Drawing conclusions from a differential inequality Let $f(x)$ be a smooth real function defined on $x>0$. It is given that:


*

*$f$ is an increasing function ($f'(x)>0$ for all $x>0$).

*$x \cdot f'(x)$ is a decreasing function.


I am trying to prove that:
$$ \lim_{x\to 0}f(x) = -\infty $$
EXAMPLE: $f(x) = -x^{-q}$, for some constant $q>0$. Then $f$ is increasing, $x\cdot f'(x) = q x^{-q}$ is decreasing, and indeed $ \lim_{x\to 0}f(x) = -\infty $. 
If this is not true, what other conditions are required to make it true?
 A: Your conjecture is correct. A proof follows below. 
By assumption
\begin{equation}
0 > (tf'(t))' = f'(t) + t f''(t), \quad t > 0.
\end{equation} 
By assumption $f'(t) > 0$, so
\begin{equation}
\frac{1}{t} < - \frac{f''(t)}{f'(t)} = -(\log(f'(t))', \quad t > 0.
\end{equation}
For $0 < x < 1$ we find by integration that
\begin{equation}
 - \log(x) = \int_x^1 \frac{1}{t} < - [\log(f'(t)]_x^1 = - \log\left( \frac{f'(1)}{f'(x)} \right) 
\end{equation}
from which it immediately follows that
\begin{equation}
x > \frac{f'(1)}{f'(x)}
\end{equation}
or equivalently
\begin{equation}
f'(x) > \frac{f'(1)}{x}.
\end{equation}
We can now show that $f(x) \rightarrow -\infty$ as $x \rightarrow 0_+$. Let $\epsilon > 0$. Then
\begin{equation}
f(1) - f(\epsilon) = \int_{\epsilon}^1 f'(x)dx > \int_{\epsilon}^1 \frac{f'(1)}{x} dx = - f'(1) \log(\epsilon) \rightarrow \infty.
\end{equation}
This completes the proof.
A: This is an alternate proof via contradiction, using limits and the greatest lower bound property. If $\lim_{x \to 0}f(x)$ is not $-\infty,$ then since $f$ is increasing for positive $x$ it would follow that $\lim_{x \to 0}f(x)=L$ where $L$ is the greatest lower bound of the set of values $f(x),\ x>0.$
From that it follows that $\lim x f(x)=0$ and so we may apply L'Hopital's rule (for $x \to 0$) to the fraction
$$\frac{x \ f(x)}{x}.$$
The denominator derivative being $1,$ the L'Hopital equivalent limit is that of
$$D[x \  f(x)]=f(x)+xf'(x)$$
as $x \to 0,$ i.e. the equivalent limit is $L+\lim_{x \to 0}[x\ f'(x)].$
Now using that the fraction we applied L'Hopital to was just $f(x)$ (whose limit is $L$), we may conclude that 
$$\lim_{x \to 0}x f'(x)=0.$$
However the other assumption of the problem is that $x\ f'(x)$ is decreasing, and combined with its limit at $0$ existing and being $0,$ we would get for positive $x$ that $x\ f'(x)<0,$ which implies $f'(x)<0$ against the other assumption that $f$ is increasing.
Note: The approaches of $x$ to zero here are all from the right, naturally; just didn't want to clutter the notation.
