Added: As Douglas Zare points out, there is no limiting distribution. There are two different limits as time runs over even or odd values. The distribution I describe below is the average of these two, and hence my $\pi_n$ means the long run average probability that the walk finds itself in state $n$.
More formally, I mean that $\pi_n=\lim_{t\to\infty} {1\over 2} \left(\mathbb{P}(X(t)=n)+\mathbb{P}(X(t+1)=n)\right)$ where $X(t)$ is the position of the walk at time $t$.
Suppose that the walk has been running for a long time and has reached its equilibrium. We need the probability of a negative step (which I'll call $q$) to be strictly greater than probability of a positive step (which I'll call $p$), otherwise the equilibrium doesn't exist.
The steady state probabilities are $\pi_n=\cases{{q-p\over 2q} & $n=0$\\[8pt] {q-p\over 2pq} \left({p\over q}\right)^n & $n\geq 1$}$
Therefore, the average position of the walk (in the long run) is $\sum_{n=0}^\infty n \pi_n ={1\over 2(q-p)}.$
When $p$ and $q$ are close together, this number is very large. But as $q$ gets close to 1, the walk spends most of its time jumping back and forth between positions 0 and 1, and its average position is close to $1/2$.