<s>[INST] Shorthand limits of matrices? Is there any relatively easy way to solve limits with matrices?
$$
\lim_{h\rightarrow \infty}
\begin{pmatrix}
1 & 0 & 0\\
\frac{1}{r} & 0 & \frac{-1}{r}\\
0 & \frac{1}{hc} & 1
\end{pmatrix}^h
=
\begin{pmatrix}
1 & 0 & 0\\
\frac{1}{r} & 0 & \frac{-1}{r}\\
\frac{1}{rc} & 0 & 1-\frac{1}{rc}
\end{pmatrix}
$$
I have painstakingly identified the solution for this particular input matrix by identifying numbers with gigantic values of h in matlab. I would prefer to not square a matrix about a hundred times and manually identify the numbers. It's like calculating the derivative of a function with limits, I rather figure out the tricks than always going through the limit.
The 4th power of the input matrix looks like this and I suppose you can make certain deductions, such as $\frac{1}{h}$ and $\frac{1}{h^2}=$ 0 and $\frac{3}{h}$ = 1.
$$
\begin{pmatrix}
1 & 0 & 0\\
\frac{1}{r} & 0 & \frac{-1}{r}\\
0 & \frac{1}{hc} & 1
\end{pmatrix}^4
=
\begin{pmatrix}
1 & 0 & 0\\
\frac{1}{r} & \frac{-1}{rhc} & \frac{-1}{r}\\
\frac{1}{rhc} & \frac{1}{hc} & 1-\frac{1}{rhc}\\
\end{pmatrix}^2
=
\begin{pmatrix}
1 & 0 & 0\\
\frac{1}{r}-\frac{1}{(rhc)^2}-\frac{1}{r^2hc} & \frac{1}{(rhc)^2}-\frac{1}{rhc} & \frac{-1}{r}+\frac{2}{r^2hc}\\
\frac{3}{rhc}-\frac{1}{(rhc)^2} & \frac{1}{hc}-\frac{2}{r(hc)^2} & 1-\frac{3}{rhc}+\frac{1}{(rhc)^2}\\
\end{pmatrix}
$$
I would like to understand how to go from the input matrix straight to the solution, I can reach the solution if I take the 4th power of the input matrix and look for the number 3. But that will become very time consuming for big matrices, on the order of 100 x 100.

Edit
The $\frac{1}{rc}$ part in the previous limit result had to receive another limit so here's the solution to that.
$$
\lim_{h\rightarrow \infty}
\begin{pmatrix}
1 & 0 & 0\\
\frac{1}{r} & 0 & \frac{-1}{r}\\
\frac{1}{rhc} & 0 & 1-\frac{1}{rhc}
\end{pmatrix}^h
=
\begin{pmatrix}
1 & 0 & 0\\
\frac{1}{r} & 0 & \frac{-1}{r}\\
1-e^\frac{-1}{rc} & 0 & e^\frac{-1}{rc}
\end{pmatrix}
$$
Again, I found the solution by squaring many times and seeing error patterns and using wolframalpha to turn error sums to e functions.
Maybe there is no simple way to find limits to matrices.
 [/INST] The problem here is the power $h$. But if the matrix is diagonalizable, there is an easy way to handle it. Let
$$H :=\begin{pmatrix}1 & 0 & 0\\
\frac{1}{r} & 0 & \frac{-1}{r}\\
0 & \frac{1}{hc} & 1
\end{pmatrix}$$
And suppose there is some matrix $Q$ with $Q^{-1}DQ = H$ where $$D =\begin{pmatrix}d_1&0&0\\0&d_2&0\\0&0&d_3\end{pmatrix}$$ is diagonal. Then
$$H^h = (Q^{-1}DQ)^h = Q^{-1}DQQ^{-1}DQQ^{-1}DQ\cdots Q^{-1}DQ = Q^{-1}D^hQ.$$
$$\lim_{h \to \infty} H^h  = \lim_{h\to \infty} Q^{-1}D^hQ$$
Because $h$ also appears in $H$, $Q$ will depend on $h$ as well. And $$D^h = \begin{pmatrix}d_1^h&0&0\\0&d_2^h&0\\0&0&d_3^h\end{pmatrix}$$
If you can find $\lim_{h \to \infty}Q$, then $\lim_{h \to \infty}Q^{-1}$ will be its inverse. And if you can also find $\lim_{h \to \infty}d_i^h$ for each $i$, then you have $\lim_{h\to\infty} D^h$, and $$\lim_{h\to \infty} H^h = \lim_{h \to \infty}Q \lim_{h\to\infty} D^h\lim_{h \to \infty}Q^{-1}$$
Which you can then compare to the other side.
</s>