We are tasked with finding the limit of the following rational function as $x \to \infty$:

$\lim_{x \to \infty} \frac{8x^{18} + 53x^{12} + 31}{31x^{18} + 3x^2 - 53}$

Step-by-Step Solution:

1. Identify the highest powers of $x$:

   • The highest power of $x$ in the numerator is $x^{18}$.
   • The highest power of $x$ in the denominator is also $x^{18}$.

2. Divide all terms by $x^{18}$: To simplify the limit as $x \to \infty$, divide every term in both the numerator and the denominator by $x^{18}$:

   $\frac{8x^{18} + 53x^{12} + 31}{31x^{18} + 3x^2 - 53} = \frac{\frac{8x^{18}}{x^{18}} + \frac{53x^{12}}{x^{18}} + \frac{31}{x^{18}}}{\frac{31x^{18}}{x^{18}} + \frac{3x^2}{x^{18}} - \frac{53}{x^{18}}}$

   This simplifies to:

   $\frac{8 + \frac{53}{x^6} + \frac{31}{x^{18}}}{31 + \frac{3}{x^{16}} - \frac{53}{x^{18}}}$

3. Evaluate the limit as $x \to \infty$: As $x \to \infty$, terms like $\frac{53}{x^6}$, $\frac{31}{x^{18}}$, $\frac{3}{x^{16}}$, and $\frac{53}{x^{18}}$ all tend to 0. Therefore, the expression simplifies to:

   $\frac{8 + 0 + 0}{31 + 0 - 0} = \frac{8}{31}$

Thus, the limit is:

$\boxed{\frac{8}{31}}$

Would you like any further details or clarification?

Related Questions:

1. How do you handle limits when the highest powers in the numerator and denominator are different?
2. What are some alternative methods for solving limits without using L'Hospital's Rule?
3. Why do terms with higher powers of $x$ in the denominator approach 0 as $x \to \infty$?
4. Can you explain why we only focus on the highest power of $x$ when finding limits at infinity for rational functions?
5. How would the limit change if the degrees of the numerator and denominator were different?

Tip: When finding limits involving rational functions as $x \to \infty$, always check the degrees of the numerator and denominator first. This will guide you to the correct simplification steps.