Chapter 9.3, Problem 40E

a.

To determine

Toshow: that $e^x$ grows faster than any polynomial $a_n{x}^n+a_{n-1}{x}^{n-1}+\mathrm{....}+a_{1}x+a_0,a_n>0$ as $x\to \infty$ .

Explanation of Solution

Given information:

The $e^x$ grows faster than aby polynomial $a_n{x}^n+a_{n-1}{x}^{n-1}+\mathrm{....}+a_{1}x+a_0,a_n>0$ as $x\to \infty$ .

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

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

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

  $\begin{array}{l}\underset{x\to \infty }{\lim }\frac{a_n{x}^n+a_{n-1}{x}^{n-1}+\mathrm{....}+a_{1}x+a_0}{e^x}=\underset{x\to \infty }{\lim }\frac{n!a_n}{e^x}\text{ [differentiate }n\text{ times]}\\ \underset{x\to \infty }{\lim }\frac{a_n{x}^n+a_{n-1}{x}^{n-1}+\mathrm{....}+a_{1}x+a_0}{e^x}=0\text{ [simplify]}\end{array}$

Thus, the $e^x$ grows faster than any polynomial $a_n{x}^n+a_{n-1}{x}^{n-1}+\mathrm{....}+a_{1}x+a_0,a_n>0$ as $x\to \infty$ .

b.

To determine

To show: that $a^x,a>1$ grows faster than any polynomial $a_n{x}^n+a_{n-1}{x}^{n-1}+\mathrm{....}+a_{1}x+a_0,a_n>0$ as $x\to \infty$ .

Explanation of Solution

Given information:

The $a^x,a>1$ grows faster than any polynomial $a_n{x}^n+a_{n-1}{x}^{n-1}+\mathrm{....}+a_{1}x+a_0,a_n>0$ as $x\to \infty$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

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

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

  $\begin{array}{l}\underset{x\to \infty }{\lim }\frac{a_n{x}^n+a_{n-1}{x}^{n-1}+\mathrm{....}+a_{1}x+a_0}{a^x}=\underset{x\to \infty }{\lim }\frac{n!{\left(\ln a\right)}^n{a}_n}{a^x}\text{ [differentiate }n\text{ times]}\\ \underset{x\to \infty }{\lim }\frac{a_n{x}^n+a_{n-1}{x}^{n-1}+\mathrm{....}+a_{1}x+a_0}{a^x}=0\text{ [simplify]}\end{array}$

Thus, the $a^x,a>1$ grows faster than any polynomial $a_n{x}^n+a_{n-1}{x}^{n-1}+\mathrm{....}+a_{1}x+a_0,a_n>0$ as $x\to \infty$ .