The problem you uploaded asks to find the limit of the rational function as $x \to \infty$:

$\lim_{x \to \infty} \frac{x^2 + 3x - 18}{x^2 + 6x}.$

Step-by-Step Solution:

1. Identify the degrees of the polynomials:

   • The numerator is $x^2 + 3x - 18$, which is a quadratic function of degree 2.
   • The denominator is $x^2 + 6x$, which is also a quadratic function of degree 2.

2. Divide both the numerator and the denominator by the highest power of $x$, which is $x^2$:

$\lim_{x \to \infty} \frac{x^2 + 3x - 18}{x^2 + 6x} = \lim_{x \to \infty} \frac{\frac{x^2}{x^2} + \frac{3x}{x^2} - \frac{18}{x^2}}{\frac{x^2}{x^2} + \frac{6x}{x^2}}.$

3. Simplify each term: $= \lim_{x \to \infty} \frac{1 + \frac{3}{x} - \frac{18}{x^2}}{1 + \frac{6}{x}}.$

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

   • As $x \to \infty$, terms with $\frac{1}{x}$ or higher powers of $x$ will approach 0. Therefore: $= \frac{1 + 0 - 0}{1 + 0} = \frac{1}{1}.$

5. Conclusion: $\lim_{x \to \infty} \frac{x^2 + 3x - 18}{x^2 + 6x} = 1.$

Would you like further clarification or a breakdown of any part of the process?

Related Questions:

1. What happens if the degree of the numerator is greater than the degree of the denominator?
2. How does the limit change if you evaluate the function as $x \to -\infty$?
3. What is the limit if a constant term is added to the denominator instead of a linear term?
4. Can you apply L'Hopital's Rule to solve this limit, and how would it work?
5. How do limits behave for rational functions where the degrees are not equal?

Tip: For rational functions, when the degree of the numerator and denominator are equal, the limit as $x \to \infty$ is determined by the ratio of the leading coefficients.