Limit of a sequence when the algebraic limit theorem 'breaks down' Background 
There is the very well-known technique of computing the limit of a sequence by taking limit on both sides whenever recurrence relation arises. For example, $b_n = \frac{\alpha^n}{n!} $ for $0 < \alpha < 1$. 
Question 
But now there is this sequence such that taking limit on both sides will yield meaningless results:

$$a_n = \frac{1}{3}a_{n-1} + \frac{2}{3}a_{n-2} \; \; \;\text{for any
integer} \; n \geq 3, $$ and $a_1 = 0$, $a_2 = 1$. 

The question is how this limit should be evaluated? (The existence of the limit is already proved by proving the sequence is a Cauchy sequence.) 
My attempt 

It can be observed that 
$$ a_n - a_{n-1} = \frac{-2}{3} a_{n-1} + \frac{2}{3} a_{n-2} =
\frac{-2}{3} (a_{n-1} - a_{n-2})$$
Then the two sides are symmetrical and hence 
$$ a_n - a_{n-1}= (\frac{-2}{3})^{n-1}(a_1-a_0) =
 (\frac{-2}{3})^{n-1}$$
But again taking limit on both sides yields $ 0 = 0$


$$a_n = \frac13a_{n-1} + \frac23a_{n-2}$$
The characteristic equation is $x^2-\frac13x-\frac23=0$.
$$3x^2-x-2=0$$
$$(3x+2)(x-1)=0$$
$$x=-\frac23,1$$
$$a_n = \alpha \left( -\frac23\right)^n+\beta$$
We have $a_1=0$ and $a_2=1$,
$$0=\alpha\left( -\frac23\right)+\beta$$
$$1=\alpha\left( -\frac23\right)^2+\beta$$
$$1=\alpha\left( -\frac23\right)\left( -\frac53\right)$$
$$\alpha=\frac9{10}, \beta=\frac23\alpha=\frac23\cdot\frac{9}{10}=\frac35$$
$$a_n = \frac{9}{10}\left( -\frac23\right)^n+\frac35$$
$$\lim_{n \to \infty}a_n = \frac35$$
