Chapter 2.2, Problem 66E

(a)

To determine

To prove: The inequalities $\frac{x-1}{x}<\frac{\mathrm{int}x}{x}\le 1$ for $x>0$ and $\frac{x-1}{x}>\frac{\mathrm{int}x}{x}\ge 1$ for $x<0$ .

Explanation of Solution

Proof:

As the function $\mathrm{int}x$ lies between $x-1$ and $x$ for every $x>0$ . So, $x-1<\mathrm{int}x\le x$ .

Divide all the sides of inequality by $x$ .

  $\begin{array}{c}\frac{x-1}{x}<\frac{\mathrm{int}x}{x}\le \frac{x}{x}\\ \frac{x-1}{x}<\frac{\mathrm{int}x}{x}\le 1\end{array}$

As the function $\mathrm{int}x$ lies between $x-1$ and $x$ for every $x<0$ . So, $x-1<\mathrm{int}x\le x$ .

Divide all the sides of inequality by $x$ .

  $\begin{array}{c}\frac{x-1}{x}>\frac{\mathrm{int}x}{x}\ge \frac{x}{x}\\ \frac{x-1}{x}>\frac{\mathrm{int}x}{x}\ge 1\end{array}$

Hence both the inequalities $\frac{x-1}{x}<\frac{\mathrm{int}x}{x}\le 1$ for $x>0$ and $\frac{x-1}{x}>\frac{\mathrm{int}x}{x}\ge 1$ for $x<0$ are true.

(b)

To determine

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

Answer to Problem 66E

The value of $\underset{x\to \infty }{\lim }\left(\frac{\mathrm{int}x}{x}\right)$ is $1$ .

Explanation of Solution

Given information:

The limit $\underset{x\to \infty }{\lim }\left(\frac{\mathrm{int}x}{x}\right)$ .

Calculation:

As proved in part(a), the inequality is $\frac{x-1}{x}<\frac{\mathrm{int}x}{x}\le 1$ for $x>0$ .

Apply limit to all the terms of inequality.

  $\begin{array}{c}\underset{x\to \infty }{\lim }\frac{x-1}{x}<\underset{x\to \infty }{\lim }\frac{\mathrm{int}x}{x}\le \underset{x\to \infty }{\lim }1\\ \underset{x\to \infty }{\lim }\frac{x-1}{x}<\underset{x\to \infty }{\lim }\frac{\mathrm{int}x}{x}\le 1\end{array}$

Evaluate the limit.

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

By Sandwich theorem, the value of $\underset{x\to \infty }{\lim }\left(\frac{\mathrm{int}x}{x}\right)$ lies between $1$ and $1$ .

Therefore, the value of $\underset{x\to \infty }{\lim }\left(\frac{\mathrm{int}x}{x}\right)$ is $1$ .

(c)

To determine

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

Answer to Problem 66E

The value of $\underset{x\to -\infty }{\lim }\left(\frac{\mathrm{int}x}{x}\right)$ is $1$ .

Explanation of Solution

Given information:

The limit $\underset{x\to -\infty }{\lim }\left(\frac{\mathrm{int}x}{x}\right)$ .

Calculation:

As proved in part(a), the inequality is $\frac{x-1}{x}>\frac{\mathrm{int}x}{x}\ge 1$ for $x<0$ .

Apply limit to all the terms of inequality.

  $\begin{array}{c}\underset{x\to -\infty }{\lim }\frac{x-1}{x}>\underset{x\to -\infty }{\lim }\frac{\mathrm{int}x}{x}\ge \underset{x\to -\infty }{\lim }1\\ \underset{x\to -\infty }{\lim }\frac{x-1}{x}>\underset{x\to -\infty }{\lim }\frac{\mathrm{int}x}{x}\ge 1\end{array}$

Evaluate the limit.

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

By Sandwich theorem, the value of $\underset{x\to -\infty }{\lim }\left(\frac{\mathrm{int}x}{x}\right)$ lies between $1$ and $1$ .

Therefore, the value of $\underset{x\to -\infty }{\lim }\left(\frac{\mathrm{int}x}{x}\right)$ is $1$ .