AP Calculus AB and BC

Limit of ln(x)/x as x Approaches Infinity

Q: Does L'Hopital's rule work on infinity over infinity, or only on 0 over 0?

Both, and the \frac{\infty}{\infty} case includes any combination of signs, so \frac{-\infty}{\infty} qualifies too. The other indeterminate forms such as 0 \cdot \infty, \infty - \infty, and 1^{\infty} have to be rewritten as a quotient first before the rule is available.

Q: Can this be done without L'Hopital's rule?

Yes, with a squeeze. For x \ge 1 the bound 0 \le \ln x \le 2\sqrt{x} holds, so dividing by x traps the function between 0 and \frac{2}{\sqrt{x}}, and both bounds go to 0. Substituting x = e^{t} also works, turning the limit into \frac{t}{e^{t}}.

Q: What does this say about the limit of x/ln(x)?

It is \infty. When a positive function has limit 0, its reciprocal is unbounded, so flipping \frac{\ln x}{x} flips the answer. Applying L'Hopital directly to \frac{x}{\ln x} gives \frac{1}{\frac{1}{x}} = x, which confirms it.

The limit of the natural log of x over x as x approaches infinity is 0. Substitution gives infinity over infinity, an indeterminate form, so L'Hopital's rule applies: differentiating top and bottom separately gives 1 over x, which goes to 0. Any positive power of x eventually outgrows the logarithm.

\lim_{x \to \infty} \frac{\ln x}{x} = 0

Settled by L'Hopital's rule.

Applying the rule once

Check the hypotheses before touching anything. Both $\ln x$ and $x$ are unbounded as $x \to \infty$ , both are differentiable for $x > 0$ , and the denominator's derivative is 1, which is never 0. The $\frac{\infty}{\infty}$ form is one of the two L'Hopital accepts.

Differentiate the numerator and the denominator separately, each on its own.

\lim_{x \to \infty} \frac{\ln x}{x} = \lim_{x \to \infty} \frac{\frac{d}{dx}[\ln x]}{\frac{d}{dx}[x]} = \lim_{x \to \infty} \frac{\frac{1}{x}}{1}

The new quotient simplifies to $\frac{1}{x}$ , which is no longer indeterminate. It has a limit you can read off.

\lim_{x \to \infty} \frac{1}{x} = 0

One application was enough, and that is the signal to stop. The numbers agree: the logarithm crawls while the denominator sprints.

$x$	$\ln x$	$\frac{\ln x}{x}$
$10$	$2.303$	$0.2303$
$1000$	$6.908$	$0.006908$
$1{,}000{,}000$	$13.82$	$0.00001382$

Why substitution fails

Both parts of the fraction are unbounded, so pushing $\infty$ through gives $\frac{\infty}{\infty}$ . That form is indeterminate because the answer depends entirely on relative speed, and the form itself records no speed at all.

Students often stall here for a different reason: they suspect $\ln x$ levels off. It does not. The logarithm is unbounded, and it does exceed any number you name, just slowly. Reaching an output of 100 takes an input of $e^{100}$ , which is a 44-digit number.

Unbounded is not the same as fast

At $x = 1{,}000{,}000$ the numerator has only reached about 13.8 while the denominator has reached a million. Both are heading to $\infty$ , and the race is not close, which is precisely what the indeterminate form cannot tell you and L'Hopital can.

The growth hierarchy this establishes

The same single application of the rule works with any positive power in the denominator, since differentiating $x^{p}$ gives $p x^{p-1}$ and the quotient becomes $\frac{1}{p x^{p}}$ .

\lim_{x \to \infty} \frac{\ln x}{x^{p}} = 0 \quad (p > 0)

So even $x^{0.01}$ eventually buries the logarithm. Pair that with the matching statement for exponentials and you get the ordering that resolves most end-behavior questions on sight.

\ln x \ll x^{p} \ll e^{x} \quad \text{as } x \to \infty

Reading it right to left gives the reciprocal results for free. Since $\frac{\ln x}{x} \to 0$ , its reciprocal is unbounded.

\lim_{x \to \infty} \frac{x}{\ln x} = \infty

The mistake students make

The signature error is differentiating the fraction with the quotient rule instead of differentiating the two pieces separately. That produces a different function and answers a different question.

\frac{d}{dx}\left[\frac{\ln x}{x}\right] = \frac{1 - \ln x}{x^{2}} \quad \text{(not what the rule asks for)}

L'Hopital replaces $\frac{f}{g}$ with $\frac{f'}{g'}$ , never with $\left(\frac{f}{g}\right)'$ . The two agree only by accident.

The second error is applying the rule again out of habit. After one pass the expression is $\frac{1}{x}$ , which is not indeterminate, so the rule no longer applies. Applying it anyway happens to return 0 here, but that is luck, not validity: on a non-indeterminate quotient the rule can return the wrong number outright, as with $\lim_{x \to 0} \frac{x+1}{x+2} = \frac{1}{2}$ , where differentiating top and bottom gives 1. Recheck the form before every application.

Not sure which technique a limit wants?

The Limit Method Chooser walks the decision from direct substitution through factoring, the conjugate, and L'Hopital, and says why each one applies or fails.

Frequently asked questions

Does L'Hopital's rule work on infinity over infinity, or only on 0 over 0?

Both, and the $\frac{\infty}{\infty}$ case includes any combination of signs, so $\frac{-\infty}{\infty}$ qualifies too. The other indeterminate forms such as $0 \cdot \infty$ , $\infty - \infty$ , and $1^{\infty}$ have to be rewritten as a quotient first before the rule is available.

Can this be done without L'Hopital's rule?

Yes, with a squeeze. For $x \ge 1$ the bound $0 \le \ln x \le 2\sqrt{x}$ holds, so dividing by $x$ traps the function between 0 and $\frac{2}{\sqrt{x}}$ , and both bounds go to 0. Substituting $x = e^{t}$ also works, turning the limit into $\frac{t}{e^{t}}$ .

What does this say about the limit of x/ln(x)?

It is $\infty$ . When a positive function has limit 0, its reciprocal is unbounded, so flipping $\frac{\ln x}{x}$ flips the answer. Applying L'Hopital directly to $\frac{x}{\ln x}$ gives $\frac{1}{\frac{1}{x}} = x$ , which confirms it.