행렬 거듭제곱 극한

$\displaystyle B ^{2} = \begin{pmatrix} 3 & 0 \\ 0 & 4 \end{pmatrix} \begin{pmatrix} 3 & 0 \\ 0 & 4 \end{pmatrix} = \begin{pmatrix} 3 ^{2} & 0 \\ 0 & 4 ^{2} \end{pmatrix}$ $\displaystyle B ^{3} = \begin{pmatrix} 3 ^{2} & 0 \\ 0 & 4 ^{2} \end{pmatrix} \begin{pmatrix} 3 & 0 \\ 0 & 4 \end{pmatrix} = \begin{pmatrix} 3 ^{3} & 0 \\ 0 & 4 ^{3} \end{pmatrix}$ … 이므로 $\displaystyle B ^{n} = \begin{pmatrix} 3 ^{n} & 0 \\ 0 & 4 ^{n} \end{pmatrix}$임을 알 수 있다. ∴ $A ^{- 1} B ^{n} A$$\displaystyle = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 3 ^{n} & 0 \\ 0 & 4 ^{n} \end{pmatrix} \begin{pmatrix} 1 & - 2 \\ 0 & 1 \end{pmatrix}$ $\displaystyle = \begin{pmatrix} 3 ^{n} & - 2 \cdot 3 ^{n} + 2 \cdot 4 ^{n} \\ 0 & 4 ^{n} \end{pmatrix}$ ∴ $\displaystyle \lim\limits _{n \rightarrow \infty} {\frac{b _{n} + c _{n}}{a _{n} + d _{n}}}$$\displaystyle = \lim\limits _{n \rightarrow \infty} {\frac{- 2 \cdot 3 ^{n} + 2 \cdot 4 ^{n}}{3 ^{n} + 4 ^{n}}}$ $\displaystyle = \lim\limits _{n \rightarrow \infty} {\frac{- 2 \cdot \left( \frac{3}{4} \right) ^{n} + 2}{\left( \frac{3}{4} \right) ^{n} + 1} = 2}$