Limit of matrix to a power and diagonalized matrix to a power give different results I have the following problem. I wish to find
$$
\lim_{n \to \infty} \psi M^n 1
$$
where
$$
\psi = \begin{bmatrix} 1/4 & 1/2 \end{bmatrix}, \quad
1=\begin{bmatrix} 1 & 1 \end{bmatrix}^T
$$
and 
$$
M=
        \begin{bmatrix}
        \left( 1-\frac{L}{n} \right)^2 
        & 2 \left(\frac{L}{n} \right) \left(1-\frac{L}{n}\right) \\
        \left(\frac{L}{n} \right) \left(1-\frac{L}{n}\right) 
        & \left( 1-\frac{L}{n} \right)^2 \\
        \end{bmatrix}
$$
I diagonalized $M=SDS^{-1}$ with the eigensystem of $M$ and found 
$$
\lim_{n \to \infty} \psi S D^n S^{-1}1
=\frac{1}{4}e^{-2L} \left(3 \cosh \left(\sqrt{2} L \right) 
+ 2 \sqrt{2} \sinh \left( \sqrt{2} L \right) \right)
$$
I found this answer both by hand and by typing it into mathematica. However, when I type the original problem into mathematica, it gives
$$
\lim_{n \to \infty} \psi M^n 1=\frac{3}{4}e^{-2L}
$$
The latter answer is honestly more what I would have expected from the problem setup, but I don't know why the two answers differ nor how to perform the latter limit by hand. Being able to do it by hand is important, because this is actually a warm up to the bigger problem I'm working on. What am I missing? Is there good resource that could help me out, preferably in pdf form?
Edit: The eigensystem of $M$ also depends on $n$, as follows:
$$
    \lambda_1 = \left(1-\frac{L}{n}\right) \left(1-\frac{L}{n} \left(1 - \sqrt{2} \right) \right),
    \qquad
    v_1 = \begin{bmatrix} \sqrt{2} & 1 \end{bmatrix}
$$
$$
    \lambda_2 = \left(1-\frac{L}{n}\right) \left(1-\frac{L}{n} \left(1 + \sqrt{2} \right) \right),
    \qquad
    v_2 = \begin{bmatrix} -\sqrt{2} & 1 \end{bmatrix}
$$
giving $D=diag(\lambda_1, \lambda_2)$ and $S=[v1,v2]$.
Edit: Fixed $\sinh$ coefficient and $3/4$ coefficient.
 A: Partial answer:
Let $A=\begin{bmatrix} 1 & -2 \\ -1 & 1\end{bmatrix} $ and let
$M(x) = (1-x) (I - xA)$.
Note that $M_n = M({L \over n})$, and $A = U \begin{bmatrix} 1+\sqrt{2} & 0 \\ 0 & 1-\sqrt{2} \end{bmatrix} U^{-1}$ for some invertible $U$.
Then $M(x)^n = (1-x)^n U \begin{bmatrix} (1-x(1+\sqrt{2}))^n & 0 \\ 0 & (1-x(1-\sqrt{2}))^n \end{bmatrix} U^{-1}$.
Hence $M_n^n =(1-{L \over n})^n U \begin{bmatrix} (1-{L \over n}(1+\sqrt{2}))^n & 0 \\ 0 & (1-{L \over n}(1-\sqrt{2}))^n \end{bmatrix} U^{-1} $ from which we get
$\lim_{n \to \infty} M_n^n = e^{-L} U \begin{bmatrix} e^{-L (1+\sqrt{2})} & 0 \\ 0 & e^{-L (1-\sqrt{2})} \end{bmatrix} U^{-1} = U \begin{bmatrix} e^{-L (2+\sqrt{2})} & 0 \\ 0 & e^{-L (2-\sqrt{2})} \end{bmatrix} U^{-1} $.
More grind gives (making the large assumption of no mistakes):
$\lim_{n \to \infty} M_n^n = e^{-2L} \begin{bmatrix} \cosh \sqrt{2}L & \sqrt{2}  \sinh \sqrt{2}L \\
{1 \over \sqrt{2}} \sinh \sqrt{2}L & \cosh \sqrt{2}L \end{bmatrix}$.
A: I got $$\frac{1}4e^{-2L}\left(3\cosh(L\sqrt{2})+2\sqrt{2}\sinh(L\sqrt{2})\right)$$ in two different ways.
