We are tasked with finding the limit:

$\lim_{{x \to \infty}} \frac{3x + 4}{x^2 + 5}$

Solution:

1. As $x \to \infty$, both the numerator and the denominator will tend to infinity, but the denominator grows much faster due to the $x^2$ term. To understand the behavior at infinity, we divide both the numerator and denominator by the highest power of $x$ in the denominator, which is $x^2$:

$\lim_{{x \to \infty}} \frac{3x + 4}{x^2 + 5} = \lim_{{x \to \infty}} \frac{\frac{3x + 4}{x^2}}{\frac{x^2 + 5}{x^2}}$

Simplifying the expression:

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

2. As $x \to \infty$, the terms involving $\frac{3}{x}, \frac{4}{x^2},$ and $\frac{5}{x^2}$ all tend to 0. Thus, the limit becomes:

$= \frac{0 + 0}{1 + 0} = 0$

Final Answer:

$\lim_{{x \to \infty}} \frac{3x + 4}{x^2 + 5} = 0$

Would you like any additional details on this solution or have any questions?

Here are 5 related questions:

1. How would the result change if the numerator had a higher degree than the denominator?
2. What is the behavior of the function as $x \to -\infty$?
3. Can you apply L'Hopital's Rule to solve this limit? Why or why not?
4. How would the limit change if the degree of the numerator and denominator were the same?
5. What is the general behavior of rational functions as $x \to \infty$?

Tip: When evaluating limits at infinity for rational functions, compare the degrees of the numerator and denominator to determine the behavior as $x \to \infty$.