To find the limit of the expression as $x \to \infty$:

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

we start by analyzing the degrees of the numerator and denominator. The numerator $2x + 1$ is a first-degree polynomial, while the denominator $x^2 + 3$ is a second-degree polynomial. Since the degree of the denominator is higher than the degree of the numerator, the limit will approach zero as $x \to \infty$.

To make this more rigorous, we can divide both the numerator and the denominator by $x^2$:

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

As $x \to \infty$, both $\frac{2}{x}$ and $\frac{1}{x^2}$ approach zero, and $\frac{3}{x^2}$ also approaches zero. Thus, the expression simplifies to:

$\lim_{x \to \infty} \frac{0 + 0}{1 + 0} = 0.$

Therefore,

$\lim_{x \to \infty} \frac{2x + 1}{x^2 + 3} = 0.$

Would you like more details or have any questions?

Here are five related questions to expand your understanding:

1. What is the general rule for finding limits of rational functions at infinity?
2. How would the limit change if the degrees of the numerator and denominator were equal?
3. What happens if the degree of the numerator is greater than that of the denominator?
4. How can L'Hôpital's Rule be applied to solve this limit?
5. What is the behavior of the function for large negative values of $x$?

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