The problem asks to compute the limit of a sequence as $n \to \infty$, given by the following expression:

[ \lim_{n \to \infty} \frac{\sqrt[3]{n} - 9n^2}{3n - \sqrt[4]{9n^8 + 1}}. ]

Step-by-Step Solution:

1. Analyze the leading terms for large $n$:

   • In the numerator $\sqrt[3]{n} - 9n^2$, the term $9n^2$ dominates as $n \to \infty$, because $n^2$ grows much faster than $n^{1/3}$.
   • In the denominator $3n - \sqrt[4]{9n^8 + 1}$, the term $\sqrt[4]{9n^8 + 1} \approx \sqrt[4]{9n^8} = 9^{1/4}n^2 = 3n^2$ dominates for large $n$.

2. Simplify the leading terms:

   • Numerator: $\sqrt[3]{n} - 9n^2 \approx -9n^2$.
   • Denominator: $3n - \sqrt[4]{9n^8 + 1} \approx 3n - 3n^2$.

3. Approximate the limit:

   Substituting these dominant terms into the limit expression, we get: $\frac{-9n^2}{3n - 3n^2}.$

4. Simplify the expression:

   Factor out $n^2$ from both the numerator and denominator: $\frac{-9n^2}{3n(1 - n)} = \frac{-9}{3(1 - n)}.$

5. Evaluate the limit as $n \to \infty$:

   As $n \to \infty$, the denominator tends to $-3n$, so the whole expression tends to: $\lim_{n \to \infty} \frac{-9}{3(1 - n)} = 0.$

Conclusion:

The limit of the given sequence is $\boxed{0}$.

Would you like a more detailed explanation, or do you have any further questions? Here are five related questions to consider:

1. What is the behavior of a function's leading term in limits as $n \to \infty$?
2. Why does the term $n^2$ dominate over $\sqrt[3]{n}$ as $n \to \infty$?
3. How can factoring the highest powers of $n$ help simplify limit problems?
4. What happens when a term in the denominator grows much faster than one in the numerator in limits?
5. How can this limit technique be applied to other sequences with mixed polynomial terms?

Tip: When calculating limits involving polynomials or roots, always look at the highest powers of $n$ in both the numerator and denominator to simplify the problem.