Here is a formal derivation of your result. The sequence you have found is a generalization of the Fibonacci sequence.
There have been many extensions of the sequence with adjustable (integer) coefficients and different (integer) initial conditions, e.g., $f_n=af_{n-1}+bf_{n-2}$. (You can look up Pell, Jacobsthal, Lucas, Pell-Lucas, and Jacobsthal-Lucas sequences.) Maynard has extended the analysis to $a,b\in\mathbb{R}$, (Ref: Maynard, P. (2008), “Generalised Binet Formulae,” $Applied \ Probability \ Trust$; available at http://ms.appliedprobability.org/data/files/Articles%2040/40-3-2.pdf.)
We have extended Maynard's analysis to include arbitrary $f_0,f_1\in\mathbb{R}$. It is relatively straightforward to show that
$$f_n=\left(f_1-\frac{af_0}{2}\right) \frac{\alpha^n-\beta^n}{\alpha-\beta}+\frac{f_0}{2} (\alpha^n+\beta^n) $$
where $\alpha,\beta=(a\pm\sqrt{a^2+4b})/2$.
The result is written in this form to underscore that it is the sum of a Fibonacci-type and Lucas-type Binet-like terms. It will also reduce to the standard Fibonacci and Lucas sequences for $a=b=1 \ \text{and} \ f_0=0, f_1=1$. Notice that there is nothing in this derivation that limits the results to integer values for the initial conditions, scaling factors, or the sequence itself.
So, specializing to your case, we can say
$$x_n=\frac{x_{n-1}}{2} +\frac{x_{n-2}}{2}$$
and
$$\alpha,\beta=\frac{\frac{1}{2}\pm\sqrt{\frac{1}{4}+2}}{2}=1,\frac{1}{2}$$
For the limit as $n\to\infty$, we note that the $\alpha$-term dominates, and is in fact, always unity, ergo
$$
\lim_{n\to\infty} x_n=\left(x_1-\frac{x_0}{4}\right)\frac{\alpha^n}{\alpha-\beta}+\frac{x_0}{2}\alpha^n=\left(x_1-\frac{x_0}{4}\right)\frac{2}{3}+\frac{x_0}{2}=\frac{2x_1+x_0}{3}
$$
as was shown previously.
This proves the OP's assertion. The advantage of this method is that it will apply to all such problems.