Logs, powers or exponentials: which grows fastest as x → ∞?

By Marius, private math tutor in Montreal · 2026-10-05

As $x \to \infty$ , $x$ grows faster than $\ln x$ , and so does every positive power of $x$ , even $x^{0.01}$ ; every power of $x$ , in turn, loses to $e^{x}$ . That one ranking settles limits such as $\frac{x^{3}}{e^{x}} \to 0$ and $\frac{\ln x}{\sqrt{x}} \to 0$ at a glance, and L'Hospital's rule proves it when a question asks for more than a glance.

The ranking

As $x \to \infty$ , logarithms, powers and exponentials all grow without bound, but at wildly different speeds. For any powers $p > 0$ and $q > 0$ :

(\ln x)^{q} ≪ x^{p} ≪ e^{x} .

Read $f ≪ g$ as " $g$ grows faster than $f$ ", and that has a precise meaning:

lim_{x \to \infty} \frac{f (x)}{g (x)} = 0 .

Some values make the gaps concrete:

At $x = 10$ : $\ln x \approx 2.3$ , $\sqrt{x} \approx 3.2$ , $x^{2} = 100$ , $e^{x} \approx 22 000$ .
At $x = 100$ : $\ln x \approx 4.6$ , $\sqrt{x} = 10$ , $x^{2} = 10 000$ , $e^{x} \approx 2.7 \times 10^{43}$ .
At $x = 1000$ : $\ln x \approx 6.9$ , $\sqrt{x} \approx 31.6$ , $x^{2} = 1 000 000$ , $e^{x} \approx 2 \times 10^{434}$ .

The ranking ignores everything except the class. The power $p$ can be tiny, the power $q$ on the logarithm can be huge, a coefficient can be a million, and the order does not change: even $x^{0.01}$ eventually passes $(\ln x)^{100}$ . Exponentials in another base behave the same way, because $a^{x} = e^{x \ln a}$ and $\ln a > 0$ whenever $a > 1$ .

Inside one class

Powers: the higher power wins. $x^{3}$ beats $x^{2}$ , since $\frac{x^{2}}{x^{3}} = \frac{1}{x} \to 0$ .
Exponentials: the faster rate wins, by a whole exponential factor. $e^{2 x}$ beats $e^{x}$ , since $\frac{e^{x}}{e^{2 x}} = e^{- x} \to 0$ .
Logarithms: a power inside the logarithm only changes a constant, because $\ln (x^{5}) = 5 \ln x$ . So $\ln (x^{5})$ and $\ln x$ grow at the same rate, and their quotient is simply $5$ .

Reading a limit at a glance

For a quotient, find the fastest-growing term in the numerator and in the denominator, and compare only those two.

The numerator's term is faster: the quotient grows without bound, as in $\frac{e^{x}}{x^{100}} \to \infty$ .
The denominator's term is faster: the quotient tends to $0$ , as in $\frac{x^{3}}{e^{x}} \to 0$ and $\frac{\ln x}{\sqrt{x}} \to 0$ .
The same term dominates both: the limit is the ratio of its two coefficients. Worked example 2 below is one of these.

This is the move you already use on rational functions, where you compare the highest powers. Here "highest power" becomes "fastest class".

Proving it when a glance is not enough

The ranking is a theorem, and L'Hospital's rule proves it: each comparison is an $\frac{\infty}{\infty}$ form, and the rule can be applied as often as the form returns. It also asks two things that every quotient below satisfies: the derivative of the bottom is never $0$ for large $x$ , and the new quotient has a limit, a number or $\infty$ . Each chain below ends on a limit that exists, and that is what justifies every step before it.

