Let $f:[0,\infty)\to\mathbb{R}$ be a positive differentiable function, and suppose that $\lim\limits _{x\to\infty}\left(\log f\right)'\left(x\right)$ exists is negative. Prove that $\int_{0}^{\infty}f\left(x\right)dx $ converges.
By differentiating $$\left(\log f\right)'\left(x\right)=\left(\frac{1}{f}\cdot f'\right)\left(x\right)=\frac{f'\left(x\right)}{f\left(x\right)}$$ All I was able to show is that $f$ is bounded and goes to $0$, and that $f'(x)$ is always negative (after some point). I know I can find some bound on $f'(x)$ as well, but wasn't able to figure out any way of using it.
any hints?
First we will notice that by the chain rule we get:
$$ \left(\log f\right)'\left(x\right)=\left(\frac{1}{f}\cdot f'\right)\left(x\right)=\frac{f'\left(x\right)}{f\left(x\right)} $$
now suppose $\lim\limits _{x\to\infty}\left(\log f\right)'\left(x\right)=-L<0$, then we can find $x_{0}$ such that for $x>x_{0}$, $$\frac{f'\left(x\right)}{f\left(x\right)}<-\frac{L}{2}$$
or $$-f'\left(x\right)>\frac{L}{2}f\left(x\right)$$
This in turn shows that for any fixed $n>x_{0}$ $$\int_{0}^{n}f\left(x\right)dx=\int_{0}^{x_{0}}f\left(x\right)dx+\frac{2}{L}\int_{x_{0}}^{n}\frac{L}{2}f\left(x\right)dx $$ $$\leq\int_{0}^{x_{0}}f\left(x\right)dx+\frac{2}{L}\int_{x_{0}}^{n}-f'\left(x\right)dx$$
then
$$\int_{0}^{x_{0}}f\left(x\right)dx+\frac{2}{L}\int_{x_{0}}^{n}-f'\left(x\right)dx=\int_{0}^{x_{0}}f\left(x\right)dx-\frac{2}{L}f\left(x\right)\bigg|_{x_{0}}^{n}$$
and
$$\lim_{n\to\infty}\int_{0}^{x_{0}}f\left(x\right)dx-\frac{2}{L}f\left(x\right)\bigg|_{x_{0}}^{n}=\int_{0}^{x_{0}}f\left(x\right)dx-\frac{2}{L}\left(\lim_{n\to\infty}f\left(n\right)-f\left(x_{0}\right)\right)$$
which converges if and only if $\lim\limits _{n\to\infty}f\left(n\right)$ converges, but as for $x>x_{0}$, $\frac{f'\left(x\right)}{f\left(x\right)}<0$, and as $f\left(x\right)>0$, it must be that $f'\left(x\right)<0$. This shows both that $f\left(x\right)$ is decreasing and bounded by $0$. Then it must converge, completing the proof.