To solve the limit

$\lim_{x \to \infty} \frac{2x^3 - 4x^2 + 8}{4x^4 + x^2 + 1},$

we can analyze the behavior of the function as $x \to \infty$ by comparing the degrees of the terms in the numerator and the denominator.

Step-by-Step Solution:

1. Identify the highest powers of $x$ in the numerator and denominator:

   • The highest power in the numerator is $x^3$.
   • The highest power in the denominator is $x^4$.

2. Divide each term in the numerator and the denominator by $x^4$, the highest power in the denominator:

   $\lim_{x \to \infty} \frac{\frac{2x^3}{x^4} - \frac{4x^2}{x^4} + \frac{8}{x^4}}{\frac{4x^4}{x^4} + \frac{x^2}{x^4} + \frac{1}{x^4}}.$

3. Simplify each term:

   $= \lim_{x \to \infty} \frac{\frac{2}{x} - \frac{4}{x^2} + \frac{8}{x^4}}{4 + \frac{1}{x^2} + \frac{1}{x^4}}.$

4. Evaluate the limit as $x \to \infty$:

   • As $x \to \infty$, all terms with $x$ in the denominator ($\frac{2}{x}, \frac{4}{x^2}, \frac{8}{x^4}, \frac{1}{x^2},$ and $\frac{1}{x^4}$) approach $0$.

   • Thus, the expression simplifies to:

     $\frac{0}{4} = 0.$

Final Answer:

$\lim_{x \to \infty} \frac{2x^3 - 4x^2 + 8}{4x^4 + x^2 + 1} = 0.$

Would you like more details or have any questions?

Related Questions:

1. What is the behavior of the limit when the degree of the numerator is higher than the denominator?
2. How would the solution change if the highest powers in both the numerator and denominator were the same?
3. Can we apply L'Hôpital's Rule to solve this problem? Why or why not?
4. How do limits involving infinity relate to horizontal asymptotes?
5. What are other methods for evaluating limits at infinity?

Tip:

When dealing with limits at infinity, always compare the degrees of the numerator and denominator for an efficient solution!