Worked example 1: $\frac{x^{3}}{e^{x}} \to 0$

Both $x^{3}$ and $e^{x}$ grow without bound, so $\frac{x^{3}}{e^{x}}$ has the form $\frac{\infty}{\infty}$ . Differentiate top and bottom: $\frac{3 x^{2}}{e^{x}}$ .
Still $\frac{\infty}{\infty}$ . Again: $\frac{6 x}{e^{x}}$ .
Still $\frac{\infty}{\infty}$ . Again: $\frac{6}{e^{x}}$ .
The numerator is now fixed and the denominator grows without bound, so $lim_{x \to \infty} \frac{x^{3}}{e^{x}} = lim_{x \to \infty} \frac{6}{e^{x}} = 0 .$

Each application lowers the power by one and leaves $e^{x}$ exactly as it was, because $e^{x}$ is its own derivative. For any whole-number power $n$ , the power is gone after $n$ steps. For a power that is not a whole number, compare it with the next whole number up: for $x \geq 1$ ,

0 < \frac{x^{2.5}}{e^{x}} \leq \frac{x^{3}}{e^{x}},

and the Squeeze Theorem finishes the job.

Logarithms against powers

Here one application is enough, for any $p > 0$ . The quotient $\frac{\ln x}{x^{p}}$ has the form $\frac{\infty}{\infty}$ , and differentiating top and bottom gives

\frac{1 / x}{p x^{p - 1}} = \frac{1}{p x^{p}} ⟶ 0,

so $\frac{\ln x}{x^{p}} \to 0$ as well.

A substitution shows why the two comparisons are really one. Put $x = e^{t}$ , so that $\ln x = t$ , and $t \to \infty$ as $x \to \infty$ :

\frac{(\ln x)^{q}}{x^{p}} = \frac{t^{q}}{e^{p t}} .

A logarithm against a power is a power against an exponential in disguise, so the first half of the ranking follows from the second.

Worked example 2: a quotient of sums

lim_{x \to \infty} \frac{2 e^{x} + x^{5}}{e^{x} + \sqrt{x}}

At a glance, $e^{x}$ dominates the top and the bottom, so the limit is the ratio of its coefficients, $2$ . To justify that, divide every term by $e^{x}$ :

\frac{2 e^{x} + x^{5}}{e^{x} + \sqrt{x}} = \frac{2 + \frac{x^{5}}{e^{x}}}{1 + \frac{\sqrt{x}}{e^{x}}} ⟶ \frac{2 + 0}{1 + 0} = 2,

because $\frac{x^{5}}{e^{x}} \to 0$ and $\frac{\sqrt{x}}{e^{x}} \to 0$ , both by the ranking.

"Eventually" is part of the statement

The common mistake: judging the ranking from a few values. The ranking describes what happens for every sufficiently large $x$ , and "sufficiently large" can be very large indeed. $x^{3}$ is bigger than $e^{x}$ for every $x$ between about $1.86$ and $4.54$ , and only after that does the exponential pull ahead for good. $x^{10}$ is ahead of $e^{x}$ from about $x = 1.1$ all the way to $x \approx 35.8$ , and $\ln x$ is bigger than $x^{0.1}$ from about $x = 3$ to about $x = 3.4 \times 10^{15}$ . A table that stops at $x = 100$ can point the wrong way. The limit decides, not the table.

y = e^{x}

and

y = x^{3}

. The cubic is ahead between the crossings at

x \approx 1.86

and

x \approx 4.54

; after the second one the exponential leads, and the gap only widens.

Check your understanding

No calculator. Easiest first; questions 5 and 6 ask you to judge an argument, and 7 and 8 ask for a proof.

Put these in order, from slowest-growing to fastest-growing as $x \to \infty$ : $x^{0.2}$ , $e^{x / 10}$ , $(\ln x)^{3}$ , $x^{5}$ .
Evaluate $lim_{x \to \infty} \frac{x^{8}}{{1.1}^{x}}$ .
Evaluate $lim_{x \to \infty} \frac{\ln (x^{8})}{\sqrt[3]{x}}$ .
Evaluate $lim_{x \to \infty} \frac{5 x^{2} + e^{x}}{x^{6} + 4 e^{x}}$ .
True or false, with a reason: "At $x = 100$ , $x^{100}$ is far bigger than $e^{x}$ , so $x^{100}$ grows faster than $e^{x}$ ."
Find the error: a student says that $lim_{x \to \infty} \frac{x^{2}}{e^{- x}} = 0$ "because the exponential always wins".
Prove that $lim_{x \to \infty} \frac{(\ln x)^{2}}{x} = 0$ , checking the form before each step.
For large $x$ , which is bigger: $x^{\ln x}$ or $e^{x}$ ? Write both as powers of $e$ , and use question 7.

