Non-Wide-Sense-Stationary process with autocorrelation that depends on time shift My question consists of two parts.
1)
 A wide sense stationary process has two properties:
a) it's mean is constant.
b) it's autocorrelation function depends on the time shift only.
My problem is that I can't think of a non wide sense stationary process that has the second condition(b) but not the first(a)
2)
Is the limit of the autocorrelation as the time shift approaches infinity is the square of the mean of the process? Is there a proof for this? 
 A: Regarding your questions:
1) Let $A$ and $B$ be independent random variables, where $A$ takes values in $\{0,\sqrt{2}\}$ with probability 0.5 each and $B$ takes values in $\{-1,1\}$ with probability 0.5 each. Notice that $\mathbb{E}[A] =\frac{\sqrt{2}}{2}$, $\mathbb{E}\left[A^2\right] = \mathbb{E}\left[B^2\right] = 1$ and $\mathbb{E}[B] = \mathbb{E}[AB] = 0$. 
Then, define the process $X_t = A\cos (t) + B\sin (t)$ for any time $t\in\mathbb{R}$. The mean of $X_t$ is $\mathbb{E}\left[X_t\right] = \mathbb{E}[A]\cos (t) + \mathbb{E}[B]\sin( t) = \frac{\sqrt{2}}{2}\cos (t)$, which is not constant with respect to the time $t$.
However, the autocorrelation of $X_t$ is 
\begin{align}
\mathbb{E}\left[X_sX_t\right] &= \mathbb{E}\left[\left(A\cos(s) + B\sin(s)\right)\left(A\cos(t) + B\sin(t)\right)\right]\\
&=\mathbb{E}\left[A^2\right]\cos(s)\cos(t) + \mathbb{E}\left[B^2\right]\sin(s)\sin(t) \\
&=\cos(s)\cos(t) + \sin(s)\sin(t)\\
&=\cos \left(s-t\right),
\end{align}
where in the last step we used the trigonometric addition formula. So the autocorrelation depends only on the time shift $s-t$.
2) I don't think that it's true in general, such as if the limit doesn't exist. For example, consider the process $X_t = \cos(t+\varphi)$, where $\varphi$ is a uniform random variable in the interval $(-\pi,\pi)$. The autocorrelation of this process is $\mathbb{E}\left[X_sX_t\right] = \frac{1}{2}\cos(s-t)$, or as a function of the time difference $\tau = s-t$, it is $\frac{1}{2}\cos(\tau)$. But the limit $\lim\limits_{\tau\rightarrow\infty}\left(\frac{1}{2}\cos(\tau)\right)$ doesn't exist.
