Limit of x² (ln x)³ as x approaches 0
Evaluate the limit: $$ L = \lim_{x \to 0^+} x^2 \ln^3(x). $$
Explanation of Method
As $x \to 0^+$, $x^2 \to 0$ while $\ln^3(x) \to -\infty$, giving the indeterminate form $0 \cdot (-\infty)$. Rather than converting to a fraction and applying L'Hôpital's Rule repeatedly (which here would require three applications), it is cleaner to substitute $t = -\ln x$. This turns the problem into the classic fact that exponentials dominate polynomials: $t^p e^{-qt} \to 0$ as $t \to \infty$ for any fixed $p, q > 0$.
- Substitute $t = -\ln x$, so that $x = e^{-t}$ and $t \to +\infty$ as $x \to 0^+$.
- Rewrite $x^2\ln^3(x)$ entirely in terms of $t$.
- Apply the standard fact that $t^3 e^{-2t} \to 0$ as $t \to \infty$.
Step 1 · Substitute $t = -\ln x$
Let $t = -\ln x$. As $x \to 0^+$, $\ln x \to -\infty$, so $t \to +\infty$. Solving for $x$:
$$ x = e^{-t}. $$
Also, $\ln x = -t$, so $\ln^3(x) = (-t)^3 = -t^3$.
Step 2 · Rewrite the expression in terms of $t$
Substituting into $x^2 \ln^3(x)$:
$$ x^2 \ln^3(x) = \left(e^{-t}\right)^2 \cdot (-t^3) = -t^3 e^{-2t}. $$
So the original limit becomes:
$$ L = \lim_{x \to 0^+} x^2\ln^3(x) = \lim_{t \to +\infty} \left(-t^3 e^{-2t}\right) = -\lim_{t \to +\infty} \frac{t^3}{e^{2t}}. $$
Step 3 · Apply the exponential-dominates-polynomial fact
For any fixed positive integer $p$ and constant $q>0$, exponential growth always outpaces polynomial growth, so:
$$ \lim_{t \to +\infty} \frac{t^{p}}{e^{qt}} = 0. $$
Here $p = 3$ and $q = 2$, so:
$$ \lim_{t \to +\infty} \frac{t^3}{e^{2t}} = 0. $$
Therefore:
$$ L = -\lim_{t \to +\infty} \frac{t^3}{e^{2t}} = -0 = 0. $$
Summary Table
| Quantity | In terms of $t=-\ln x$ | Behavior as $t \to +\infty$ |
|---|---|---|
| $x^2$ | $e^{-2t}$ | decays exponentially |
| $\ln^3(x)$ | $-t^3$ | grows like $-t^3$ |
| $x^2\ln^3(x)$ | $-t^3 e^{-2t}$ | $\to 0$ |
Additional Notes
- The substitution $t=-\ln x$ converts any limit of the form $\lim_{x\to0^+} x^p (\ln x)^q$ into $\lim_{t\to\infty} \pm t^q e^{-pt}$, reducing it to the single well-known fact that exponentials beat polynomials at infinity.
- The same fact can be proved directly with L'Hôpital's Rule, but it would require applying the rule three times in a row here (once for each power of $t$), since each differentiation only reduces the polynomial degree by one.
- More generally, $\lim_{x\to 0^+} x^p \ln^q(x) = 0$ for every fixed $p>0$ and every integer $q \geq 0$ — polynomial decay in $x$ always wins over any power of a logarithmic blow-up.
Result
The value of the limit is: