Chapter 9.3, Problem 39E

a.

To determine

Tostate: that why $e^x$ grows faster than $x^n$ as $x\to \infty$ for any positive integer $n$ .

Answer to Problem 39E

The $e^x$ grows faster than $x^n$ as $x\to \infty$ for any positive integer $n$ .

Explanation of Solution

Given information:

The $e^x$ grows faster than $x^n$ as $x\to \infty$ for any positive integer $n$ .

Concept used: the definition of faster, slower, same rate growth,

1. $f$ grows faster than $g$ as $x\to \infty$if

  $\underset{x\to \infty }{\lim }\frac{f\left(x\right)}{g\left(x\right)}=\infty$

2. $f$ grows slower than $g$ as $x\to \infty$if

  $\underset{x\to \infty }{\lim }\frac{f\left(x\right)}{g\left(x\right)}=0$

3. $f$ and $g$ grow at same rate as $x\to \infty$if

  $\underset{x\to \infty }{\lim }\frac{f\left(x\right)}{g\left(x\right)}=L\ne 0$

to find the function grows faster, find the limit of ratio of the functions,

  $\begin{array}{l}\underset{x\to \infty }{\lim }\frac{x^n}{e^x}=\underset{x\to \infty }{\lim }\frac{n!}{e^x}\text{ [differentiate }n\text{ times]}\\ \underset{x\to \infty }{\lim }\frac{x^n}{e^x}=0\text{ [simplify]}\end{array}$

Thus, the $e^x$ grows faster than $x^n$ as $x\to \infty$ for any positive integer $n$ .

b.

To determine

To state: that why $a^x,a>1$ grows faster than $x^n$ as $x\to \infty$ for any positive integer $n$ .

Answer to Problem 39E

The $a^x,a>1$ grows faster than $x^n$ as $x\to \infty$ for any positive integer $n$ .

Explanation of Solution

Given information:

The $a^x,a>1$ grows faster than $x^n$ as $x\to \infty$ for any positive integer $n$ .

Concept used: the definition of faster, slower, same rate growth,

1. $f$ grows faster than $g$ as $x\to \infty$if

  $\underset{x\to \infty }{\lim }\frac{f\left(x\right)}{g\left(x\right)}=\infty$

2. $f$ grows slower than $g$ as $x\to \infty$if

  $\underset{x\to \infty }{\lim }\frac{f\left(x\right)}{g\left(x\right)}=0$

3. $f$ and $g$ grow at same rate as $x\to \infty$if

  $\underset{x\to \infty }{\lim }\frac{f\left(x\right)}{g\left(x\right)}=L\ne 0$

to find the function grows faster, find the limit of ratio of the functions,

  $\begin{array}{l}\underset{x\to \infty }{\lim }\frac{x^n}{a^x}=\underset{x\to \infty }{\lim }\frac{n!{\left(\ln a\right)}^n}{a^x}\text{ [differentiate }n\text{ times]}\\ \underset{x\to \infty }{\lim }\frac{x^n}{a^x}=0\text{ [simplify]}\end{array}$

Thus, the $a^x,a>1$ grows faster than $x^n$ as $x\to \infty$ for any positive integer $n$ .