Chapter 9.3, Problem 24E

To determine

Todetermine: that whether the function grows faster than $x^2$ , at the same rate as $x^2$ or slower than $x^2$ as $x\to \infty$ .

Answer to Problem 24E

The function $2^x$ grows faster than function $x^2$ .

Explanation of Solution

Given information:

The functionis $2^x$ .

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 determine the function grows faster, slower or same rate, find the limit of the ration of functions,

  $\begin{array}{l}\underset{x\to \infty }{\lim }\frac{2^x}{x^2}=\underset{x\to \infty }{\lim }\frac{2^x}{2{\left(\ln 2\right)}^2}\text{ [differentiate]}\\ \underset{x\to \infty }{\lim }\frac{2^x}{x^2}=\infty \text{ [simplify]}\end{array}$

By the definition of faster, slower or same rate growth of functions ,

The function $2^x$ grows faster thanfunction $x^2$ .