Learning Outcomes
- Describe the relative growth rates of functions
Suppose the functions ff and gg both approach infinity as x→∞x→∞. Although the values of both functions become arbitrarily large as the values of xx become sufficiently large, sometimes one function is growing more quickly than the other. For example, f(x)=x2f(x)=x2 and g(x)=x3g(x)=x3 both approach infinity as x→∞x→∞. However, as shown in the following table, the values of x3x3 are growing much faster than the values of x2x2.
xx | 10 | 100 | 1000 | 10,000 |
f(x)=x2f(x)=x2 | 100 | 10,000 | 1,000,000 | 100,000,000 |
g(x)=x3g(x)=x3 | 1000 | 1,000,000 | 1,000,000,000 | 1,000,000,000,0001,000,000,000,000 |
In fact,
As a result, we say x3x3 is growing more rapidly than x2x2 as x→∞x→∞. On the other hand, for f(x)=x2f(x)=x2 and g(x)=3x2+4x+1g(x)=3x2+4x+1, although the values of g(x)g(x) are always greater than the values of f(x)f(x) for x>0x>0, each value of g(x)g(x) is roughly three times the corresponding value of f(x)f(x) as x→∞x→∞, as shown in the following table. In fact,
xx | 10 | 100 | 1000 | 10,000 |
f(x)=x2f(x)=x2 | 100 | 10,000 | 1,000,000 | 100,000,000 |
g(x)=3x2+4x+1g(x)=3x2+4x+1 | 341 | 30,401 | 3,004,001 | 300,040,001 |
In this case, we say that x2x2 and 3x2+4x+13x2+4x+1 are growing at the same rate as x→∞x→∞.
More generally, suppose ff and gg are two functions that approach infinity as x→∞x→∞. We say gg grows more rapidly than ff as x→∞x→∞ if
On the other hand, if there exists a constant M≠0M≠0 such that
we say ff and gg grow at the same rate as x→∞x→∞.
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 lnxlnx, x2x2, and exex
For each of the following pairs of functions, use L’Hôpital’s rule to evaluate limx→∞(f(x)g(x))limx→∞(f(x)g(x)).
- f(x)=x2f(x)=x2 and g(x)=exg(x)=ex
- f(x)=lnxf(x)=lnx and g(x)=x2g(x)=x2
Watch the following video to see the worked solution to Example: Comparing the Growth Rates of lnxlnx, x2x2, and exex.
Try It
Compare the growth rates of x100x100 and 2x2x.
Using the same ideas as in the last example. it is not difficult to show that exex grows more rapidly than xpxp for any p>0p>0. In Figure 5 and the table below it, we compare exex with x3x3 and x4x4 as x→∞x→∞.

Figure 5. The exponential function exex grows faster than xpxp for any p>0p>0. (a) A comparison of exex with x3x3. (b) A comparison of exex with x4x4.
xx | 5 | 10 | 15 | 20 |
x3x3 | 125 | 1000 | 3375 | 8000 |
x4x4 | 625 | 10,000 | 50,625 | 160,000 |
exex | 148 | 22,026 | 3,269,017 | 485,165,195 |
Similarly, it is not difficult to show that xpxp grows more rapidly than lnxlnx for any p>0p>0. In Figure 6 and the table below it, we compare lnxlnx with 3√x3√x and √x√x.

Figure 6. The function y=lnxy=lnx grows more slowly than xpxp for any p>0p>0 as x→∞x→∞.
xx | 10 | 100 | 1000 | 10,000 |
lnxlnx | 2.303 | 4.605 | 6.908 | 9.210 |
3√x3√x | 2.154 | 4.642 | 10 | 21.544 |
√x√x | 3.162 | 10 | 31.623 | 100 |
Candela Citations
- 4.8 L'Hopital's Rule. Authored by: Ryan Melton. License: CC BY: Attribution
- Calculus Volume 1. Authored by: Gilbert Strang, Edwin (Jed) Herman. Provided by: OpenStax. Located at: https://openstax.org/details/books/calculus-volume-1. License: CC BY-NC-SA: Attribution-NonCommercial-ShareAlike. License Terms: Access for free at https://openstax.org/books/calculus-volume-1/pages/1-introduction