Prove that a differentiable function f with $f'(x) \geq f(x)^2$ is non positive and has limit $0$ Let $f:(0,\infty) \to \mathbb{R}$, with $f'(x) \geq f(x)^2$, be a differentiable function. I must prove that $f(x)\leq 0$ and $\lim_{x \to \infty}f(x) = 0$
As $f'(x) \geq f(x)^2 \geq 0$, $f$ is non decreasing. So, if I have $\lim_{x \to \infty} = 0$, it is easy to prove that $f(x) \leq 0$.
But I'm stuck on proving $\lim_{x \to \infty} = 0$
I'd appreciate some help.
 A: Since $f' \geq 0$, $f$ is an increasing function and hence $\lim_{x \to \infty} f(x) = L$ exists, but may possibly be infinite. If $L>0$, then there is some $y$ with $f(y)>0$. Then $f'(x)/f^2(x) \geq 1$ for all $x>y$. Then since $1/f$ is also differentiable on the interval $[y,\infty]$, for each $x>y$  the mean value theorem implies we have 
$$\frac{\frac{1}{f(y)}-\frac{1}{f(x)}}{x-y} = \frac{f'(c_x)}{f(c_x)^2} \geq 1$$
for some $c_x$ depending on $x$ between $x$ and $y$.
In particular we find
$$
\frac{1}{f(y)}-\frac{1}{f(x)} \geq x-y.
$$
The right side tends to $\infty$ as $x \to \infty$, but the left side is at most $\frac{1}{f(y)}$, a contradiction.
Thus it must be that $L \leq 0$. In particular, $L$ is finite (it clearly can't be $-\infty$). Now take the liminf as $x \to \infty$ on both sides of
$f'(x) \geq f(x)^2$ to find
$$
\liminf_{x \to \infty} f'(x) \geq L^2.
$$
If $L \neq 0$, then this would say that $f'(x) > L^2-\epsilon$ for all $x$ large enough, which would in particular show that $f$ eventually becomes strictly positive, again leading to a contradiction. Thus it must be that $L=0$.
