Unclear problem with $n$-th power matrix and limit 
Find $$\lim\limits_{n \to \infty} \frac{A_n}{D_n}$$ where
$$\begin{pmatrix}    
    19 & -48 \\
    8 & -21 \\
    \end{pmatrix} ^{\! n} =  
    \begin{pmatrix}
    A_n & B_n \\
    C_n & D_n \\
    \end{pmatrix}$$


$n$ - is the power of a matrix, but what is $A_n, B_n, C_n, D_n$ then? Is it a corresponding element of a matrix in the $n$-th power?
How is this type of problem called? And what is the way to solve that problem?
 A: $$A:=\begin{pmatrix}19&-48\\8&-21\end{pmatrix}=\begin{pmatrix}2&3\\1&1\end{pmatrix}\begin{pmatrix}-5&0\\0&3\end{pmatrix}\begin{pmatrix}2&3\\1&1\end{pmatrix}^{-1}=PDP^{-1}$$
So $A^n=PDP^{-1}PDP^{-1}\cdots PDP^{-1}=PD^nP^{-1}$
That is, $$A^n=\begin{pmatrix}2&3\\1&1\end{pmatrix}\begin{pmatrix}(-5)^n&0\\0&3^n\end{pmatrix}\begin{pmatrix}2&3\\1&1\end{pmatrix}^{-1}=\begin{pmatrix}-2 (-5)^n + 3^{1 + n}& 6 (-5)^n - 2\times 3^{1 + n}\\-(-5)^n + 3^n& 
 3 (-5)^n - 2\times 3^n\end{pmatrix}$$
A: Here is an unconventional approach: we have
$$
M = \pmatrix{19 & -48\\ 8 & -21}.
$$
We find that the eigenvalues satisfy
$$
\det(M - xI) = x^2 + 2x - 15 = 0 \implies x = -5,3.
$$
By the Cayley Hamilton theorem, the powers of $M$ satisfy the recurrence
$$
M^n + 2M^{n-1} -15 M^{n-2} = 0
$$
From the theory of constant coefficient homogeneous linear difference equations, it follows that $M^n$ has the form
$$
M^n = (-5)^n P + 3^n Q
$$
for some matrices $P,Q$. We can solve for $P,Q$ using the "initial conditions" of $n=0,1$. We have
$$
P + Q = M^0 = \pmatrix{1&0\\0&1}, \quad (-5)P + 3Q = M^1 = \pmatrix{19 & -48\\8 & -21}.
$$
Subtracting the second equation from $3$ times the first yields
$$
3P - (-5)P + 0Q = 3\pmatrix{1&0\\0&1} - \pmatrix{19 & -48\\8 & -21} \implies\\
8P = \pmatrix{-16 & -48\\8 & 24} \implies P = \pmatrix{-2&-6\\1&3}.
$$
We use the first equation to find
$$
Q = \pmatrix{1&0\\0&1} - P = \pmatrix{3&6\\-1&-2}.
$$
So, we have
$$
M^n = (-5)^n \pmatrix{-2&-6\\1&3} + 3^n\pmatrix{3&6\\-1&-2}.
$$
