Step 1: Understanding the Concept:
We need to find the limit of a sequence involving exponential terms. The standard technique for limits of this form (a fraction with sums of exponentials) is to divide both the numerator and the denominator by the fastest-growing term.
Step 2: Key Formula or Approach:
1. Identify the term with the largest base in the expression. In this case, it's \(3^n\).
2. Divide both the numerator and the denominator by this term.
3. Use the property that for any \(|r| < 1\), \(\lim_{n \to \infty} r^n = 0\).
Step 3: Detailed Explanation:
The expression is \(\frac{2^{n+1} + 3^{n+1}}{2^n + 3^n}\).
The term with the largest base is \(3^n\). We will divide the numerator and the denominator by \(3^n\).
\[ \lim_{n \to \infty} \frac{\frac{2^{n+1} + 3^{n+1}}{3^n}}{\frac{2^n + 3^n}{3^n}} = \lim_{n \to \infty} \frac{\frac{2 \cdot 2^n}{3^n} + \frac{3 \cdot 3^n}{3^n}}{\frac{2^n}{3^n} + \frac{3^n}{3^n}} \]
\[ = \lim_{n \to \infty} \frac{2 \left(\frac{2}{3}\right)^n + 3}{\left(\frac{2}{3}\right)^n + 1} \]
Now, we evaluate the limit. Since \(|\frac{2}{3}| < 1\), we have:
\[ \lim_{n \to \infty} \left(\frac{2}{3}\right)^n = 0 \]
Substitute this result into the expression:
\[ = \frac{2(0) + 3}{0 + 1} = \frac{3}{1} = 3 \]
Step 4: Final Answer:
The limit of the sequence is 3, which corresponds to option (A).