The elements of the set $\;g+W^{1,2}(\mathbb R;\mathbb R^n)\;$ have limits at infinity Let $\;g:\mathbb R \to \mathbb R^n\;$ be a smooth function such that $\;\vert g(x)-l_{+} \vert \le e^{-kx}\;$ and $\;\vert g(x)-l_{-} \vert \le e^{kx}\;$ where $\;k \gt 0\;$ and $\;\vert \cdot \vert\;$ stands for the Euclidean norm. 
I wonder if the assumption $\;f \in g+W^{1,2}(\mathbb R;\mathbb R^n)\;$ is enough in order to deduce: $\;\lim_{x\to \pm \infty} f(x)=l_{\pm}\;$
If I could show that $\;f\;$ has the same asymptotic behavior as $\;g\;$ , then I would have the desirable limits. 
Thus I considered $\;h \in W^{1,2}(\mathbb R;\mathbb R^n)\;$ and wrote: 
$\;\vert f(x)-l_{+} \vert =\vert g(x)-l_{+} + h(x) \vert \le \vert g(x)-l_{+} \vert +\vert h(x) \vert \le e^{-kx} + \vert h(x) \vert \;$
Similarly, $\;\vert f(x)-l_{-} \vert \le e^{kx} + \vert h(x) \vert\;$
Now, I believe I should somehow take advantage of the fact that $\;h\in W^{1,2}(\mathbb R;\mathbb R^n)\;$ so as to find upper bound for $\;h\;$. However at this point, I 've been stuck.
 A: First, the usual caveat: Sobolev spaces consist of equivalence sets of functions. Changing function on a set of measure zero can certainly destroy its limit at infinity. However, a function $h\in W^{1, 2}(\mathbb{R}) $ has an absolutely continuous representative, and this is the object we normally think of when dealing with $h$. 
Then the answer is affirmative: $h$ tends to $0$ at infinity. This can be shown easily if you know that $\|h\|_\infty \le C\|h\|_{W^{1, 2}}$, and that smooth compactly supported functions are dense in $W^{1,2}(\mathbb{R})$. But here is a proof from scratch. 
Step 1: For any $\epsilon>0$  there exists $M$ such that $\int_{|x|>M}|h'(x)|^2 \,dx <\epsilon$ and $\int_{|x|>M}|h(x)|^2 \,dx <\epsilon$. This is by the integrability of $|h'|^2$ and $|h|^2$. 
Step 2: if $M < x < y $ and $y-x \le 1$, then $|h(x) - h(y)|<\sqrt{\epsilon}$. This is by the Fundamental Theorem of Calculus combined with Cauchy-Schwarz: 
$$
|h(x)-h(y)|\le \int_{x}^y |h'(t)|\,dt \le \sqrt{y-x}\sqrt{\int_{x}^y |h'(t)|^2\,dt} < \sqrt{\epsilon}
$$
Step 3: if $x > M$, then $|h(x)| \le 2\sqrt{\epsilon}$. Suppose not. WLOG $h(x) > 2\sqrt{\epsilon}$. By step 2, $h(y) > \sqrt{\epsilon}$ for all $y$ in $[x, x+1]$. Hence, $\int_{x}^{x+1} |h(t)|^2\,dt > \epsilon$, contradicting the choice of $M$ in step 1.
