trying to prove the following convergence result So, this is propably some standard result from integral calculus:
Let $f:\mathbb{R} \mapsto \mathbb{R}$,  $f \geq 0$ such that
$\int^\infty_0 f < \infty$,
and
$|\frac{d}{dx} f| \leq C$ for all x some constant C.
Then $\lim\limits_{x \to \infty} f(x) = 0$. 
I got nowhere with mean value theorems, and i just cant seem to find it. can anyone help?
thanks.
 A: Let $\let\epsilon\varepsilon\epsilon>0$ and assume there is a sequence  $(x_n)_{n=0}^\infty$  with $f(x_n)\ge \epsilon$ and $x_n\to \infty$. Show that $\int_{x_n-\frac\epsilon C}^{x_n+\frac\epsilon C}f(t)\,\mathrm dt>\frac{\epsilon^2}C$.
We may assume wlog. that $x_{n+1}>x_n+\frac{2\epsilon}C$ and conlude $\int_0^{x_n} f(t)\,\mathrm dt\ge n\cdot\frac{\epsilon^3} C\to\infty$.
A: Hint: If $|f(x)| \geq \epsilon$ holds for some $\epsilon > 0$ for infinitely many arbitrarily large $x$, then choose an increasing subsequence $x_n$ of such $x$ that are at least some constant distance away from each other, e.g. $x_{n+1} - x_n > 1$. Then since the derivative of $f$ is bounded, you will be able to lower bound the absolute value of a contribution to the integral centered at $x_n$, over some fixed length interval centered there. As a further hint, take your intervals to be $[x_n - x_0, x_n + x_0]$ for some fixed positive $x_0$ where $x_0 < 1/2$ and $x_n < \epsilon / C$.
A: We have $\lim_{M \to \infty} \int_M^\infty f = 0$, and
$f(x) \ge f(x_0) - C |x-x_0|$
If $f(x_0) >0$, then the set $\{x | f(x)\ge 0 \}$ contains the set
$[x_0-{ f(x_0)\over C}, x_0+{ f(x_0)\over C}]$ and so
$\int_{x_0}^\infty f \ge { f(x_0)\over C} f(x_0)$. It is easy to see that this holds for all $x_0$, hence we see that $f(x_0) \to 0 $.
