Prove the following: $f \in L^2(\mathbb{R})$ and $f'$ bounded $\Rightarrow$ $\lim_{|x| \to \infty}f(x)=0$.
In general, is it also true that $f \in H^1(\mathbb {R} ^n)$ $\Rightarrow \lim_{|x| \to \infty} f(x)=0$ (where $H^1$ denotes the Sobolev space $W^{1,2}$)?
You can prove the first bit by slightly tweaking David Mitra's answer to this question (note that $f'$ being bounded implies that $f$ is uniformly continuous).
Not sure about the second bit though.
The second part is not true when $n\ge 2$. Integrating in polar coordinates, you can check that the unbounded function $u(x)=\log\log(1/|x|)$ is in $W^{1,2}$ in small neighborhood of $0$. (If $n\ge 3$, a simpler example can be given: $u(x)=|x|^{-p}$ with $0<p<(n-2)/2$.)
Therefore, for every $n$ there is $r_n>0$ such that the norm on $u$ in the ball $|x|<r_n$ is less than $2^{-n}$. Let $u_n$ be such restriction, adjusted by adding a constant so that it vanishes when $|x|=r_n$. Extend $u_n$ by zero to $\mathbb R^n$; we still have $u_n\in W^{1,2}$, with the norm $<2^{-n}$.
Finally, let $v(x)=\sum_{n=1}^\infty u_n(x-ne_1)$ where $e_1$ is a basis vector (or any other nonzero vector, for that matter). Then $v\in W^{1,2}(\mathbb R^n)$ but $\limsup_{|x|\to \infty} v(x)=\infty$.
The same works for every Sobolev space that contains unbounded functions. (For spaces that don't contain unbounded functions, an embedding theorem gives uniform continuity, from where the conclusion $\lim_{|x|\to \infty} f(x)=0$ follows.)
