Exponential Growth Compared With Polynomial Growth

Why Compare Growth Rates?

At infinity, knowing the individual behavior of two functions does not always tell us the behavior of their quotient. A polynomial such as \(x^8\) grows without bound, but an exponential such as \(2^x\) grows faster: eventually, the exponential is larger by an arbitrarily large factor. This comparison explains why quotients like \(x^8/2^x\) approach zero, even though both numerator and denominator grow without bound.

The key distinction is not simply whether a function grows, but how quickly it grows relative to another function. This tutorial establishes the exponential-versus-polynomial comparison and applies it to mixed quotients.

The Basic Comparison

For \(a>1\), the function \(a^x\) is an increasing exponential. For \(p\geq 0\), \(x^p\) is a polynomial power on \(x>0\); \(p\) need not be an integer for the comparison below. The central result concerns the limit as \(x\) tends to positive infinity.

The second limit follows from the first because, for positive \(x\), the two quotients are reciprocals. The proof below establishes the first limit directly. It uses the binomial theorem for integer powers, together with an integer power larger than \(p\).

Why the Comparison Holds

Write \(a=1+h\), where \(h=a-1>0\). First consider a positive integer \(n\). Choose an integer \(k>p\). The binomial theorem gives a sum of nonnegative terms, so, when \(n\geq 2k\), we can keep just its \(k\)-th term:

Since \(n\geq 2k\), each factor \(n-j\), for \(0\leq j\leq k-1\), is at least \(n/2\). Therefore

Now take any real \(x\geq 1\) and let \(n=\lfloor x\rfloor\), the greatest integer less than or equal to \(x\). Then \(n\leq x<n+1\). Because \(a>1\), \(a^x\geq a^n\); because \(p\geq 0\), \(x^p\leq(n+1)^p\). For sufficiently large \(x\), the bound just proved applies, giving

As \(x\to\infty\), the integer \(n=\lfloor x\rfloor\) also tends to infinity. Since \(k>p\), the last expression tends to zero: its numerator grows like \(n^p\), while its denominator contains \(n^k\). The squeeze theorem therefore gives \(x^p/a^x\to 0\). Taking reciprocals of these positive quantities gives \(a^x/x^p\to+\infty\). This proves both claims.

Useful Consequences

The theorem applies directly to a polynomial divided by an exponential. A polynomial of degree \(m\) is bounded in magnitude by a constant times \(x^m\) for sufficiently large positive \(x\). Consequently, for \(a>1\), any polynomial \(P\) satisfies

It also applies when the exponential has a positive constant multiplying \(x\) in its exponent. For \(c>0\), set \(t=cx\). Then \(x^p=t^p/c^p\), so the fixed factor \(1/c^p\) does not change the limit. In particular, \(a^{cx}\) outgrows \(x^p\) as \(x\to\infty\).

If \(0<a<1\), the direction is reversed: write \(a=1/b\), where \(b=1/a>1\). Then \(a^x=1/b^x\), and the theorem implies

At negative infinity, an exponential with base \(a>1\) decays faster than any fixed inverse power. Setting \(t=-x\) gives \(a^x/|x|^p=1/(a^t t^p)\) for \(x<0\), and this tends to zero as \(t\to\infty\).

Worked Quotients

Comparing Several Terms

For sums of terms, identify which terms dominate before simplifying a quotient. If a numerator and denominator contain the same exponential factor, divide through by that factor and use the theorem to show that polynomial terms divided by it vanish. If the exponentials have different bases, dividing by one of them produces a ratio of exponentials. For example, \(5^x/2^x=(5/2)^x\), which tends to infinity because \(5/2>1\).

These comparisons require attention to coefficients and possible cancellation. The largest-looking terms do not determine a quotient if they cancel exactly. For instance, \(4^x-4^x+x^3=x^3\): the exponential terms disappear, leaving only a polynomial. After simplifying, identify the terms that actually remain and compare those.

A Common Mistake

A frequent error is to assume that a power in the denominator must dominate an exponential because the power has a large exponent. For example, this reasoning is incorrect:

The theorem gives the opposite conclusion. Since \(2^x/x^{10}\) is the reciprocal of \(x^{10}/2^x\), and \(x^{10}/2^x\to 0\), it follows that

The correction is to compare growth rates, not merely inspect which expression looks complicated or has the larger written exponent. For a fixed positive power and a base greater than one, the exponential eventually wins.

When This Comparison Is Useful

Exponential-versus-polynomial comparisons are a reliable way to resolve quotients at infinity that initially have both numerator and denominator growing without bound. They justify discarding polynomial terms after division by a growing exponential, and they help distinguish exponentials with different bases. The method gives a limit only after the relevant terms and signs have been checked; cancellation, a zero denominator, or a negative coefficient can change how the quotient must be analyzed.