The given limit is:

$\lim_{{x \to \infty}} \frac{-3x^4 + 4x^3 + 8x^2 - 8x}{4x^3 - 7x^2 + 4x - 3}$

Step-by-Step Solution:

1. Identify the highest powers of $x$:

   • In the numerator, the highest power of $x$ is $x^4$.
   • In the denominator, the highest power of $x$ is $x^3$.

2. Divide both the numerator and denominator by $x^4$:

$\lim_{{x \to \infty}} \frac{-3x^4 + 4x^3 + 8x^2 - 8x}{4x^3 - 7x^2 + 4x - 3} = \lim_{{x \to \infty}} \frac{\frac{-3x^4}{x^4} + \frac{4x^3}{x^4} + \frac{8x^2}{x^4} - \frac{8x}{x^4}}{\frac{4x^3}{x^4} - \frac{7x^2}{x^4} + \frac{4x}{x^4} - \frac{3}{x^4}}$

3. Simplify each term:

   • Numerator: $\frac{-3x^4}{x^4} + \frac{4x^3}{x^4} + \frac{8x^2}{x^4} - \frac{8x}{x^4} = -3 + \frac{4}{x} + \frac{8}{x^2} - \frac{8}{x^3}$
   • Denominator: $\frac{4x^3}{x^4} - \frac{7x^2}{x^4} + \frac{4x}{x^4} - \frac{3}{x^4} = \frac{4}{x} - \frac{7}{x^2} + \frac{4}{x^3} - \frac{3}{x^4}$

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

   • As $x \to \infty$, all terms involving $\frac{1}{x^n}$ (for $n \geq 1$) approach 0.
   • So, the numerator approaches $-3$ and the denominator approaches $0$.

5. Conclusion: The numerator approaches a non-zero value $-3$, while the denominator approaches 0, which indicates that the limit tends to infinity (specifically negative infinity).

$\lim_{{x \to \infty}} \frac{-3x^4 + 4x^3 + 8x^2 - 8x}{4x^3 - 7x^2 + 4x - 3} = -\infty$

Would you like further explanation or have any questions?

Related Questions:

1. How does dividing by the highest power of $x$ help in solving limits?
2. What is the significance of a rational function's degrees in determining the behavior at infinity?
3. What happens if the highest powers in the numerator and denominator are equal?
4. How does the limit change if the leading coefficient in the numerator was positive?
5. Can L'Hopital's Rule be applied to this problem? Why or why not?

Tip: When evaluating limits involving rational functions as $x \to \infty$, focusing on the highest degree terms simplifies the analysis.