The given limit is:

$\lim_{x \to \infty} \frac{(\ln x)^2}{\sqrt[3]{x}}.$

We want to determine the behavior of this expression as $x \to \infty$.

Solution

1. Examine the Growth Rates:

   • The numerator, $(\ln x)^2$, grows very slowly compared to $x$ (polynomial or exponential functions).
   • The denominator, $\sqrt[3]{x} = x^{1/3}$, grows unboundedly but slower than $x$ itself.

   This suggests that the denominator dominates the numerator as $x \to \infty$, and the entire fraction might approach $0$.

2. Formal Calculation Using L'Hôpital's Rule:
   Since this is an indeterminate form $\frac{\infty}{\infty}$, we can apply L'Hôpital's Rule.

   • Differentiate the numerator $(\ln x)^2$ with respect to $x$: $\frac{d}{dx} \left[ (\ln x)^2 \right] = 2 \ln x \cdot \frac{1}{x} = \frac{2 \ln x}{x}.$

   • Differentiate the denominator $\sqrt[3]{x} = x^{1/3}$ with respect to $x$: $\frac{d}{dx} \left[ x^{1/3} \right] = \frac{1}{3} x^{-2/3}.$

   Apply L'Hôpital's Rule: $\lim_{x \to \infty} \frac{(\ln x)^2}{\sqrt[3]{x}} = \lim_{x \to \infty} \frac{\frac{2 \ln x}{x}}{\frac{1}{3} x^{-2/3}}.$

3. Simplify the Result:
   Combine terms: $\frac{\frac{2 \ln x}{x}}{\frac{1}{3} x^{-2/3}} = \frac{2 \ln x}{x} \cdot \frac{3}{x^{-2/3}} = 6 \ln x \cdot x^{-1 + 2/3} = 6 \ln x \cdot x^{-1/3}.$

   As $x \to \infty$:

   • $\ln x \to \infty$, but it grows much slower than any power of $x$.
   • $x^{-1/3} \to 0$.

   Therefore: $6 \ln x \cdot x^{-1/3} \to 0.$

Final Answer:

$\lim_{x \to \infty} \frac{(\ln x)^2}{\sqrt[3]{x}} = 0.$

Let me know if you want further clarifications or steps!

Related Questions:

1. What happens if $(\ln x)^3$ replaces $(\ln x)^2$ in the numerator?
2. How does the limit change if the denominator is $x^{1/2}$ instead of $x^{1/3}$?
3. Can you explain L'Hôpital's Rule in general?
4. How does the behavior of $\ln x$ compare to $x^a$ for small powers $a$?
5. What if the numerator were $\ln x$ instead of $(\ln x)^2$?

Tip:

For indeterminate forms like $\frac{\infty}{\infty}$, L'Hôpital's Rule is often a powerful tool.