}$

Calculation:

  $\begin{array}{c}\underset{x\to \infty }{\lim }\frac{x-1}{x}=\underset{x\to \infty }{\lim }\left(1-\frac{1}{x}\right)\\ =1-0\\ =1\\ =\underset{x\to \infty }{\lim }1\end{array}$

To use the Sandwich Theorem and component (a),

  $\underset{x\to \infty }{\lim }\frac{\text{ int }x}{x}=1$

(c)

To determine

To find: The value of $\underset{x\to -\infty }{\lim }\frac{\mathrm{int}(x)}{x}$

Answer to Problem 66E

The limit is 1.

Explanation of Solution

Given information: The limit is $\underset{x\to -\infty }{\lim }\frac{\mathrm{int}(x)}{x}$

Calculation:

The given limit is:

  $\underset{x\to -\infty }{\lim }\frac{\mathrm{int}(x)}{x}$

To apply the Squeeze theorem, prove that the extremities have a right end behavior limit and that it is the same,

  $\underset{x\to -\infty }{\lim }1=1$

And

  $\begin{array}{c}\underset{x\to -\infty }{\lim }\frac{x-1}{x}=\underset{x\to \infty }{\lim }\frac{x}{x}\\ =1\end{array}$

As 1 is an end behavior model for $(x-1)/x$

As the extremities have same limit ,then so will the same function, and therefore,

  $\underset{x\to -\infty }{\lim }\frac{\mathrm{int}(x)}{x}=1$

Therefore, the required $\underset{x\to -\infty }{\lim }\frac{\mathrm{int}(x)}{x}=1$ .