The question asks to find the horizontal asymptote of the function $f(x) = \frac{2^x + 5}{e^x + 1}$ for $x > 0$.

To determine the horizontal asymptote, we need to compute the limit of $f(x)$ as $x \to \infty$.

Step-by-step approach:

$f(x) = \frac{2^x + 5}{e^x + 1}$

1. As $x \to \infty$, both $2^x$ and $e^x$ grow exponentially, and the terms $5$ and $1$ become negligible in comparison. Therefore, we can simplify the expression by focusing on the dominant exponential terms: $f(x) \approx \frac{2^x}{e^x}$

2. Now, we compute the limit of this simplified expression: $\lim_{x \to \infty} \frac{2^x}{e^x} = \lim_{x \to \infty} \left( \frac{2}{e} \right)^x$

Since $\frac{2}{e} < 1$, the term $\left( \frac{2}{e} \right)^x$ approaches 0 as $x \to \infty$.

Thus, the horizontal asymptote is: $\boxed{y = 0}$

This corresponds to option $\text{A}$, which is the correct answer. The reasoning behind the incorrect selection of option $\text{B}$ (which claimed $y = \frac{2}{e}$) likely came from misinterpreting the behavior of the exponential functions.

Would you like a more detailed explanation of exponential growth, or do you have any questions on this topic?

Related Questions:

1. What are the rules for finding horizontal asymptotes for rational functions?
2. How do exponential functions compare in growth rates?
3. How can you determine vertical asymptotes?
4. What happens to functions involving different bases of exponentials as $x \to \infty$?
5. How do polynomial and exponential terms behave at large values of $x$?

Tip: When evaluating limits at infinity, focus on the terms that grow fastest and how they compare to one another!