AP Calculus AB and BC

Limit of (ln x)^2/x at Infinity Is 0

Q: Does the answer change if the exponent is 3 instead of 2?

No, it is still 0; it just takes three passes. Each application peels one power off the logarithm, so \frac{(\ln x)^{n}}{x} needs n rounds and ends at \frac{n!}{x}, which goes to 0 for every fixed n.

Q: Is (ln x)^2 the same as ln(x^2)?

No. The power law gives \ln\left(x^2\right) = 2\ln x, a constant multiple of the logarithm, while (\ln x)^2 is the logarithm squared. They agree only at x = 1 and x = e^{2}. Both quotients happen to have limit 0, but only one of them needs two passes.

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

Yes. Substituting x = e^{t} turns it into \frac{t^{2}}{e^{t}}, and an exponential outgrows any polynomial. A squeeze also works: for x \ge 1 the bound \ln x \le 4x^{1/4} gives (\ln x)^2 \le 16\sqrt{x}, so the function is trapped between 0 and \frac{16}{\sqrt{x}}.

The limit of (ln x)^2 over x as x approaches infinity is 0. Substitution gives infinity over infinity. One round of L'Hopital's rule leaves 2 ln x over x, still infinity over infinity, and a second round leaves 2 over x, which goes to 0. Squaring a logarithm does not help it keep up with x.

\lim_{x \to \infty} \frac{(\ln x)^2}{x} = 0

Settled by L'Hopital's rule applied twice.

Two passes to strip the logarithm

The numerator is a composition, $u^2$ with $u = \ln x$ , so its derivative needs the chain rule: $\frac{d}{dx}\left[(\ln x)^2\right] = 2\ln x \cdot \frac{1}{x}$ . The denominator differentiates to $1$ .

\lim_{x \to \infty} \frac{(\ln x)^2}{x} = \lim_{x \to \infty} \frac{2\ln x \cdot \frac{1}{x}}{1} = \lim_{x \to \infty} \frac{2\ln x}{x}

One power of the logarithm is gone, but the new quotient is still $\frac{\infty}{\infty}$ , so the rule is available again. The second pass is the easy one, since $2\ln x$ differentiates to $\frac{2}{x}$ .

\lim_{x \to \infty} \frac{2\ln x}{x} = \lim_{x \to \infty} \frac{\frac{2}{x}}{1} = \lim_{x \to \infty} \frac{2}{x} = 0

Two passes were enough because each one peels off one power of $\ln x$ while leaving the denominator untouched. The values fall slowly, which is why a table alone rarely convinces anyone.

$x$	$(\ln x)^2$	$\frac{(\ln x)^2}{x}$
$10$	$5.302$	$0.5302$
$1000$	$47.72$	$0.04772$
$1{,}000{,}000$	$190.9$	$0.0001909$

You are allowed to stop after one pass

The first application already leaves $\frac{2\ln x}{x}$ , and $\lim_{x \to \infty}\frac{\ln x}{x} = 0$ is a standard result worth quoting. Doing that finishes the problem a step early and earns the same credit. The second pass is the honest route when that result is not yet in hand.

Why substitution fails

Both parts grow without bound, so pushing $\infty$ through the fraction returns $\frac{\infty}{\infty}$ . That form records that a race is happening and nothing about who wins.

The real obstacle is the instinct that squaring should tip the race toward the top. Squaring does make $\ln x$ larger, but it does not change what kind of function it is. At $x = 1{,}000{,}000$ the squared logarithm has reached about $191$ while the denominator has reached a million, and the gap keeps widening.

Squaring buys the logarithm one extra round of L'Hopital's rule, which is exactly what the two passes above spend.

The pattern the two passes belong to

Nothing depends on the exponent being $2$ . Each application lowers the power of the logarithm by one, so $(\ln x)^n$ over $x$ takes $n$ passes and arrives at $\frac{n!}{x}$ .

\lim_{x \to \infty} \frac{(\ln x)^{n}}{x^{p}} = 0 \qquad (n > 0,\; p > 0)

A substitution shows the same thing without any bookkeeping. Put $x = e^{t}$ , so $t \to \infty$ , and the quotient becomes a polynomial over an exponential, a comparison most students already trust.

\frac{(\ln x)^2}{x} = \frac{t^{2}}{e^{t}} \longrightarrow 0 \qquad (x = e^{t})

The mistakes students make

Reading $(\ln x)^2$ as $\ln\left(x^2\right)$ . The second one is $2\ln x$ , a different function: at $x = 10$ the first is about $5.30$ and the second about $4.61$ .
Differentiating $(\ln x)^2$ as $\frac{2}{x}$ , or as $2\ln x$ . The chain rule gives $2\ln x \cdot \frac{1}{x}$ , and dropping it collapses two passes into one and hides why the rule has to run twice.
Using the quotient rule on the whole fraction. L'Hopital's rule replaces $\frac{f}{g}$ with $\frac{f'}{g'}$ , never with the derivative of the quotient.
Deciding the squared logarithm wins because it is squared. Every fixed power of $\ln x$ loses to $x$ , and the limit is still $0$ for $(\ln x)^{100}$ .
Applying the rule a third time out of habit. After the second pass the expression is $\frac{2}{x}$ , which is not indeterminate, so the rule no longer applies.

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 the answer change if the exponent is 3 instead of 2?

No, it is still $0$ ; it just takes three passes. Each application peels one power off the logarithm, so $\frac{(\ln x)^{n}}{x}$ needs $n$ rounds and ends at $\frac{n!}{x}$ , which goes to $0$ for every fixed $n$ .

Is (ln x)^2 the same as ln(x^2)?

No. The power law gives $\ln\left(x^2\right) = 2\ln x$ , a constant multiple of the logarithm, while $(\ln x)^2$ is the logarithm squared. They agree only at $x = 1$ and $x = e^{2}$ . Both quotients happen to have limit $0$ , but only one of them needs two passes.

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

Yes. Substituting $x = e^{t}$ turns it into $\frac{t^{2}}{e^{t}}$ , and an exponential outgrows any polynomial. A squeeze also works: for $x \ge 1$ the bound $\ln x \le 4x^{1/4}$ gives $(\ln x)^2 \le 16\sqrt{x}$ , so the function is trapped between $0$ and $\frac{16}{\sqrt{x}}$ .