The problem asks to evaluate the limit:

$\lim_{x \to \infty} \frac{3x^5 - 1}{2x^3 - 5x^2 - 1}$

Solution:

As $x \to \infty$, the highest-degree terms dominate. So, we focus on the leading terms:

$\lim_{x \to \infty} \frac{3x^5}{2x^3} = \lim_{x \to \infty} \frac{3x^5}{2x^3} = \lim_{x \to \infty} \frac{3x^2}{2}$

As $x \to \infty$, $x^2 \to \infty$, so:

$\lim_{x \to \infty} \frac{3x^2}{2} = \infty$

Final Answer:

$\lim_{x \to \infty} \frac{3x^5 - 1}{2x^3 - 5x^2 - 1} = \infty$

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

Here are 5 related questions to expand your understanding:

1. What is the limit of $\frac{4x^4 + 2}{x^2 - 1}$ as $x \to \infty$?
2. How do you determine horizontal asymptotes using limits?
3. Evaluate $\lim_{x \to -\infty} \frac{2x^6 - x}{5x^6 + 3}$.
4. How does L'Hôpital's Rule apply to rational functions with infinity limits?
5. Find the limit $\lim_{x \to \infty} \frac{x^3 + 2}{x^5 - 7x}$.

Tip: When evaluating limits at infinity, focus on the highest-degree terms in the numerator and denominator to simplify the process.