Growth Rates of Functions

Learning Outcomes

Describe the relative growth rates of functions

Suppose the functions $f$ and $g$ both approach infinity as $x\to \infty$ . Although the values of both functions become arbitrarily large as the values of $x$ become sufficiently large, sometimes one function is growing more quickly than the other. For example, $f(x)=x^2$ and $g(x)=x^3$ both approach infinity as $x\to \infty$ . However, as shown in the following table, the values of $x^3$ are growing much faster than the values of $x^2$ .

Comparing the Growth Rates of $x^2$ and $x^3$
$x$	10	100	1000	10,000
$f(x)=x^2$	100	10,000	1,000,000	100,000,000
$g(x)=x^3$	1000	1,000,000	1,000,000,000	$1,000,000,000,000$

In fact,

$\underset{x\to \infty }{\lim}\dfrac{x^3}{x^2}=\underset{x\to \infty}{\lim} x=\infty$ or, equivalently,

$\underset{x\to \infty}{\lim}\dfrac{x^2}{x^3}=\underset{x\to \infty }{\lim}\dfrac{1}{x}=0$

As a result, we say $x^3$ is growing more rapidly than $x^2$ as $x\to \infty$ . On the other hand, for $f(x)=x^2$ and $g(x)=3x^2+4x+1$ , although the values of $g(x)$ are always greater than the values of $f(x)$ for $x>0$ , each value of $g(x)$ is roughly three times the corresponding value of $f(x)$ as $x\to \infty$ , as shown in the following table. In fact,

$\underset{x\to \infty }{\lim}\dfrac{x^2}{3x^2+4x+1}=\dfrac{1}{3}$

Comparing the Growth Rates of $x^2$ and $3x^2+4x+1$
$x$	10	100	1000	10,000
$f(x)=x^2$	100	10,000	1,000,000	100,000,000
$g(x)=3x^2+4x+1$	341	30,401	3,004,001	300,040,001

In this case, we say that $x^2$ and $3x^2+4x+1$ are growing at the same rate as $x\to \infty$ .

More generally, suppose $f$ and $g$ are two functions that approach infinity as $x\to \infty$ . We say $g$ grows more rapidly than $f$ as $x\to \infty$ if

$\underset{x\to \infty }{\lim}\dfrac{g(x)}{f(x)}=\infty$ or, equivalently,

$\underset{x\to \infty }{\lim}\dfrac{f(x)}{g(x)}=0$

On the other hand, if there exists a constant $M \ne 0$ such that

$\underset{x\to \infty }{\lim}\dfrac{f(x)}{g(x)}=M$ ,

we say $f$ and $g$ grow at the same rate as $x\to \infty$ .

Next we see how to use L’Hôpital’s rule to compare the growth rates of power, exponential, and logarithmic functions.

example: Comparing the Growth Rates of $\ln x$ , $x^2$ , and $e^x$

For each of the following pairs of functions, use L’Hôpital’s rule to evaluate $\underset{x\to \infty }{\lim}\left(\dfrac{f(x)}{g(x)}\right)$ .

$f(x)=x^2$ and $g(x)=e^x$
$f(x)=\ln x$ and $g(x)=x^2$

Show Solution

Since $\underset{x\to \infty }{\lim} x^2=\infty$ and $\underset{x\to \infty }{\lim} e^x= \infty$ , we can use L’Hôpital’s rule to evaluate $\underset{x\to \infty }{\lim}\left[\frac{x^2}{e^x}\right]$ . We obtain
$\underset{x\to \infty }{\lim}\frac{x^2}{e^x}=\underset{x\to \infty }{\lim}\frac{2x}{e^x}$

Since $\underset{x\to \infty }{\lim}2x=\infty$ and $\underset{x\to \infty }{\lim}e^x=\infty$ , we can apply L’Hôpital’s rule again. Since

$\underset{x\to \infty }{\lim}\frac{2x}{e^x}=\underset{x\to \infty }{\lim}\frac{2}{e^x}=0$ ,

we conclude that

$\underset{x\to \infty }{\lim}\frac{x^2}{e^x}=0$

Therefore, $e^x$ grows more rapidly than $x^2$ as $x\to \infty$ (See Figure 3 and the table below).

Figure 3. An exponential function grows at a faster rate than a power function.

Growth rates of a power function and an exponential function.

$x$ 5 10 15 20

$x^2$ 25 100 225 400

$e^x$ 148 22,026 3,269,017 485,165,195
Since $\underset{x\to \infty }{\lim} \ln x=\infty$ and $\underset{x\to \infty }{\lim} x^2=\infty$ , we can use L’Hôpital’s rule to evaluate $\underset{x\to \infty }{\lim}\frac{\ln x}{x^2}$ . We obtain
$\underset{x\to \infty }{\lim}\frac{\ln x}{x^2}=\underset{x\to \infty }{\lim}\frac{1/x}{2x}=\underset{x\to \infty }{\lim}\frac{1}{2x^2}=0$

Thus, $x^2$ grows more rapidly than $\ln x$ as $x\to \infty$ (see Figure 4 and the table below).

Figure 4. A power function grows at a faster rate than a logarithmic function.

Growth rates of a power function and a logarithmic function

$x$ 10 100 1000 10,000

$\ln x$ 2.303 4.605 6.908 9.210

$x^2$ 100 10,000 1,000,000 100,000,000

Watch the following video to see the worked solution to Example: Comparing the Growth Rates of $\ln x$ , $x^2$ , and $e^x$ .

Closed Captioning and Transcript Information for Video

Try It

Compare the growth rates of $x^{100}$ and $2^x$ .

Hint

Apply L’Hôpital’s rule to $\frac{x^{100}}{2^x}$

Show Solution

The function $2^x$ grows faster than $x^{100}$ .

Using the same ideas as in the last example. it is not difficult to show that $e^x$ grows more rapidly than $x^p$ for any $p>0$ . In Figure 5 and the table below it, we compare $e^x$ with $x^3$ and $x^4$ as $x\to \infty$ .

This figure has two figures marked a and b. In figure a, the functions y = ex and y = x3 are graphed. It is obvious that ex increases more quickly than x3. In figure b, the functions y = ex and y = x4 are graphed. It is obvious that ex increases much more quickly than x4, but the point at which that happens is further to the right than it was for x3.

Figure 5. The exponential function $e^x$ grows faster than $x^p$ for any $p>0$ . (a) A comparison of $e^x$ with $x^3$ . (b) A comparison of $e^x$ with $x^4$ .

An exponential function grows at a faster rate than any power function
$x$	5	10	15	20
$x^3$	125	1000	3375	8000
$x^4$	625	10,000	50,625	160,000
$e^x$	148	22,026	3,269,017	485,165,195

Similarly, it is not difficult to show that $x^p$ grows more rapidly than $\ln x$ for any $p>0$ . In Figure 6 and the table below it, we compare $\ln x$ with $\sqrt[3]{x}$ and $\sqrt{x}$ .

This figure shows y = the square root of x, y = the cube root of x, and y = ln(x). It is apparent that y = ln(x) grows more slowly than either of these functions.

Figure 6. The function $y=\ln x$ grows more slowly than $x^p$ for any $p>0$ as $x\to \infty$ .

A logarithmic function grows at a slower rate than any root function
$x$	10	100	1000	10,000
$\ln x$	2.303	4.605	6.908	9.210
$\sqrt[3]{x}$	2.154	4.642	10	21.544
$\sqrt{x}$	3.162	10	31.623	100

Module 4: Applications of Derivatives