Finding limits of diagonalised matrix I diagonalised 
$A=$
$\left[\begin{array}[c]{rr} 0.6 & 0.9\\ 0.4 & 0.1\end{array}\right]$ 
and got 
$SAS^-$$^1$$= V = (-1/13)$
$\left[\begin{array}[c]{rr} -1 & -1\\ -4 & 9\end{array}\right]$ 
$\left[\begin{array}[c]{rr} 1 & 0\\ 0 & -0.3\end{array}\right]$ 
$\left[\begin{array}[c]{rr} 9 & 1\\ 4 & -1\end{array}\right]$
now I am supposed to somehow deduce what matrix $V^k$ approaches as $k$ approaches $∞$. How am I supposed to figure that out?
Also, I am supposed to find the limit for $SV^kS^-$$^1$ when $k$ approaches $∞$ and specify what the columns of this matrix portray. 
What came to mind is that when all $|λ|<1$ then $A^k$ approaches $0$, and $λ_1=1$ and $λ_2 = -0.3$. I'm not really sure if that's of any use though. I'm somewhat lost and any help is much appreciated!

Your solution looks close.
Write the matrix as fractions:
$$A=\left[\begin{array}[c]{rr} \dfrac{6}{10} & \dfrac{9}{10}\\ \dfrac{4}{10} & \dfrac{1}{10}\end{array}\right]$$
I have my eigenvalues and eigenvectors swapped from the order you show yours in.
Diagonalization yields:
$$A = PJP^{-1} = 
\left(\begin{array}{cc}
 -1 & \frac{9}{4} \\ 1 & 1 \\\end{array} \right)\left(\begin{array}{cc}
 -\frac{3}{10} & 0 \\ 0 & 1 \\\end{array}\right) \left(\begin{array}{cc}
 -\frac{4}{13} & \frac{9}{13} \\ \frac{4}{13} & \frac{4}{13} \\ \end{array}
\right)$$
This yields:
$$A^k = PJ^kP^{-1} = \left(
\begin{array}{cc}
 \frac{1}{13} 2^{2-k} \left(-\frac{3}{5}\right)^k+\frac{9}{13} & ~\frac{9}{13}-\frac{1}{13} \left(-\frac{1}{10}\right)^k 3^{k+2} \\
 \frac{4}{13}-\frac{1}{13} \left(-\frac{3}{5}\right)^k 2^{2-k} & ~\frac{1}{13} 3^{k+2} \left(-\frac{1}{10}\right)^k+\frac{4}{13} \\
\end{array}
\right)$$
Note: Once we have diagonalized the matrix, we have:
$$J^k = \left(\begin{array}{cc}
 -\frac{3}{10} & 0 \\ 0 & 1 \\\end{array}\right)^k = \left(\begin{array}{cc}
 \left(-\frac{3}{10}\right)^k & 0 \\ 0 & (1)^k \\\end{array}\right)$$
Now, as far as the limit goes, do you see what happens to the $k$ terms as $k$ approaches infinity? They approach zero, so we are left with:
$$\displaystyle \lim_{k \to \infty} A^k = \left(
\begin{array}{cc}
 \frac{9}{13}  & ~\frac{9}{13} \\
 \frac{4}{13} & ~\frac{4}{13}  \\
\end{array}
\right) $$
An easier approach would have been to take the limit of the diagonalized matrix first, and then multiply out, which yields:
$$\displaystyle \lim_{k \to \infty} J^k = \left(
\begin{array}{cc}
 0 & 0 \\
 0 & 1  \\
\end{array}
\right) $$
