The given limit expression is:

\[
\lim_{{x \to \infty}} \frac{2 - 3x}{1 - 5x}
\]

### Step-by-Step Solution:

1. **Simplify the expression:**
   As \( x \to \infty \), the terms involving \( x \) (i.e., \(-3x\) and \(-5x\)) will dominate the constants \(2\) and \(1\). Therefore, we want to focus on the highest degree terms in the numerator and denominator.
   
   We divide both the numerator and denominator by \( x \) (the highest power of \( x \)):

   \[
   \lim_{{x \to \infty}} \frac{\frac{2}{x} - 3}{\frac{1}{x} - 5}
   \]

2. **Evaluate the limit as \( x \to \infty \):**
   As \( x \to \infty \), the terms \(\frac{2}{x}\) and \(\frac{1}{x}\) approach 0. Thus, the expression simplifies to:

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

### Final Answer:
\[
\lim_{{x \to \infty}} \frac{2 - 3x}{1 - 5x} = \frac{3}{5}
\]

Do you want further details or have any questions?

Here are 5 related questions for further exploration:
1. What happens to the limit if the degree of the numerator and denominator are different?
2. How do limits behave when \( x \) approaches negative infinity?
3. What is the significance of dividing by the highest power of \( x \) when solving limits at infinity?
4. Can you explain what happens if both numerator and denominator approach 0 as \( x \to \infty \)?
5. How does L'Hôpital's Rule apply to limits at infinity?

**Tip:** Always simplify complex limit expressions by factoring out the highest powers of \( x \) to evaluate the behavior as \( x \to \infty \).

Discover how to solve the limit of the rational function lim x→∞ (2 - 3x) / (1 - 5x) by focusing on the highest degree terms. This step-by-step solution demonstrates the simplification and evaluation techniques used in limit problems involving infinity, suitable for Grades 11-12.

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