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We are asked to find the limit:

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

### Step 1: Identify Dominant Terms
As \(x\) approaches infinity, the terms with the highest powers of \(x\) in both the numerator and the denominator will dominate. The expression can be approximated by focusing on the dominant terms.

- In the numerator, \( -3x \) dominates over \(2\).
- In the denominator, \( -5x \) dominates over \(1\).

### Step 2: Simplify the Expression
Now, simplify the expression by dividing both the numerator and the denominator by \(x\):

\[
\frac{2 - 3x}{1 - 5x} = \frac{\frac{2}{x} - 3}{\frac{1}{x} - 5}
\]

As \(x \to \infty\), both \(\frac{2}{x}\) and \(\frac{1}{x}\) approach zero. This simplifies the expression to:

\[
\frac{0 - 3}{0 - 5} = \frac{-3}{-5} = \frac{3}{5}
\]

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

Let me know if you'd like any further clarification or details!

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Here are 5 related questions for further exploration:

1. What happens if we change the signs in the numerator or denominator?
2. How does the limit behave if the powers of \(x\) in the numerator and denominator are different?
3. What is the limit of \(\frac{2 + 3x}{1 + 5x}\) as \(x \to \infty\)?
4. Can you apply L'Hopital's Rule to find the limit instead of dividing by the highest power of \(x\)?
5. How would the limit change if we were considering \(x \to -\infty\)?

**Tip**: When finding limits involving rational expressions at infinity, always focus on the highest degree terms in both the numerator and the denominator!

What is the limit of (2 - 3x) / (1 - 5x) as x approaches infinity?

Learn how to find the limit of (2 - 3x) / (1 - 5x) as x approaches infinity. This problem focuses on dominant terms in rational functions and how to simplify expressions to find limits at infinity.

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