$f$ is differential in $(0,\infty)$, $\lim_{x\to \infty}f(x) = 0$.
I need to prove:
$f'$ is strictly growing in $(0,\infty) \implies f'(x)<0$, for every $x\in(0,\infty)$.
I tried using the Fundamental theorem of calculus on a closed interval in $(0,\infty)$ but could not find how to link it to the fact that $\lim_{x\to \infty}f(x) = 0$.
How can I prove that?
Edit: Thanks to Jhon if I assume by contradiction, I get that $f'>0$ starting in some $x_0 \in (0,\infty)$ and that f is strictly growing starting at some point $y \in (x_0, \infty)$. How can I get to a contradiction using this information?
Suppose by contradiction that there exists an $x_0\in (0,+\infty)$ such that $f'(x_0)\ge0$. Since $f'$ is monotone growing in $(0,+\infty)$, you must have $$ f'(y)> f'(x_0)\ge 0 $$ for all $y> x_0$. This means that $f$ is monotone growing in $[x_0,+\infty)$. On the other hand, $f'$ is also monotone growing by hypothesis. Hence $f'$ is monotone growing and strictly positive in $[y,+\infty]$. This implies that $f$ is not only strictly growing, but growing faster and faster, which contradicts $\lim_{x\rightarrow+\infty}f(x)=0.$
Note: the statement is true only if by "monotone growing" you mean "strictly growing". Otherwise the function $f=0$ is a counterexample as pointed out by martini