I have a function $f(x)$ and its first derivative, which is continuous, $f'(x)$. I know that $\lim_{x\to\infty}f'(x)=0$. Also $f'(x)>0$ for all $x$.
I also have another function $p(x)$ which is a continuous probability density function of a distribution with a well defined mean. So $p$ has also $\lim_{x\to\infty}p(x)=0$.
I also know that $\lim_{x\to\infty}\frac{p(x)}{f'(x)}=k>0$.
From the above I can conclude that in a neighborhood of $\infty$ $f'(x)\simeq A+\frac{1}{k}p(x)$. As $\lim_{x\to\infty}p(x)=\lim_{x\to\infty}f'(x)=0$, it has to be that $A=0$, so $f'(x)\simeq \frac{1}{k}p(x)$.
Can I integrate this ``equation'' and conclude that $\lim_{x\to\infty}f(x)<\infty$?
If you assume $p(x)$ is a nonnegative function such that $\int_{-\infty}^{\infty} p(x)dx = 1$, and $f(x)$ is a continuously differentiable function such that $$ \lim_{x\rightarrow\infty} \frac{p(x)}{f'(x)} = k \quad (Eq 1)$$ for some positive real number $k$, then you can conclude that $\lim_{x\rightarrow\infty} f(x)$ exists and is finite.
Proof: From (Eq 1) we know there exists a real number $y$ such that $f'(x)\neq 0$ for all $x \geq y$, and: $$ \frac{p(x)}{f'(x)} \geq k/2 \quad \forall x \geq y \quad (Eq 2)$$ Since $k/2>0$, it follows that $p(x)\neq 0$ for all $x \geq y$, and so $p(x)>0$ for all $x \geq y$, and hence $f'(x)>0$ for all $x \geq y$. Hence $f(x)$ is eventually nondecreasing and has a well defined (possibly infinite) limit. Since $f'(x)>0$ for all $x \geq y$, we rearrange (Eq 2) to get: $$ f'(x) \leq (2/k)p(x) \quad \forall x \geq y $$ Integrating this gives, for all $t>y$: $$ \int_y^{t} f'(x)dx\leq (2/k)\int_y^t p(x) \leq (2/k) $$ and so: $$ f(t) - f(y) \leq 2/k \quad \forall t > y $$ Thus: $$\lim_{t\rightarrow\infty} f(t) \leq f(y) + 2/k < \infty $$