Let's solve the given limit:

$\lim_{{x \to \infty}} \frac{2x^5 + 9x^4 - 4x^3}{26 - 8x^5}$

Step 1: Factor out the highest power of $x$ in the numerator and the denominator.

The highest power of $x$ in both the numerator and denominator is $x^5$. Factor $x^5$ out:

$\lim_{{x \to \infty}} \frac{x^5\left(2 + \frac{9}{x} - \frac{4}{x^2}\right)}{x^5\left(\frac{26}{x^5} - 8\right)}$

Step 2: Simplify the expression.

Since $x^5$ is common in both the numerator and the denominator, it cancels out:

$\lim_{{x \to \infty}} \frac{2 + \frac{9}{x} - \frac{4}{x^2}}{\frac{26}{x^5} - 8}$

Step 3: Apply the limit.

As $x \to \infty$, the terms with $\frac{9}{x}$, $\frac{4}{x^2}$, and $\frac{26}{x^5}$ approach zero:

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

Final Answer:

$\lim_{{x \to \infty}} \frac{2x^5 + 9x^4 - 4x^3}{26 - 8x^5} = -\frac{1}{4}$

Would you like more details or have any questions?

Here are five related questions you might find interesting:

1. How does the degree of the polynomial in the numerator and denominator affect the limit as $x$ approaches infinity?
2. What happens if the highest power of $x$ in the denominator is higher than in the numerator?
3. Can limits involving infinity always be evaluated by factoring out the highest power of $x$?
4. How would the solution change if the sign of the highest power of $x$ in the denominator were positive?
5. How can L'Hôpital's Rule be applied to similar limit problems?

Tip: When dealing with limits approaching infinity, always factor out the highest power of $x$ to simplify the expression and make the limit easier to evaluate.