Question

\(\displaystyle \lim_{n\to \infty}\frac{2^{n+1}+3^{n+1}}{2^n+3^n}=\)

• A. \(3\)
• B. \(2\)
• C. \(1\)
• D. \(0\)

Hint:

For limits involving \(a^n\) and \(b^n\), divide by the largest base power.

Answer 1

The Correct Option is A

Concept:
When powers with different bases are present and \(n\to\infty\), the term with the larger base dominates. Here, \(3^n\) grows faster than \(2^n\).

Step 1: Write the limit. \[ L=\lim_{n\to\infty}\frac{2^{n+1}+3^{n+1}}{2^n+3^n} \]

Step 2: Divide numerator and denominator by \(3^n\). \[ L=\lim_{n\to\infty} \frac{\frac{2^{n+1}}{3^n}+\frac{3^{n+1}}{3^n}} {\frac{2^n}{3^n}+\frac{3^n}{3^n}} \] \[ = \lim_{n\to\infty} \frac{2\left(\frac{2}{3}\right)^n+3} {\left(\frac{2}{3}\right)^n+1} \]

Step 3: Use the fact that \(\left(\dfrac23\right)^n\to 0\). Since \[ \left|\frac{2}{3}\right|<1 \] we have \[ \left(\frac{2}{3}\right)^n\to 0 \] Therefore, \[ L=\frac{2(0)+3}{0+1} \] \[ L=3 \]

Step 4: Final answer. \[ \boxed{3} \]