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The limit you're trying to evaluate is:

\[
\lim_{x \to -\infty} \frac{4x^2 - x}{2x^3 - 5}
\]

### Solution:
To solve this limit as \( x \) approaches negative infinity, we can factor out the highest power of \( x \) in both the numerator and the denominator. The highest power of \( x \) in the numerator is \( x^2 \), and in the denominator, it's \( x^3 \).

#### Step 1: Factor \( x^2 \) from the numerator and \( x^3 \) from the denominator:
\[
\frac{4x^2 - x}{2x^3 - 5} = \frac{x^2(4 - \frac{1}{x})}{x^3(2 - \frac{5}{x^3})}
\]

#### Step 2: Simplify the expression:
\[
\frac{4x^2 - x}{2x^3 - 5} = \frac{\frac{x^2}{x^3}(4 - \frac{1}{x})}{\frac{x^3}{x^3}(2 - \frac{5}{x^3})} = \frac{\frac{4 - \frac{1}{x}}{x}}{2 - \frac{5}{x^3}}
\]

#### Step 3: Evaluate the limit as \( x \to -\infty \):
As \( x \) approaches negative infinity, the terms \(\frac{1}{x}\) and \(\frac{5}{x^3}\) both approach 0. Therefore, the expression simplifies to:

\[
\lim_{x \to -\infty} \frac{\frac{4 - 0}{x}}{2 - 0} = \lim_{x \to -\infty} \frac{\frac{4}{x}}{2} = \lim_{x \to -\infty} \frac{4}{2x}
\]

\[
\lim_{x \to -\infty} \frac{2}{x} = 0
\]

### Conclusion:
The limit is:

\[
\lim_{x \to -\infty} \frac{4x^2 - x}{2x^3 - 5} = 0
\]

Would you like any details on any step or have any questions?

### Follow-up Questions:
1. How would the limit change if \( x \) approached positive infinity instead?
2. What is the importance of factoring out the highest power of \( x \) when calculating limits at infinity?
3. How does the degree of the numerator and denominator affect the outcome of limits at infinity?
4. Can L'Hôpital's rule be used to solve this limit? How?
5. What happens to the limit if the terms in the denominator had a different power?

### Tip:
When evaluating limits at infinity, always consider the highest degree term in the numerator and the denominator. This can simplify the process and lead you directly to the limit.

Calculate the limit: lim_{x -> -∞} (4x^2 - x) / (2x^3 - 5)

Understand how to solve the limit of the expression (4x^2 - x) / (2x^3 - 5) as x approaches negative infinity. This problem involves key calculus concepts, including limits at infinity and the importance of factorization. Suitable for students in Grades 11-12.

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