This is close to your complicated example. I get a different coefficient for $\sinh(L\sqrt{2}).$ (Your coefficient was $\sqrt{2}.$)
In general, if $M=\begin{pmatrix}a&2b\\b&a\end{pmatrix},$ it has eigenvectors $\begin{pmatrix}\pm\sqrt{2}\\1\end{pmatrix}$ with eigenvalues $a\pm b\sqrt{2}.$ We can then write $$\begin{pmatrix}1\\1\end{pmatrix}=\left(\frac{1}{2}+\frac{\sqrt{2}}4\right)\begin{pmatrix}\sqrt{2}\\1\end{pmatrix}+\left(\frac{1}{2}-\frac{\sqrt{2}}4\right)\begin{pmatrix}-\sqrt{2}\\1\end{pmatrix}$$
So $$M^n 1 =\left(\frac{1}{2}+\frac{\sqrt{2}}4\right)\left(a+b\sqrt{2}\right)^n\begin{pmatrix}\sqrt{2}\\1\end{pmatrix}+\left(\frac{1}{2}-\frac{\sqrt{2}}4\right)\left(a-b\sqrt{2}\right)^n\begin{pmatrix}-\sqrt{2}\\1\end{pmatrix}$$
and hence:
$$\psi M^n 1=\left(\frac{3}{8}+\frac{\sqrt{2}}{4}\right)\left(a+b\sqrt{2}\right)^n+\left(\frac{3}{8}-\frac{\sqrt{2}}{4}\right)\left(a-b\sqrt{2}\right)^n$$
Here, we can write $$M_n =  a_n\begin{pmatrix}a_n&2(1-a_n)\\1-a_n&a_n\end{pmatrix}$$ where $a_n=1-\frac{L}{n}.$ So $b_n=1-a_n=\frac{L}{n}.$
Then $$\psi M_n^n 1= a_n^n\left[\left(\frac{3}{8}+\frac{\sqrt{2}}{4}\right)\left(a_n+b_n\sqrt{2}\right)^n+\left(\frac{3}{8}-\frac{\sqrt{2}}{4}\right)\left(a_n-b_n\sqrt{2}\right)^n\right]$$
We know $a_n^n\to e^{-L}.$
And $$\left(a_n+b_n\sqrt{2}\right)^n = \left(1+\frac{L(\sqrt{2}-1)}{n}\right)^n\to e^{L(\sqrt 2-1)}$$ and $$\left(a_n-b_n\sqrt{2}\right)^n = \left(1+\frac{L(-\sqrt{2}-1)}{n}\right)^n\to e^{L(-\sqrt 2-1)}$$
So we get that:
$$\begin{align}\left(\frac{3}{8}+\frac{\sqrt{2}}{4}\right)\left(a_n+b_n\sqrt{2}\right)^n+\left(\frac{3}{8}-\frac{\sqrt{2}}{4}\right)\left(a_n-b_n\sqrt{2}\right)^n&\to \left(\frac{3}{8}+\frac{\sqrt{2}}{4}\right)e^{L(\sqrt{2}-1)}+\left(\frac{3}{8}-\frac{\sqrt{2}}{4}\right)e^{L(-\sqrt{2}-1)}\\
&=e^{-L}\left(\frac{3}{4}\cosh(L\sqrt{2})+\frac{\sqrt{2}}{2}\sinh(L\sqrt{2})\right)\end{align}$$
So I get the limit is:
$$e^{-2L}\left(\frac{3}{4}\cosh(L\sqrt{2})+\frac{\sqrt{2}}{2}\sinh(L\sqrt{2})\right)$$

Another approach.
Note that if $a_n=1-L/n$ and $\frac{1-a_n}{a_n}=\frac{L/n}{1-L/n}=\frac{1}{n/L-1}$. Then  $$\begin{align}M_n&=a_n^2\left(I+\frac{1}{n/L-1}\begin{pmatrix}0&2\\1&0\end{pmatrix}\right)
\end{align}$$
Now, $a_n^{2n}\to e^{-2L}.$
But there is no reason to expect $$\left(I+\frac{1}{n/L-1}\begin{pmatrix}0&2\\1&0\end{pmatrix}\right)^n$$ to contribute nothing to this formula.
Indeed, I'd expect it to converge to $$\exp\left(L\begin{pmatrix}0&2\\1&0\end{pmatrix}\right)$$
3here $\exp(A)=\sum_{k=0}^{\infty}\frac{1}{k!}A^k$ is the matrix exponential.
 This will be true at least when $L$ is an integer and and restricting to $n$ which are multiples of $L.$ 
Now, taking $A=\begin{pmatrix}0&2\\1&0\end{pmatrix},$ we have $A^2=2I$ and thus: $$\begin{align}\exp(LA)&=\left(\sum_{k=0}^{\infty} \frac{1}{(2k)!}2^kL^{2K}\right)I+\left(\sum_{k=0}^{\infty}\frac{1}{(2k+1)!}L^{2k+1}2^k\right)A\\
&=\cosh(\sqrt{2}L)I + \frac{1}{\sqrt{2}}\sinh(L\sqrt{2})A
\end{align}$$
Then $\psi I 1=\frac{3}{4}$ and $\psi A 1=1$ so you get the limit is:
$$e^{-2L}\left(\frac{3}{4}\cosh(L\sqrt{2})+\frac{1}{\sqrt{2}}\sinh(L\sqrt{2}\right)$$
which is $$\frac{1}{4}e^{-2L}\left(3\cosh(L\sqrt{2})+2\sqrt{2}\sinh(L\sqrt{2})\right)$$
