Chapter 9.3, Problem 29E

To determine

Todetermine: the order of functions from slowest-growing to fastest-growing as $x\to \infty$ .

Answer to Problem 29E

The order of function from slowest-growing to fastest-growing as $x\to \infty$ is $e^{\frac{x}{2}},e^x,{\left(\ln x\right)}^x,x^x$ .

Explanation of Solution

Given information:

The functions are $e^x,x^x,{\left(\ln x\right)}^x,e^{\frac{\frac{x}{2}}{}}$ .

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 order, first find the limit of ratio of the functions,

  $\begin{array}{l}\underset{x\to \infty }{\lim }\frac{e^{\frac{x}{2}}}{e^x}=\underset{x\to \infty }{\lim }\frac{1}{e^{\frac{x}{2}}}\text{ [simplify]}\\ \underset{x\to \infty }{\lim }\frac{e^{\frac{x}{2}}}{e^x}=0\text{ [simplify]}\end{array}$

Thus, by the definition of the faster, slower, same rate growth, of the functions,

The function $e^{\frac{x}{2}}$ grows slower than function $e^x$ .

  $\underset{x\to \infty }{\lim }\frac{e^x}{{\left(\ln x\right)}^x}=0$

Thus, by the definition of the faster, slower, same rate growth, of the functions,

The function ${\left(\ln x\right)}^x$ grows slower than function $e^{\frac{x}{2}}$ .

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

Thus, by the definition of the faster, slower, same rate growth, of the functions,

The function $x^x$ grows slower than function ${\left(\ln x\right)}^x$ .

Thus, the order is $e^{\frac{x}{2}},e^x,{\left(\ln x\right)}^x,x^x$ .