Existence of function satisfying a certain limit Does there exist a real function $f$ such that $\lim_{x\to+\infty}f(x)=+\infty$ and such that $$\lim_{x\to+\infty}\frac{f(x+1)f(x)}{x(f(x+1)-f(x))}=0?$$
I tried powers, logarithms, exponentials in all sort of forms, but it seems if $f$ grows 'faster' than $x$ the top of the limit makes it go to infinity, while if it grow 'slower' then the bottom part makes it go to infinity.
On the other hand I've not been able to prove that for every such $f$ the limit is not zero.
 A: There is no solution to your problem if $f$ is assumed to be 
continuous or monotonic.
I'll demonstrate that for the case that $f$ is continuous,
but the same proof works under the assumption that $f$ is monotonic.
Assume that $f : \Bbb R \to \Bbb R$ is continuous with
$$ \tag{1}
 \lim_{x \to \infty} |f(x)| =  \infty \text{ and }
  \lim_{x\to+\infty}\frac{f(x+1)f(x)}{x(f(x+1)-f(x))} = 0 \, .
$$
$f$ must have constant sign ultimately, so without loss of generality
we can assume that $f$ is strictly positive and
$  \lim_{x \to \infty} f(x) =  +\infty $.
Then $g := 1/f$ is continuous and positive, with 
$$ \tag{2}
 \lim_{x \to \infty} g(x) = 0 \text{ and } 
 \lim_{x \to \infty} \, x \cdot \bigl|g(x) - g(x+1) \bigr| = \infty \, .
$$
It follows that for $x \ge x_0$,
$$ 
 \bigl|g(x) - g(x+1) \bigr| > \frac 1x \, .
$$
In particular $g(x) - g(x+1)$ is not zero and therefore
(since $g$ is continuous) of constant sign for $x \ge x_0$.
Now $g(x) - g(x+1) < 0$ would imply that $g(x_0 + k) > g(x_0)$
all $k \in \Bbb N$, which is a contradiction to 
$ \lim_{x \to \infty} g(x) = 0$.
Therefore we have
$$
 g(x) - g(x+1)  > \frac 1x \text{ for } x \ge x_0 \, ,
$$
Repeated application of this inequality gives
$$
\begin{aligned}
 g(x_0) &> \frac {1}{x_0} + g(x_0+1) \\
 &> \frac {1}{x_0} + \frac {1}{x_0+1} + g(x_0+2) \\
  & ... \\
  &> \frac {1}{x_0} + \frac {1}{x_0+1} + ...  +\frac {1}{x_0+k} + g(x_0+k+1) \\
 &>  \frac {1}{x_0} + \frac {1}{x_0+1} + ... +\frac {1}{x_0+k}
\end{aligned}
$$
for any $k \in \Bbb N$.
This is a contradiction because for  $k \to \infty$, the right hand side diverges to infinity.

So a solution to your problem cannot be continuous or monotonic,
but here is an example: Define $g(x)$ for $x \ge 1$ by
$$
 g(x) = \begin{cases}
 \frac 1n & \text{if $n \le x < n+1$ and $n \in \Bbb N$ is even} \\
 \frac 1n - \frac{1}{\sqrt n}& \text{if $n \le x < n+1$ and $n \in \Bbb N$ is odd} \\
\end{cases}
$$
It can be verified that $g$ satisfies $(2)$, so that
$f := 1/g$ satisfies $(1)$.
A: I'm only making the proof for the case when $f$ is 'nicely' increasing, but I'm confident that it can be adapted for any $f$ such that $f\to\infty$ (it's just a bit more technical).
Let's suppose that $\lim_{x\to+\infty}\frac{f(x+1)f(x)}{x(f(x+1)-f(x))}=0$.


*

*If for big enough $x$, we have $a>0$ such that $a\le f(x+1)-f(x)$ : then $f(x+1)f(x)=o(xa)$ which is absurd since $f(x+1)-f(x)\ge a$ implies that $f$'s behavior when $x\to\infty$ is at least asymptotic to $ax$, if it's not growing even faster.

*Thus for big enough $x$, $f(x+1)-f(x)\to0$. (there could of course be cases where this isn't the case, but they're borderline cases, hence why I supposed that the function was nicely increasing).
In this case, $f(x+1)f(x)\sim f^2(x)$
Thus necessarily $\frac{f^2(x)}x\to0$. However, $f(x+1)-f(x)\sim\frac{f^2(x+1)-f^2(x)}{2f(x)}$, thus $\frac{f(x+1)f(x)}{x(f(x+1)-f(x))}\sim2\frac{f^2(x)}{x(f^2(x+1)-f^2(x))}f(x)$. Since $f^2(x)\to0$, $f^2(x)$ is at most asymptotically $x(f^2(x+1)-f^2(x))$.
Thus $\boxed{\lim_{x\to+\infty}\frac{f(x+1)f(x)}{x(f(x+1)-f(x))}\neq0}$
A: This quite similar to Hippalectryon's approach (and similarly incomplete); just writing to give another approach to think about solving.
Consider the Taylor expansion of such an $f$
$$f(x+1)\approx f(x) + f'(x)$$
Substituting, we get
$$\frac{f(x)}x \left(\frac{f(x)}{f'(x)}+1\right)$$
Just intuitively, if this limit is to go to 0, we need at least one (and hopefully both) multiplicand to do so as well.
Keeping in mind that $\lim_\limits{x\rightarrow\infty}f(x)=\infty$, we need $f$ to reach infinity "slower" than $x$, but also for its derivative to be keeping pace with itself.
I don't think these opposing forces can be reconciled. If the right factor is to be finite, we need $f$ to approach a line, but it's easy to see that such $f$ go to $\infty$. We reach the same conclusion by examining the left factor.
If we allow the left factor to be infinite, we must have $f'(x) < 1$ for $x$ sufficiently large, but this bounds the right factor above 0.
The remaining possibility is that the right factor is infinite but the left approaches 0 sufficiently quick as to annihilate the right, which suggests L'Hopital; indeed we can get a sort of L'Hopital form by rearranging the original expression (as hinted by Martin R):
$$\frac{\frac1x}{\frac1{f(x)}-\frac1{f(x+1)}}$$
I couldn't really get anything out of taking the numerator/denominator derivatives here, though.
