Rates of Growth

We first considered exponential functions because of their rapid growth. Now we would like to say more precisely how fast they grow by comparing them to some other functions. In particular, we will show that the function $e^x$ grows much more rapidly than any power function $x^p$ where $ p $ is a positive exponent.

First, we'll create a measurement of the relative rates of growth by considering the ratio of to . That means we will consider the function

$f(x) = \frac{x^p}{e^x}$

Notice that when this function is small, it means that the denominator, , is larger than the numerator, . We will show that we can make this ratio as small as we like by increasing the values of $ x $ that we consider. This will be our first principle: to study what happens to this ratio $ f(x) $ as becomes very large.

Statement of Results

As mentioned above, we will show you that the exponential function grows much faster than the power function as becomes very large. We will write this as
$\lim_{x\to\infty} \frac{x^p}{e^x} = 0$
which is another of expressing the fact that the ratio between the two functions can be made as small as we like by considering large values of . What is remarkable is that this is true for any power function. We saw early in the course that the graphs of power functions can become very steep when the exponent is large. Our result here is saying that an exponential function will always eventually be much larger than that power function.
In addition, it is true that logarithms grow much more slowly than power functions. That is to say
$\lim_{x\to\infty}\frac{\ln(x)}{x^p} = 0$
for any power function. We will not explicitly show this to you but you may wish to think about it as following naturally from the fact that the natural logarithm function is the inverse of the exponential function.
To accomplish our aim, we are going to use some of the techniques we have already developed. However, you may find that this is a little more difficult than what we have done up to this point. If so, that's all right---you may view this section as optional. Others may gain something from seeing a slightly more sophisticated application of our ideas.

The derivative of

Since we are interested in how this function grows, let's compute the derivative of $f(x) = \frac{x^p}{e^x}$ :
$\begin{eqnarray*} f^\prime(x) & = & \frac{px^{p-1}e^x - x^pe^x}{e^{2x}} \\ \\ & = & \frac{px^{p-1} - x^p}{e^x} \\ & = & (\frac{p}{x} - 1) \frac{x^p}{e^x} \end{eqnarray*}$
Remember that we are interested in what happens for very large values of and for large values of , the term $\frac px$ is very small. That means that the derivative of $ f(x) $ is approximately equal to $ -f(x)
$ :
$f^\prime(x) \approx -f(x)$
This is interesting because it suggests that is behaving very much like a decaying exponential. If we knew this, it would be enough to conclude that becomes very close to zero for large values of .
In fact, if we consider values of such that $2p \leq x$ , then we have $\frac px \leq \frac 12$ and so
$f^\prime(x) \leq -\frac 12 f(x)$
Notice that if we had equality, we could conclude that $ f(x)
$ was $Ce^{-x/2}$ for some constant $ C $ . Since we do not have equality, we cannot assume this but it gives us a hint as to how to proceed as we'll see shortly.
This should remind you of a differential equation in that it is a relation between the derivative of $ f(x) $ and the value of . We call such relationships differential inequalities and we may use them to derive information about the function just as we may when presented with a differential equation.

Comparison with an exponential

The differential inequality gives us the feeling that $ f(x)
$ might be related to the exponential function
$g(x) = Ce^{-x/2}$
Notice that $g^\prime(x) = -\frac 12 g(x)$ . Let us choose the constant $C = \frac{(2p)^p}{e^{3p}}$ . This seems strange, but the reason we do it is so that $
g(2p) = f(2p) $ . Remember that our inequality for $f^\prime(x)$ holds only for $x \geq 2p$ and so we are setting up $ g(x) $ so that the two functions agree at the start of this region.
How can we compare $ f(x) $ and ? By considering their ratio
$h(x) = \frac{f(x)}{g(x)}$
Notice that $h(2p) = \frac{f(2p)}{g(2p)} = 1$ . Also
$\begin{eqnarray*} h^\prime(x) & = & \frac{f^\prime(x)g(x) - f(x) g^\prime(x)}{[g(x)]^2} \\ \\ & \leq & \frac 12\frac{f(x)g(x) - f(x)g(x)}{[g(x)]^2} = 0 \end{eqnarray*}$
In other words, $h^\prime(x) \leq 0$ which means that $ h(x) $ is a decreasing function when $x \geq 2p$ . Since $ h(2p) = 1 $ , we may conclude that
$h(x) \leq 1 ~~~\mbox{ or }~~~f(x) \leq g(x)$
when $x \geq 2p$ . In other words, we know that
$0 \leq \frac{x^p}{e^x} \leq Ce^{-x/2}$
Now as becomes very large, the exponential $Ce^{-x/2}$ becomes very close to 0. Since the function $\frac{x^p}{e^x}$ is trapped between 0 and the exponential, it too must be very close to 0 and so we conclude that
$\lim_{x\to\infty} \frac{x^p}{e^x} = 0$
which is what we set out to see at the top of the page.

You may have found this argument a little confusing. That's all right if you did. It is our intention merely to give you a feeling for how one might use the tools we have developed so far---computing derivatives and understanding their implications for increasing and decreasing functions---to reach a slightly deeper result.