TITLE: Limit of sequence in which each term is defined by the average of preceding two terms
QUESTION [100 upvotes]: We have a sequence of numbers $x_n$ determined by the equality 
$$x_n = \frac{x_{n-1} + x_{n-2}}{2}$$
The first and zeroth term are $x_1$ and $x_0$.The following limit must be expressed in terms of $x_0$ and $x_1$
$$\lim_{n\rightarrow\infty} x_n $$
The options are: 
A)$\frac{x_0 + 2x_1}{3}$
B)$\frac{2x_0 + 2x_1}{3}$
C)$\frac{2x_0 + 3x_1}{3}$
D)$\frac{2x_0 - 3x_1}{3}$
Since it was a multiple choice exam I plugged $x_0=1$ and $x_1=1$. Which means that all terms of this sequence is $1$,i.e, $$x_n=1, n\in \mathbb{N} $$
From this I concluded that option A was correct.I could not find any way to solve this one hence I resorted to this trick. What is the actual method to find the sequence's limit?

REPLY [4 votes]: First, the sequence $(x_n)$ has clearly a limit $\ell$ because 
$$
\left|x_{n+1}-x_{n}\right|=\frac{|x_1-x_0|}{2^n}.
$$
Now, sum the $n-1$ equations $2x_{i+2}=x_{i+1}+x_i$ for $i=0,1,\ldots,n-2$ and we obtain
$$
2x_n+x_{n-1}=2x_1+x_0.
$$
Passing to the limit we reach
$$
\ell=\frac{2x_1+x_0}{3}.
$$