Does $f, f' \in L^1([0, \infty))$ imply that $\lim_{x \to \infty} xf(x) = 0$? Does $\int_0^\infty |f(x)| \, dx$ and $\int_0^\infty |f'(x)| \, dx$ being finite imply that $\lim_{x \to \infty} xf(x) = 0$? 
(Context: I am working through an analytic number theory textbook. In a lemma, the book assumes the conditions on $f$ and $f'$. The term $xf(x)$ appears from an integration by parts, but is tossed out, which must mean that $\lim_{x\to\infty} xf(x) = 0$. However, I've tried to prove this and was not successful. I also cannot think of any counterexamples.)

Here's a construction of a counterexample. The idea is based on a simple construction for series that is given on p. 350 of the centenary edition of G. H. Hardy's A Course of Pure Mathematics. The idea in Hardy's example is to construct a sequence $\{a_n\}$ of positive numbers such that $\sum a_n < \infty$ but $na_n \not\to 0$ as $n\to\infty$; to do this he simply puts $a_n = 1/n^2$ unless $n$ is a perfect square, in which case he puts $a_n = 1/n$.
To adapt this idea, choose a smooth function $\rho$, with $0\leq \rho \leq 1$, that is supported in $(-2,2)$ and that satisfies $\rho(x)=1$ for all $x \in (-1,1)$. Let $M$ be a bound for $|\rho'|$. Put
$$
\begin{align*}
\eta(x) & = \sum_{n = 2}^\infty \rho(x - n^2).
\end{align*}
$$
Basically, $\eta$ is identically $1$ in a neighborhood of each squared integer, and vanishes at each $x$ for which $(x-2,x+2)$ contains no squared integer. In fact, since the supports of $\rho(x-n^2)$ are disjoint, we have $0\leq \eta\leq 1$ and $|\eta'|\leq M$. 
Finally, define
$$
f(x) = {\eta(x)\over 1 + x}.
$$
for $x \geq 0$. Obviously $f(x)\geq 0$ for all $x$. Also, $f$ is integrable, since
$$
\begin{align}
\int_0^\infty f(x)\,dx &\leq \sum_{n = 2}^\infty \int_{n^2 - 2}^{n^2 + 2}{dx\over 1 + x}  \leq \sum_{n = 2}^\infty{4\over n^2 - 1}.
\end{align}
$$
Now
$$
\begin{align}
f'(x) & = {\eta'(x)\over 1+x} + O\left({1\over (1+x)^2}\right)
\end{align}
$$
as $x \to \infty$. Since $|\eta'|\leq M$, we get
$$
\begin{align*}
\int_0^\infty{|\eta'(x)|\over 1+x}\,dx & = \sum_{n = 2}^\infty \int_{n^2-2}^{n^2+2}{|\eta'(x)|\over 1+x}\,dx  \leq M\sum_{n = 2}^\infty {4\over n^2-1},
\end{align*}
$$
and it follows that $f'(x)$ is integrable over $(0,\infty)$ as well.
On the other hand, $xf(x) \not\to0$ as $x\to\infty$, since
$$
n^2f(n^2) = {n^2\over 1 + n^2} \to 1
$$
as $n\to\infty$. Thus $f$ provides a counterexample.
