Given, $\lim_{x \to \infty} \frac{2x + 3^{x+1}}{(2x+3)^2 + (2x+4)^3}$
Step-by-Step Solution
Key Concept: Identify dominant terms in exponential and polynomial expressions separately, then apply L'Hôpital's rule or algebraic manipulation.
For the limit $\lim_{x \to \infty} \frac{2x + 3^{x+1}}{(2x+3)^2 + (2x+4)^3}$, we identify the dominant terms. In the numerator, $3^{x+1}$ grows exponentially and dominates $2x$. In the denominator, $(2x+4)^3$ is a cubic polynomial and dominates $(2x+3)^2$. Dividing by $3^x$ in numerator and $(2x)^3 = 8x^3$ in denominator: $\lim_{x \to \infty} \frac{3 \cdot 3^x}{8x^3}$ still shows exponential dominance. Factoring more carefully with $x = \frac{1}{x}$ substitution approach, the limit simplifies to $\sqrt{2}$.
Correct Answer: 1