Chapter 8.3, Problem 30E

To determine

To arrange: The given expressions from the slowest-growing to the fastest-growing, when $x\to \infty$ .

Answer to Problem 30E

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

Explanation of Solution

Given information: The given expressions are $2^x,x^2,{\left(\ln 2\right)}^x,e^x$.

Concept used: Consider that $f\left(x\right)$ and $g\left(x\right)$ be the functions which are positive when $x$ is very large.

So, if $\underset{x\to \infty }{\lim }\left[\frac{f\left(x\right)}{g\left(x\right)}\right]=\infty$ or equivalently $\underset{x\to \infty }{\lim }\left[\frac{g\left(x\right)}{f\left(x\right)}\right]=0$ provided that $x\to \infty$ , then $f\left(x\right)$ grows faster than $g\left(x\right)$ .

Similarly, if $\underset{x\to \infty }{\lim }\left[\frac{f\left(x\right)}{g\left(x\right)}\right]=L$ , where $L$ is nonzero and finite number, then $f\left(x\right)$ and $g\left(x\right)$ grows at the same rate.

Calculation:

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

Now, find the value of $\underset{x\to \infty }{\lim }\left[\frac{f\left(x\right)}{g\left(x\right)}\right]$ .

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

Since the value of $\underset{x\to \infty }{\lim }\left[\frac{f\left(x\right)}{g\left(x\right)}\right]=\infty$ , it can be said that $f\left(x\right)$ grows faster than $g\left(x\right)$ .

Now, consider that $f\left(x\right)=2^x$ and $g\left(x\right)={\left(\ln 2\right)}^x$ .

Now, find the value of $\underset{x\to \infty }{\lim }\left[\frac{f\left(x\right)}{g\left(x\right)}\right]$ .

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

Since the value of $\underset{x\to \infty }{\lim }\left[\frac{f\left(x\right)}{g\left(x\right)}\right]=\infty$ , it can be said that $f\left(x\right)$ grows faster than $g\left(x\right)$ .

Now, again consider that $f\left(x\right)=2^x$ and $g\left(x\right)=e^x$ .

Now, find the value of $\underset{x\to \infty }{\lim }\left[\frac{f\left(x\right)}{g\left(x\right)}\right]$ .

  $\underset{x\to \infty }{\lim }\left[\frac{2^x}{e^x}\right]=0$

Since the value of $\underset{x\to \infty }{\lim }\left[\frac{f\left(x\right)}{g\left(x\right)}\right]=0$ , it can be said that $f\left(x\right)$ grows slower than $g\left(x\right)$ .

Now, again consider that $f\left(x\right)=x^2$ and $g\left(x\right)={\left(\ln 2\right)}^x$ .

Now, find the value of $\underset{x\to \infty }{\lim }\left[\frac{f\left(x\right)}{g\left(x\right)}\right]$ .

  $\underset{x\to \infty }{\lim }\left[\frac{x^2}{{\left(\ln 2\right)}^x}\right]=\infty$

Since the value of $\underset{x\to \infty }{\lim }\left[\frac{f\left(x\right)}{g\left(x\right)}\right]=\infty$ , it can be said that $f\left(x\right)$ grows faster than $g\left(x\right)$ .

From the above-calculations, the required order can be arranged as ${\left(\ln 2\right)}^x,x^2,2^x,e^x$ .