Chapter 8.4, Problem 1QQ

To determine

Which of the given functions out of the four given functions grows faster than $x^2$ as $x\to \infty$ .

Answer to Problem 1QQ

Option (D)

Explanation of Solution

Given information: Functions $x^2$ , $e^{-x}$ , $7x+10$ , $2x^2-3x$ and $0.1x^3$ are given.

Concept used: For any two functions $f\left(x\right)$ and $g\left(x\right)$ :

• $f\left(x\right)$ grows slower than $g\left(x\right)$ as $x\to \infty$ if $\underset{x\to \infty }{\lim }\frac{f\left(x\right)}{g\left(x\right)}=0$ .
• $f\left(x\right)$ grows faster than $g\left(x\right)$ as $x\to \infty$ if $\underset{x\to \infty }{\lim }\frac{f\left(x\right)}{g\left(x\right)}=\infty$ .
• $f\left(x\right)$ and $g\left(x\right)$ grow at same rate as $x\to \infty$ if $\underset{x\to \infty }{\lim }\frac{f\left(x\right)}{g\left(x\right)}=L\ne 0$ .

provided $f\left(x\right)$ and $g\left(x\right)$ are positive where $x$ is sufficiently large and L is a finite number other than zero.

Calculation:

Option(A):

Consider $f\left(x\right)=x^2$ and $g\left(x\right)=e^{-x}$ .

Determine $\underset{x\to \infty }{\lim }\frac{f\left(x\right)}{g\left(x\right)}$ for the considered functions.

  $\begin{array}{c}\underset{x\to \infty }{\lim }\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to \infty }{\lim }\frac{x^2}{e^{-x}}\\ =\underset{x\to \infty }{\lim }{e}^x{x}^2\\ =\underset{x\to \infty }{\lim }\left(e^x\right)\cdot \underset{x\to \infty }{\lim }\left(x^2\right)\\ =\infty \cdot \infty \\ =\infty \end{array}$

Since the obtained limit is equal to $\infty$ , $x^2$ grows faster than $e^{-x}$ or in other words $e^{-x}$ grows slower than $x^2$ .

As a result, option (A) is the not the correct option.

Option (B):

Consider $f\left(x\right)=x^2$ and $g\left(x\right)=7x+10$ .

Determine $\underset{x\to \infty }{\lim }\frac{f\left(x\right)}{g\left(x\right)}$ for the considered functions.

  $\begin{array}{c}\underset{x\to \infty }{\lim }\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to \infty }{\lim }\frac{x^2}{7x+10}\\ =\underset{x\to \infty }{\lim }\left(\frac{x}{7+\frac{10}{x}}\right)\\ =\frac{\underset{x\to \infty }{\lim }\left(x\right)}{\underset{x\to \infty }{\lim }\left(7+\frac{10}{x}\right)}\\ =\frac{\infty }{7}\\ =\infty \end{array}$

Since the obtained limit is equal to $\infty$ , $x^2$ grows faster than $7x+10$ or $7x+10$ grows slower than $x^2$ .

As a result, option (B) is the not the correct option.

Option (C):

Consider $f\left(x\right)=x^2$ and $g\left(x\right)=2x^2-3x$ .

Determine $\underset{x\to \infty }{\lim }\frac{f\left(x\right)}{g\left(x\right)}$ for the considered functions.

  $\begin{array}{c}\underset{x\to \infty }{\lim }\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to \infty }{\lim }\frac{x^2}{2x^2-3x}\\ =\underset{x\to \infty }{\lim }\frac{x^2}{x^2\left(2-\frac{3}{x}\right)}\\ =\underset{x\to \infty }{\lim }\frac{1}{\left(2-\frac{3}{x}\right)}\\ =\underset{x\to \infty }{\lim }{\left(2-\frac{3}{x}\right)}^{-1}\\ ={\left(\underset{x\to \infty }{\lim }2-\underset{x\to \infty }{\lim }\frac{3}{x}\right)}^{-1}\\ ={\left(2-3\underset{x\to \infty }{\lim }\frac{1}{x}\right)}^{-1}\\ ={\left(1-3\frac{\underset{x\to \infty }{\lim }1}{\underset{x\to \infty }{\lim }x}\right)}^{-1}\\ ={\left(1-3\left(\frac{1}{\infty }\right)\right)}^{-1}\\ ={\left(1-3\left(0\right)\right)}^{-1}\\ =1\\ =L\ne 0\end{array}$

Since the obtained limit is equal to 1 which is a finite number other than zero, $x^2$ grows as the same rate as $2x^2-3x$ .

As a result, option (C) is the not the correct choice.

Option (D):

Consider $f\left(x\right)=x^2$ and $g\left(x\right)=0.1x^3$ .

Determine $\underset{x\to \infty }{\lim }\frac{f\left(x\right)}{g\left(x\right)}$ for the considered functions.

  $\begin{array}{c}\underset{x\to \infty }{\lim }\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to \infty }{\lim }\frac{x^2}{0.1x^3}\\ =\underset{x\to \infty }{\lim }\left(\frac{1}{0.1x^2}\right)\\ =\underset{x\to \infty }{\lim }\left(\frac{10}{x^2}\right)\\ =\frac{\underset{x\to \infty }{\lim }\left(10\right)}{\underset{x\to \infty }{\lim }\left(x^2\right)}\\ =0\end{array}$

Since the obtained limit is equal to 0, $x^2$ grows slower than $0.1x^3$ or in other words $0.1x^3$ grows faster than $x^2$ .

As a result, option (D) is the correct choice.