Find all the $\bf{x}\in\mathbb{R}^3$, which has a property that $\displaystyle \lim_{ n \to \infty } A^n\bf{x} = \bf{x}$ $A$ is a matrix with elements
$$\begin{pmatrix}
\frac{2}{3} & -\frac{2}{3}& -\frac{1}{3} \\
-\frac{2}{3}& 1 & \frac{2}{3} \\
\frac{4}{3} & 0 & -\frac{1}{3}
\end{pmatrix}$$
Find all the $\bf{x}\in\mathbb{R}^3$, which has a property that $$\displaystyle \lim_{ n \to \infty } A^n\bf{x} = \bf{x}$$

A's eigenvalues are $-\frac{1}{3}$,1 and $\frac{2}{3}$ and eigenvectors are $\left(
  \begin{array}{c}
    0 \\
    -\frac{1}{2} \\
    1
  \end{array}
\right),\left(
  \begin{array}{c}
    1 \\
    -1 \\
    1
  \end{array}
\right),\left(
  \begin{array}{c}
    \frac{3}{4} \\
    -\frac{1}{2} \\
    1
  \end{array}
\right)$ for each.
I tried to get $A^n$ at first then got $$\begin{pmatrix}
(\frac{2}{3})^n & -2+\frac{2^{n+1}}{3}& 0 \\
-\frac{2+2^{n+1}}{3^{n+1}}& \frac{1-2^{n+2}}{3^{n+1}}-2 & \frac{2}{3^n} \\
\frac{4+2^{n+2}}{3^{n+1}} & \frac{2-2^{n+3}}{3^{n+1}} & -\frac{1}{3^n}
\end{pmatrix}$$
Then I tried find $$\displaystyle \lim_{ n \to \infty }(\frac{2}{3})^nx_1+(-2+\frac{2^{n+1}}{3})x_2=x_1 \\\displaystyle \lim_{ n \to \infty }-\frac{2+2^{n+1}}{3^{n+1}}x_1+(\frac{1-2^{n+2}}{3^{n+1}}-2)x_2+\frac{2}{3^n}x_3=x_2\\\ \displaystyle \lim_{ n \to \infty }\frac{4+2^{n+2}}{3^{n+1}}x_1+\frac{2-2^{n+3}}{3^{n+1}}x_2-\frac{1}{3^n}x_3=x_3$$
But I could not lead meaningful conclusion from these equations but $(0,0,0)$.
Where did I get wrong?

The answer of @AryamanMaithani explains very well how you can solve this question, but it uses the fact that the matrix is diagonalizable. I just would like to add a small precision, that you can prove directly and more easily that :
for every matrix $A \in \mathcal{M}_n(\mathbb{R})$ (diagonalizable or not),
$$\forall x \in \mathbb{R}^n, \text{ } \lim_{n\rightarrow +\infty} A^nx =x \text{ } \Longleftrightarrow \text{ }Ax=x$$
(so the set $\lbrace x \in \mathbb{R}^n \text{ }|\text{ } \lim_{n\rightarrow +\infty} A^nx =x \rbrace$ is always equal to the eigenspace associated to the eigenvalue $1$)
Indeed, if $A^nx \rightarrow x$, then you have $A^{n+1}x \rightarrow Ax$ and $A^{n+1}x \rightarrow x$, so $Ax=x$, so $x$ is an eigenvector associated to the eigenvalue $1$. Conversely, if $Ax=x$, it is immediate that $A^nx=x$ for all $n$, and then $A^nx \rightarrow x$ when $n$ tends to $+\infty$.
