Compare common growth rates
Which curve is higher far out
The chalk curve is \(2^x\), and the faint curve is \(x^3\). The readouts give the heights at the orange points and the ratio \(2^x / x^3\).
At \(x = 5\), \(2^x\) is 32 and \(x^3\) is 125, so the cube is higher and the ratio is about 0.26.
Move the points to the right. The curves meet near \(x = 9.94\). At \(x = 10\), \(2^x\) is 1,024 and \(x^3\) is 1,000, so the ratio is about 1.02.
Past \(x = 11\), the chalk curve is above the top of the figure. At \(x = 14\), \(2^x\) is 16,384 and \(x^3\) is 2,744, so the ratio is about 5.97, and it continues to grow.
The exponential is lower for a long stretch but higher after the curves meet, so a comparison of growth must use large x.
Logarithm, power, exponential
The faint curve is a logarithm, and the two chalk curves are a power of x and an exponential. Each one rises without bound.
Move the orange points to the right. Near the left edge, the order of the three heights changes from place to place.
Far to the right, the logarithm is the lowest, the power is in the middle, and the exponential is the highest.
This far order holds for every logarithm, every positive power of x and every exponential \(a^x\) with \(a > 1\).
Notation
A function g dominates a function f when the ratio \(f/g\) approaches 0 as x grows, so f becomes small next to g.
\[\lim_{x \to \infty} \frac{f(x)}{g(x)} = 0\]The sign \(\ll\) means "is dominated by". The ranking holds for every power \(x^p\) with \(p > 0\) and every base \(a > 1\).
\[\ln x \;\ll\; x^{p} \;\ll\; a^{x}\]In a sum, the dominant term decides the behavior for large x. A higher power dominates a lower one, so \(x^4 - 15x^2\) behaves like \(x^4\) because the faint term approaches 0.
\[\frac{x^4 - 15x^2}{x^4} = 1 - \cq{\frac{15}{x^2}} \to \cy{1}\]In a ratio, divide the top and the bottom by the dominant term \(\cy{D}\) of the bottom. The bottom then approaches a nonzero number.
\[\frac{f}{g} = \frac{f/\cy{D}}{g/\cy{D}}\]
When it fails
Dominance describes large x only. In the red rectangle, from \(x = 2\) to \(x = 9\), the faint curve \(x^3\) is above \(2^x\), so this stretch alone gives the wrong ranking.
The stretch is too short, because the curves meet near \(x = 9.94\) and \(2^x\) is higher after that.
Example 1: a power over an exponential
Find \(\displaystyle\lim_{x \to \infty} \frac{x^5}{3^x}\).
The top \(x^5\) is a power, and the bottom \(3^x\) is an exponential with base \(3 > 1\). The curve shows their ratio, which rises to a height of about 13 near \(x = 4.5\).
An exponential dominates every power, so the bottom becomes far larger than the top. The ratio is about 1.69 at x = 10, about 0.0009 at x = 20, and about 0.0000001 at x = 30.
\[x^5 \ll 3^x\]The ratio approaches 0, so the curve approaches the x-axis, now orange.
\[\lim_{x \to \infty} \frac{x^5}{3^x} = \cy{0}\]
Example 2: a large coefficient
Which dominates as x grows: \(20x^2\) or \(1.5^x\)?
At \(x = 15\), the faint curve \(20x^2\) is far higher than the chalk curve \(1.5^x\).
\[20 \cdot 15^2 = 4500 \qquad 1.5^{15} \approx 438\]A tempting wrong move: The power is far higher, so the power dominates.
\[20x^2 \;\cr{\gg}\; 1.5^x\]The red bar marks the error: it compares the curves at one input, but dominance concerns large x. The chalk curve grows by a factor of 1.5 for each step of 1 in x. At \(x = 30\), \(1.5^x\) is about 191,751, and \(20x^2\) is only 18,000.
An exponential with a base greater than 1 dominates every power, whatever the coefficient, so \(1.5^x\) dominates. The curves meet near \(x = 22.8\), and the chalk curve is higher after that.
\[\lim_{x \to \infty} \frac{20x^2}{1.5^x} = \cy{0}\]
Practice
Find \(\displaystyle\lim_{x \to \infty} \frac{3^x + x^4}{2 \cdot 3^x - x}\).
Find the dominant term in the top and in the bottom. Then move the point far to the right to test your answer.
Solution
In the top, \(3^x\) dominates \(x^4\). In the bottom, \(3^x\) dominates \(x\). Divide the top and the bottom by \(3^x\).
\[\frac{1 + \cq{x^4/3^x}}{2 - \cq{x/3^x}}\]The exponential dominates both powers, so \(x^4/3^x\) and \(x/3^x\) approach 0, and the curve approaches the dashed orange line.
\[\frac{1 + 0}{2 - 0} = \cy{\tfrac12}\]
Questions
The graph shows \(\ln x\) (faint), \(\sqrt{x}\) and \(1.05^x\) (the steeper chalk curve) for x from 0 to 50.
Beyond which input does the curve \(1.05^x\) stay above both of the other curves?
- Beyond about x = 37
- Beyond about x = 24
- Beyond about x = 10
- At every input
Answer
Beyond about x = 37
The curve \(1.05^x\) is below \(\sqrt{x}\) near \(x = 25\) and above it near \(x = 50\). What happens for larger x?
- It stays above, because an exponential dominates a power
- It falls below again, because the power was higher first
- The curves cross again and again
- The power catches up, because it never stops rising
Answer
It stays above, because an exponential dominates a power
Which limit, as x grows, equals 0?
- \(\dfrac{\ln x}{\sqrt{x}}\)
- \(\dfrac{x^2}{\ln x}\)
- \(\dfrac{1.01^x}{x^{100}}\)
- \(\dfrac{5x^3 + 2}{x^3}\)
Answer
\(\dfrac{\ln x}{\sqrt{x}}\)
Summary
You can rank a logarithm below every positive power of x, and a power below every exponential with a base greater than 1.
It holds because the dominated function becomes small next to the dominant one as x grows, even if it is higher for small x.
In a limit at infinity that mixes logarithms, powers and exponentials, the dominant term decides the result.
This is one of four skills in topic 1.15 of Which & Why: AP Calculus.