We are tasked with finding the limit of the following expression as $t \to -\infty$:

$\lim_{t \to -\infty} \frac{2t^2 + t}{t^3 - 8t + 1}$

Step-by-Step Solution:

1. Degree of the Polynomial:

   • The degree of the numerator $2t^2 + t$ is 2.
   • The degree of the denominator $t^3 - 8t + 1$ is 3.

2. Behavior as $t \to -\infty$:

   • For large $|t|$, the term with the highest degree dominates. Thus, for the numerator, $2t^2$ dominates, and for the denominator, $t^3$ dominates.

3. Simplifying the Limit: We can divide both the numerator and denominator by $t^3$, the highest degree in the denominator:

   $\lim_{t \to -\infty} \frac{\frac{2t^2 + t}{t^3}}{\frac{t^3 - 8t + 1}{t^3}} = \lim_{t \to -\infty} \frac{\frac{2}{t} + \frac{1}{t^2}}{1 - \frac{8}{t^2} + \frac{1}{t^3}}$

4. Evaluate the Limit: As $t \to -\infty$, all terms with $\frac{1}{t}, \frac{1}{t^2}, \frac{1}{t^3}$ approach 0. This simplifies the expression to:

   $\lim_{t \to -\infty} \frac{0}{1} = 0$

Final Answer:

$\lim_{t \to -\infty} \frac{2t^2 + t}{t^3 - 8t + 1} = 0$

Would you like further details or have any additional questions?

Here are 5 related questions to expand on this concept:

1. How does the behavior change if the degrees of the numerator and denominator are the same?
2. What would happen if we were asked to evaluate this limit as $t \to +\infty$?
3. How do higher-degree terms impact the limits of rational functions as $t \to \infty$?
4. Could you apply L'Hopital's Rule to solve this limit, and would it yield the same result?
5. What would the limit be if the degrees of the numerator exceeded the denominator?

Tip: When finding limits at infinity for rational functions, focus on the highest degree terms to simplify the process.