If the recurrence relation is linear, homogeneous and has constant coefficients, here is a systematic way to solve it. Similar method will work if you have linear, inhomogeneous equation with constant coefficients. (For instance if the inhomogeneity is a constant, the inhomogeneity can be removed if possible by defining $b_n = a_n + c$ and figuring out $c$ such that the equation in $b_n$ is homogeneous. If it is not possible, which happens when the sum of the coefficients of $a_n$ add up to zero, then there is a linear growth term added to $a_n$)
First obtain the characteristic equation. To do this, assume $a_n = m^n$.
Plug it in to get a quadratic in $m$ in this case. Solve for $m$. (If you have $a_n$ depending on $a_{n-1},a_{n-2},\ldots,a_{n-p}$, you will get a $p^{th}$ order polynomial.)
Get the two roots say $m_1, m_2$. (In general, the $p$ roots).
Now the general solution is given by the linear combination of the roots namely $a_n = c_1 m_1^n + c_2 m_2^n$. (In general, $a_n = \displaystyle \sum_{k=1}^p c_k m_k^n$)
Solve for $c_1$ and $c_2$ (In general, $c_1,c_2,\ldots,c_p$) using the initial conditions
For your recurrence, the corresponding equation becomes, $2m^2-m-1 = 0 \Rightarrow (2m+1)(m-1) = 0 \Rightarrow m = -\frac{1}{2},1$
Hence, $a_n = c (-\frac{1}{2})^n + d$
If $a_0 =0, a_1 =1$, we get $c + d = 0$ and $d - \frac{c}{2} = 1$
$d = \frac{2}{3}$ and $c = -\frac{2}{3}$
Hence, $a_n = \frac{2}{3} - \frac{2}{3} (-\frac{1}{2})^{n}$
Hence, $\displaystyle \lim_{n \rightarrow \infty} a_n = \frac{2}{3}$
If $a_0 =\alpha, a_1 =\beta$, we get $c + d = \alpha$ and $d - \frac{c}{2} = \beta$
$c = \frac{2}{3}(\alpha-\beta)$, $d = \frac{\alpha + 2 \beta}{3}$
Hence, $a_n = \frac{\alpha + 2 \beta}{3} + \frac{2}{3}(\alpha-\beta) (\frac{-1}{2})^n$
Hence, $\displaystyle \lim_{n \rightarrow \infty} a_n = \frac{\alpha + 2 \beta}{3}$
This methodology is analogous to plugging in $y=e^{mx}$ when you want to solve a linear, homogeneous ODE with constant coefficients. You get the constants for the ODE using the boundary conditions. Here you have a difference equation instead of a differential equation.