Limit of (ln x)² / x at infinity
Evaluate the limit: $$ L = \lim_{x \to +\infty} \frac{\ln^2 x}{x}. $$
Explanation of Method
As $x\to+\infty$, both $\ln^2x\to\infty$ and $x\to\infty$, giving $\dfrac\infty\infty$. Differentiating $\ln^2x$ brings down a factor of $\frac1x$ (via the chain rule), which after one application of L'Hôpital's Rule leaves a single power of $\ln x$ over $x$ — the same type of indeterminate form, but simpler. A second application removes the remaining logarithm entirely, leaving a limit that clearly tends to $0$.
- Apply L'Hôpital's Rule once, differentiating $\ln^2x$ with the chain rule.
- Recognize the new limit is still $\frac\infty\infty$, and apply L'Hôpital's Rule a second time.
- Evaluate the resulting limit.
Step 1 · Apply L'Hôpital's Rule once
Since $\ln^2x\to\infty$ and $x\to\infty$ as $x\to+\infty$, L'Hôpital's Rule applies. Differentiating $\ln^2x$ with the chain rule:
$$ \frac{d}{dx}\left(\ln^2x\right) = 2\ln x\cdot\frac{d}{dx}(\ln x) = \frac{2\ln x}{x}, \qquad \frac{d}{dx}(x) = 1. $$
So:
$$ \lim_{x\to+\infty}\frac{\ln^2x}{x} = \lim_{x\to+\infty}\frac{2\ln x}{x}. $$
Step 2 · Apply L'Hôpital's Rule a second time
The new limit $\dfrac{2\ln x}{x}$ still has the form $\dfrac\infty\infty$. Differentiating again:
$$ \frac{d}{dx}(2\ln x) = \frac2x, \qquad \frac{d}{dx}(x) = 1. $$
So:
$$ \lim_{x\to+\infty}\frac{2\ln x}{x} = \lim_{x\to+\infty}\frac{2/x}{1} = \lim_{x\to+\infty}\frac2x. $$
Step 3 · Evaluate the final limit
As $x\to+\infty$, $\dfrac2x\to0$. Therefore:
$$ L = \lim_{x\to+\infty}\frac{\ln^2x}{x} = 0. $$
Summary Table
| Application | Resulting expression | Form |
|---|---|---|
| Original | $\dfrac{\ln^2x}{x}$ | $\frac\infty\infty$ |
| After 1st L'Hôpital | $\dfrac{2\ln x}{x}$ | $\frac\infty\infty$ |
| After 2nd L'Hôpital | $\dfrac2x$ | $\to0$ |
Additional Notes
- This confirms the general principle that any power of a logarithm grows slower than any positive power of $x$: for every fixed exponent $k$ and every $p>0$, $\lim_{x\to+\infty}\dfrac{\ln^kx}{x^p}=0$ — logarithms sit at the very bottom of the "growth hierarchy," below every polynomial.
- Each differentiation of a power $\ln^kx$ reduces the power by one (multiplying by $\frac1x$ via the chain rule), so a $k$-th power of $\ln x$ over $x$ always resolves after exactly $k$ applications of L'Hôpital's Rule, just as here $k=2$ took exactly two applications.
- Combined with the earlier result $\lim_{x\to+\infty}\frac{x^2}{e^x}=0$, this places $\ln^2x \ll x \ll e^x$ in terms of growth rate as $x\to\infty$ — a useful mental hierarchy: logarithmic $\ll$ polynomial $\ll$ exponential.
Result
The value of the limit is: