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Vertical asymptotes are a key feature in graph analysis and occur when a function approaches infinity as the input gets closer to a certain value. This typically happens when the denominator of a rational function equals zero and the numerator does not. In simpler terms, the function becomes undefined and shoots off to infinity, creating a vertical asymptote.

For our function, we examine the denominator \( \ln(2 + e^{3x}) \) to find vertical asymptotes. We set it to zero: \( \ln(2 + e^{3x}) = 0 \), meaning \( 2 + e^{3x} = 1 \). Solving \( e^{3x} = -1 \) leads to no solution since \( e^{3x} \) is always positive. Hence, there are no vertical asymptotes in this function. It's a good example of an instance where you might initially assume there could be vertical asymptotes but later conclude otherwise based on algebraic analysis.

Horizontal asymptotes describe the behavior of a function as the input grows very large or very small. Specifically, they are horizontal lines that the graph of the function approaches as \( x \to \infty \) or \( x \to -\infty \).

To find horizontal asymptotes for our function \( f(x) = \frac{\ln(1 + e^x)}{\ln(2 + e^{3x})} \), we evaluate the limits at infinity. As \( x \to \infty \), both numerator and denominator are dominated by exponential terms, respectively leading to \( \ln(1 + e^x) \approx x \) and \( \ln(2 + e^{3x}) \approx 3x \). Thus \( \lim_{{x \to \infty}} f(x) = \frac{x}{3x} = \frac{1}{3} \). At \( x \to -\infty \), exponential terms \( e^x \) and \( e^{3x} \) approach zero, so the expression simplifies to \( \ln(1) = 0 \) and \( \ln(2) \). Therefore, \( \lim_{{x \to -\infty}} f(x) = \frac{0}{\ln(2)} = 0 \). Consequently, the graph has horizontal asymptotes at \( y = \frac{1}{3} \) and \( y = 0 \). These asymptotes give a clear picture of how the function behaves at extreme values of \( x \).

Limits are foundational in calculus, describing how a function behaves as the input approaches a certain value. Understanding limits helps in determining asymptotic behavior, which reveals the end behavior of a graph.

For the given function, limits are crucial in identifying horizontal asymptotes. When finding \( \lim_{{x \to \infty}} \) and \( \lim_{{x \to -\infty}} \), the function simplifies due to the dominance of exponential expressions. The approach involves approximating the logarithms based on their dominating exponential terms, and understanding that properties of limits allow us to predict that the values \( \frac{1}{3} \) and \( 0 \) represent the horizontal asymptotes, corresponding to the equation \( \frac{x}{3x} \) as \( x \to \infty \) and constant \( \ln(2) \) for \( x \to -\infty \). This clear explanation of limits not only aids in solving the problem at hand, but also reinforces how limits govern the predictive behavior of function graphs.