Let `A = [(1,0),(2,3)] and A^(n) = [(a, b),(c,d)]` then `lim_(n to oo) (b + c)/(a + d)` =

Explanation

To solve the problem, we need to find the limit as \( n \) approaches infinity of the expression \( \frac{b + c}{a + d} \) where \( A^n = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \) for the matrix \( A = \begin{pmatrix} 1 & 0 \\ 2 & 3 \end{pmatrix} \).

### Step 1: Calculate \( A^2 \)

First, we calculate \( A^2 \):
\[ A^2 = A \cdot A = \begin{pmatrix} 1 & 0 \\ 2 & 3 \end{pmatrix} \cdot \begin{pmatrix} 1 & 0 \\ 2 & 3 \end{pmatrix} \]
Calculating the elements:
- First row, first column: \( 1 \cdot 1 + 0 \cdot 2 = 1 \)
- First row, second column: \( 1 \cdot 0 + 0 \cdot 3 = 0 \)
- Second row, first column: \( 2 \cdot 1 + 3 \cdot 2 = 2 + 6 = 8 \)
- Second row, second column: \( 2 \cdot 0 + 3 \cdot 3 = 0 + 9 = 9 \)

Thus, we have:
\[ A^2 = \begin{pmatrix} 1 & 0 \\ 8 & 9 \end{pmatrix} \]

### Step 2: Calculate \( A^3 \)

Next, we calculate \( A^3 \):
\[ A^3 = A^2 \cdot A = \begin{pmatrix} 1 & 0 \\ 8 & 9 \end{pmatrix} \cdot \begin{pmatrix} 1 & 0 \\ 2 & 3 \end{pmatrix} \]
Calculating the elements:
- First row, first column: \( 1 \cdot 1 + 0 \cdot 2 = 1 \)
- First row, second column: \( 1 \cdot 0 + 0 \cdot 3 = 0 \)
- Second row, first column: \( 8 \cdot 1 + 9 \cdot 2 = 8 + 18 = 26 \)
- Second row, second column: \( 8 \cdot 0 + 9 \cdot 3 = 0 + 27 = 27 \)

Thus, we have:
\[ A^3 = \begin{pmatrix} 1 & 0 \\ 26 & 27 \end{pmatrix} \]

### Step 3: Identify the Pattern

From our calculations, we can observe a pattern:
- The first row remains \( (1, 0) \).
- The second row seems to follow a pattern related to powers of 3.

We can conjecture that:
\[ A^n = \begin{pmatrix} 1 & 0 \\ 3^{n-1} \cdot 2 & 3^n \end{pmatrix} \]

### Step 4: Identify \( a, b, c, d \)

From the matrix \( A^n \), we identify:
- \( a = 1 \)
- \( b = 0 \)
- \( c = 2 \cdot 3^{n-1} \)
- \( d = 3^n \)

### Step 5: Calculate \( \frac{b + c}{a + d} \)

Now we compute:
\[ \frac{b + c}{a + d} = \frac{0 + 2 \cdot 3^{n-1}}{1 + 3^n} \]
This simplifies to:
\[ \frac{2 \cdot 3^{n-1}}{1 + 3^n} \]

### Step 6: Simplify the Limit as \( n \to \infty \)

As \( n \to \infty \):
\[ \frac{2 \cdot 3^{n-1}}{1 + 3^n} = \frac{2 \cdot 3^{n-1}}{3^n(1/3^n + 1)} = \frac{2}{3(1/3^n + 1)} \]
As \( n \to \infty \), \( \frac{1}{3^n} \to 0 \):
\[ \frac{2}{3(0 + 1)} = \frac{2}{3} \]

### Final Answer

Thus, the limit is:
\[ \lim_{n \to \infty} \frac{b + c}{a + d} = \frac{2}{3} \]