Solution:

We have the recurrence $E_{n} = \frac{1}{2}\left(E_{n-1} + 1\right) + \frac{1}{2}\left(E_{n-2} + 1\right)$, or $E_{n} = 1 + \frac{1}{2}\left(E_{n-1} + E_{n-2}\right)$, for $n \geq 2$.

Let $F_{n} = E_{n} - \frac{2}{3} n$. By directly plugging this into the recurrence for $E_{n}$, we get the recurrence $F_{n} = \frac{1}{2}\left(F_{n-1} + F_{n-2}\right)$. The roots of the characteristic polynomial of this recurrence are $1$ and $-\frac{1}{2}$, so $F_{n} = A + B\left(-\frac{1}{2}\right)^{n}$ for some $A$ and $B$ depending on the initial conditions. But clearly we have $E_{0} = 0$ and $E_{1} = 1$ so $F_{0} = 0$ and $F_{1} = \frac{1}{3}$ so $A = \frac{2}{9}$ and $B = -\frac{2}{9}$.

Hence, $E_{n} = \frac{2}{3} n + \frac{2}{9} - \frac{2}{9}\left(-\frac{1}{2}\right)^{n}$, so $\lim_{n \rightarrow \infty}\left(E_{n} - \frac{2}{3} n - \frac{2}{9}\right) = 0$. Hence $\left(\frac{2}{3}, \frac{2}{9}\right)$ is the desired pair.