Show that the improper integral converges 
Let $f[0, \infty)\to\mathbb{R}$, differentiable, positive and $\lim_{x\to\infty} (\log f)'(x) = L < 0$. Prove that $\int_0^\infty f < \infty$.

So $\lim_{x\to\infty} (\log f)'(x) = L$ implies that there's an $M$ such that for every $x>M$:$$\frac{f'(x)}{f(x)}  < 0$$
Since $f$ is positive, this is implying that for $x>M$, $f$ is monotonically decreasing, strictly.
So I made some progress but got stuck here. We just covered Abel's/Dirichlet's tests but I can't see how to apply it here.
Help would be appreciated!
EDIT
Inspired by Claudeh5:
Let $\varepsilon>0$. Then, from continuity, there's an $N$ such that for every $x>N$:
$$L-\varepsilon \le (\ln f)'(x) \le L+\varepsilon$$ 
Now, we integrate from $0$ to $x$ and get:
$$(L-\varepsilon)x \le \ln(f(x)) - \ln(f(0)) \le (L+\varepsilon)x $$
Denote $C:=\ln(f(0))$.
$$(L-\varepsilon)x + C \le \ln(f(x)) \le (L+\varepsilon)x + C $$
Finally, we take exponent:
$$e^{(L-\varepsilon)x + C} \le f(x) \le e^{(L+\varepsilon)x + C}$$
So it seems that $\lim_{x\to\infty} f(x) = 0$. Isn't it?
 A: We can view that as $f'(x)\~Lf(x)$. Solving the differential equation obtained by replacing $\~$ with $=$ we get $ce^{Lx}$. Now this curve is indeed integrable at infinity, se if we show $f$ is asymptotic to it we are done. Certainly the ratios of the derivatives and functions are. Integrating, we get the logs will differ by $o(x)$: $\log f=Lx+o(x)$. So $\frac{f}{e^{Lx}}=e^{o(x)}$. But then $f$ is $e^{Lx}e^{o(x)}$, which is eventually at most $e^{Lx+x}$. So at least if $L>-1$, we are done. But surely $e^{o(x)}\leq e^{kx}$ for all positive $k$, eventually, so for all strictly negative $L$ we will find a $k:L+k\leq0$, which solves our problem.
A: We have $\lim_{x \to \infty}(\ln f)' = \lim_{x \to \infty}\frac{f'}{f} = L <0$ then  $\forall \epsilon>0 \exists x_0$,   $x>x_0$ then (integrating the inequality $|\frac{f'}{f} -L| \le \epsilon$ for x>x_0) $|\ln f(x)-Lx|\le \epsilon x$ and  so $f$ is integrable at $+\infty$ because $\exp((L-\epsilon)x) \le |f(x)| \le \exp((L+\epsilon)x)$. 
(evidently we suppose $\epsilon <|L|$ ) 
A: For $x>0,$ consider $ (\ln f(x))/(Lx).$ Because $Lx \to -\infty,$ L'Hopital is open to us. The quotient of derivatives is $(\ln f)'(x)/L \to 1.$ Therefore $ (\ln f(x))/(Lx) \to 1.$ So for large $x,$ $(\ln f(x))/(Lx) > 1/2.$ Because $L < 0,$ we get $\ln f(x) < Lx/2$ for large $x.$ This gives $f(x) < e^{Lx/2}$ for large $x$ and again using $L<0,$ we get $\int_0^\infty f < \infty$ as desired.
