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To solve the given limit, we need to analyze the behavior of the function as \( x \) approaches infinity:

\[
\lim_{{x \to \infty}} \left( \frac{\sqrt[3]{2x^2 - x^3}}{3x^2 - 2\sqrt{3x^2} - 5} \right)
\]

### Step-by-Step Solution:

1. **Simplify the Numerator and Denominator:**

   First, simplify the numerator:
   \[
   \sqrt[3]{2x^2 - x^3} = \sqrt[3]{-x^3 (1 - \frac{2x^2}{x^3})} = \sqrt[3]{-x^3 (1 - \frac{2}{x})}
   \]
   As \( x \to \infty \), \( \frac{2}{x} \) approaches 0. Thus:
   \[
   \sqrt[3]{2x^2 - x^3} \approx \sqrt[3]{-x^3} = -x
   \]

   Next, simplify the denominator:
   \[
   3x^2 - 2\sqrt{3x^2} - 5 = 3x^2 - 2\sqrt{3}x - 5
   \]

2. **Evaluate the Dominant Terms:**

   For large \( x \), the dominant terms in the numerator and denominator will dictate the behavior of the fraction:
   \[
   \frac{\sqrt[3]{2x^2 - x^3}}{3x^2 - 2\sqrt{3x^2} - 5} \approx \frac{-x}{3x^2} = \frac{-1}{3x}
   \]

3. **Take the Limit:**

   As \( x \to \infty \):
   \[
   \lim_{{x \to \infty}} \frac{-1}{3x} = 0
   \]

### Conclusion:
\[
\lim_{{x \to \infty}} \left( \frac{\sqrt[3]{2x^2 - x^3}}{3x^2 - 2\sqrt{3x^2} - 5} \right) = 0
\]

Would you like any additional details or have any questions about this solution?

Here are 5 questions you might ask next:
1. How do we handle limits involving square roots or cube roots?
2. What are the general steps for evaluating limits at infinity?
3. How can L'Hôpital's rule be applied to more complex limits?
4. What are some common pitfalls when simplifying algebraic expressions?
5. How do we determine the dominant term in a polynomial as \( x \to \infty \)?

**Tip:** When simplifying limits, always identify the dominant terms that grow the fastest as \( x \to \infty \). This can often simplify the process significantly.

Learn how to evaluate the limit as x approaches infinity for a complex mathematical expression involving cube roots, polynomials, and limits. This detailed solution provides step-by-step guidance on simplifying the expression and applying limit rules effectively. Suitable for advanced high school students and college-level calculus learners.

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