Show the answers

$(\ln x)^{3}$ , then $x^{0.2}$ , then $x^{5}$ , then $e^{x / 10}$ . Any power of a logarithm loses to any positive power of $x$ , the higher power beats the lower one, and an exponential with a positive rate, even $\frac{1}{10}$ , beats every power.
$0$ . ${1.1}^{x} = e^{x \ln 1.1}$ is an exponential with the positive rate $\ln 1.1$ , so it beats $x^{8}$ , however slowly it starts.
$0$ . $\ln (x^{8}) = 8 \ln x$ is still a logarithm, and $\sqrt[3]{x} = x^{1 / 3}$ is a positive power, which grows faster.
$\frac{1}{4}$ . $e^{x}$ dominates the top and the bottom; dividing every term by it gives $\frac{5 x^{2} / e^{x} + 1}{x^{6} / e^{x} + 4} \to \frac{0 + 1}{0 + 4}$ .
False. The first half is true: $x^{100} = 10^{200}$ at $x = 100$ , while $e^{100} < 3^{100} < 10^{100}$ . But growing faster is about the limit, and $\frac{x^{100}}{e^{x}} \to 0$ . $e^{x}$ overtakes $x^{100}$ at about $x = 647$ and stays ahead from then on.
$e^{- x}$ tends to $0$ ; it does not grow at all, so the ranking says nothing about it. In fact $\frac{x^{2}}{e^{- x}} = x^{2} e^{x}$ , which grows without bound. The ranking compares functions that grow, so an exponential $e^{k x}$ counts only when $k > 0$ .
The form is $\frac{\infty}{\infty}$ . One application gives $\frac{2 \ln x \cdot \frac{1}{x}}{1} = \frac{2 \ln x}{x}$ , still $\frac{\infty}{\infty}$ ; a second gives $\frac{2 / x}{1} = \frac{2}{x} \to 0$ . Substituting $x = e^{t}$ works too: it turns the quotient into $\frac{t^{2}}{e^{t}}$ , which tends to $0$ .
$e^{x}$ . Since $x = e^{\ln x}$ , $x^{\ln x} = e^{(\ln x)^{2}}$ , so compare the exponents $(\ln x)^{2}$ and $x$ . By question 7, $x - (\ln x)^{2} = x (1 - \frac{(\ln x)^{2}}{x})$ , a factor growing without bound times a factor tending to $1$ , so it tends to $\infty$ , and $\frac{e^{x}}{x^{\ln x}} = e^{x - (\ln x)^{2}} \to \infty$ .

Practise it

The free CEGEP Calculus I Limits with Trig, Exponential and Log Functions worksheet has limits at infinity with exponentials, logarithms and arctangent, where dividing by the dominant term starts. The proofs above use L'Hospital's rule, which comes later in Calculus I and has its own free Indeterminate Forms and L'Hospital's Rule worksheet.

Want one-on-one help with this? I offer CEGEP Science calculus and linear algebra tutoring in Montreal, in person or online. Sessions start with a short written diagnostic, and a written recap goes to whoever books after every session.

Logs, powers or exponentials: which grows fastest as x → ∞?

The ranking

Inside one class

Reading a limit at a glance

Proving it when a glance is not enough

Worked example 1: $\frac{x^{3}}{e^{x}} \to 0$

Logarithms against powers

Worked example 2: a quotient of sums

"Eventually" is part of the statement

Check your understanding

Practise it

Free worksheets on this topic

Ready to improve your grades?

Logs, powers or exponentials: which grows fastest as x → ∞?

The ranking

Inside one class

Reading a limit at a glance

Proving it when a glance is not enough

Worked example 1: x3ex→0

Logarithms against powers

Worked example 2: a quotient of sums

"Eventually" is part of the statement

Check your understanding

Practise it

Free worksheets on this topic

Ready to improve your grades?

Worked example 1: $\frac{x^{3}}{e^{x}} \to 0$