We are given the following limit expression:

[ \lim_{n \to \infty} \sqrt[3]{n} \cdot \left( \sqrt[3]{n^2} - \sqrt[3]{n(n-1)} \right) ]

Step-by-Step Solution

1. Rewrite the expression: Let's simplify the cube roots:

   • $\sqrt[3]{n} = n^{1/3}$
   • $\sqrt[3]{n^2} = n^{2/3}$
   • $\sqrt[3]{n(n-1)} = \sqrt[3]{n^2 - n} = n^{2/3} \cdot \sqrt[3]{1 - \frac{1}{n}}$

   Substituting these into the limit expression:

   $\lim_{n \to \infty} n^{1/3} \left( n^{2/3} - n^{2/3} \cdot \sqrt[3]{1 - \frac{1}{n}} \right)$

2. Factor out $n^{2/3}$:

   $\lim_{n \to \infty} n^{1/3} \cdot n^{2/3} \left( 1 - \sqrt[3]{1 - \frac{1}{n}} \right)$

   This simplifies further to:

   $\lim_{n \to \infty} n \left( 1 - \sqrt[3]{1 - \frac{1}{n}} \right)$

3. Expand the cube root using a series approximation:

   For large $n$, $\sqrt[3]{1 - \frac{1}{n}}$ can be approximated by a Taylor expansion:

   $\sqrt[3]{1 - \frac{1}{n}} \approx 1 - \frac{1}{3n} + O\left(\frac{1}{n^2}\right)$

   Substituting this approximation into the expression:

   $\lim_{n \to \infty} n \left( 1 - \left(1 - \frac{1}{3n}\right) \right)$

   Simplifying:

   $\lim_{n \to \infty} n \cdot \frac{1}{3n} = \frac{1}{3}$

Thus, the value of the limit is:

$\boxed{\frac{1}{3}}$

Let me know if you would like more details or further explanations! Here are some related questions:

1. How do Taylor expansions help in simplifying limits involving roots?
2. What other approximations are useful for limits as $n \to \infty$?
3. How does the cube root function behave near infinity?
4. Can this method be applied to higher-order roots or powers?
5. How would you handle limits involving different types of logarithms?

Tip: When approaching limits, approximating complicated expressions with series expansions can significantly simplify the evaluation.