$f$ continuously differentiable, bounded derivative on $[a, \infty]$, $\int_0^{\infty}|f(x)|dx$ converges, prove $\lim_{x\rightarrow\infty}f(x)=0$ 
given $a \in \mathbb{R}$ and $f$, a continuously differentiable function with a bounded derivative on $[a, \infty]$, such that $\int_0^{\infty}|f(x)|dx$ converges, prove that $\lim_{x\rightarrow\infty}f(x)=0$
hint: observe $\int f(x)f'(x)$

I was thinking maybe $\int f(x)f'(x)dx = \frac{f^2(x)}{2}$ but I don't see how it helps. more hints are welcome
 A: We can in fact prove a stronger statement :

If $f$ is uniformly continuous and $\int_0^\infty |f(x)|dx < \infty$ then $\lim_{x \to \infty} f(x) = 0$.

To see that this is stronger, show that a continuously differentiable function with bounded derivative is uniformly continuous.
Proof : Suppose not. Then there is an $\epsilon > 0$, such that there exists a sequence $r_i \to \infty$ such that $|f(r_i)| > \epsilon$ for all $i$. By uniform continuity, there exists a $\delta > 0$ such that on each of the intervals $[r_i -\delta,r_i+\delta]$, the value of $|f|$ stays above $|\frac{\epsilon}{2}|$.
We note :
$$
\int_{0}^\infty |f| \geq \int_{\cup [r_i-\delta,r_i+\delta]}\frac{|\epsilon|}2 dx
$$ 
and now conclude that $\int_0^\infty |f|dx = \infty$ because that union has infinite measure (why? Actually, we first choose the $\delta$ then the $r_i$ so that the intervals are disjoint, of course their total measure must be infinite). 
This contradiction shows that the limit must exist and be zero.
A: If you want to use the hint. Then here is my idea.
$f$ has bounded derivative then there exists $M >0$ such that $\mid f'(x) \mid <M$ for all $x \geq a$. Choose any $\epsilon >0$ we prove that there exists $N>0$ such that $f^2(x) <\epsilon$ for all $x >N$. 
Since $\int_0^{\infty}|f(x)|dx$ bounded, for $\frac{\epsilon}{4M}$ there exists $K>a$ such that $\int_K^{\infty}|f(x)|dx<\frac{\epsilon}{4M}$.
Using Mean Value Theorem, there exists $N \in (K,K+1)$ such that $\mid f(N)\mid = \int_K^{K+1}|f(x)|dx<\frac{\epsilon}{4M}$
We have for all $x > N$ $$\mid \int_N^{x} f(t)f'(t)dt\mid \leq  \int_N^{x} \mid f(t)\mid \mid f'(t)\mid dt \leq M\int_N^{x} \mid f(x)\mid dx \leq \frac{\epsilon}{4}$$
So
$$\mid f^2(x)-f^2(N) \mid \leq \frac{\epsilon}{2}$$
which implies $f^2(x) \leq \frac{\epsilon}{2}+\frac{\epsilon^2}{16M^2}<\epsilon$ for all $x>N$. 
We conclude $\lim_{x \to \infty}f(x)=0.$
