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!

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!