This is a practice problem for a midterm in a real analysis undergrad class, but I tagged it as homework anyway.
Let $x_0\in R$ and define $x_{n+1}=\frac{1}{2}(3-x_n)$ for $n\geq 0$. Prove that $x_n\to 1$ as $n\to \infty$.
To start I know that if $x_n$ is convergent then for sufficiently large $k$, $x_k$ is within the epsilon neighborhood of $x_{k+1}$. So if $x_n$ is convergent the limit is the solution to $C=\frac12(3-C)$ or $C=1$. Now if I can prove $x_n$ is convergent, it must converge to $1$, but I cannot figure out how to prove it. Any help would be great. My midterm is tomorrow unfortunately.