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We are tasked with finding the limit:

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

### Step-by-Step Solution:

1. **Analyze the behavior as \( x \to -\infty \):**
   - For large negative \( x \), the term \( x^2 \) dominates \( -x \) inside the square root.
   - So, \( \sqrt{x^2 - x} \approx \sqrt{x^2} = |x| \). But since \( x \) is negative, \( |x| = -x \).

2. **Simplify the expression:**
   - We approximate the square root as:
   \[
   \sqrt{x^2 - x} \approx -x \quad \text{as} \quad x \to -\infty.
   \]
   - Now rewrite the original limit:
   \[
   \lim_{x \to -\infty} \frac{3x}{-x}.
   \]
   - This simplifies to:
   \[
   \lim_{x \to -\infty} -3 = -3.
   \]

Therefore, the limit is:

\[
\boxed{-3}.
\]

Would you like a more detailed explanation or have any questions?

---

Here are 5 related questions for further practice:
1. What is the limit of \(\lim_{x \to +\infty} \frac{3x}{\sqrt{x^2 - x}}\)?
2. How would the limit change if the square root expression were \( \sqrt{x^2 + x} \)?
3. Find \(\lim_{x \to -\infty} \frac{2x^2}{\sqrt{x^4 - 1}}\).
4. What is the limit of \(\lim_{x \to -\infty} \frac{x^2}{\sqrt{x^2 + 4x}}\)?
5. How would the result change if the numerator was \(5x\) instead of \(3x\)?

**Tip:** When finding limits at infinity, focus on dominant terms, as lower-degree terms become negligible!

Find the limit as x approaches negative infinity of (3x) / (sqrt(x^2 - x)).

Learn how to find the limit as x approaches negative infinity for the function (3x) / (sqrt(x² - x)) using asymptotic analysis. This example demonstrates how to simplify square root expressions and apply limits for large negative x. Ideal for advanced high school or college calculus students.

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