Is there any function which grows 'slower' than its derivative? Does a function $f: \mathbb{R} \rightarrow \mathbb{R}$ such that $f'(x) > f(x) > 0$ exist?
Intuitively, I think it can't exist.
I've tried finding the answer using the definition of derivative:


*

*I know that if $\lim_{x \rightarrow k} f(x)$ exists and is finite, then $\lim_{x \rightarrow k} f(x) = \lim_{x \rightarrow k^+} f(x) = \lim_{x \rightarrow k^-} f(x)$

*Thanks to this property, I can write:
$$\begin{align}
& f'(x) > f(x) > 0 \\
& \lim_{h \rightarrow 0^+} \frac{f(x + h) - f(x)}h > f(x) > 0 \\
& \lim_{h \rightarrow 0^+} f(x + h) - f(x) > h f(x) > 0 \\
& \lim_{h \rightarrow 0^+} f(x + h) > (h + 1) f(x) > f(x) \\
& \lim_{h \rightarrow 0^+} \frac{f(x + h)}{f(x)} > h + 1 > 1
\end{align}$$


*This leads to the result $1 > 1 > 1$ (or $0 > 0 > 0$ if you stop earlier), which is false.


However I guess I made serious mistakes with my proof. I think I've used limits the wrong way. What do you think?
 A: Let's keep working with your reasoning.  If $f'(x) > f(x)$, then we have that for $h > 0$ sufficiently small,
$$\frac{f(x+h) - f(x)}{h} > f(x)$$
so
$$f(x+h) - f(x) > hf(x)$$
$$\frac{f(x+h)}{f(x)} > h+1$$
As you found.  The problem occurs when you take the limit.  The correct way to take the limit is
$$\lim_{h \to 0^+} \frac{f(x+h)}{f(x)} \ge \lim_{h \to 0^+} h+1$$
$$1 \ge 1$$
which is not a contradiction.  To see why we have to take the limit like this, let's look at the functions $f(x) = x$, $g(x) = 1$.  Now, we can certainly argue that for $x < 1$, $f(x) < g(x)$; however,
$$\lim_{x\to1^-}f(x) = 1 = \lim_{x\to1^-}g(x)$$

As others have already pointed out, what you are trying to prove is false: $f(x) = e^{2x}$ is a counterexample.
A: expanded from David's comment 
$f' > f$ means $f'/f > 1$ so $(\log f)' > 1$.  Why not take $\log f > x$, say $\log f = 2x$, or $f = e^{2x}$.
Thus $f' > f > 0$ since $2e^{2x} > e^{2x} > 0$.
added: Is there a sub-exponential solution?
From $(\log f)'>1$ we get
$$
\log(f(x))-\log(f(0))  > \int_0^x\;1\;dt = x
$$
so
$$
\frac{f(x)}{f(0)} > e^x
$$
and thus
$$
f(x) > C e^x
$$
for some constant $C$ ... it is not sub-exponential.
A: Let $f(x) =e^x + C $ where $C<0$ then $f'(x)-f(x)=-C>0$ or $f'(x)>f(x)$ for all x - a trivial example. In this example $f(x)$ can be uniformly >, or < $f(x)$ depending on thee value of $C$.
A: There are multiple mistakes in your proof (i.e. dividing by $f(x)$ is not necessarily okay, since it is not necessarily positive). The most major is that you treat the variable in the limit as if it were not "bound". That is, if you have something like
$$\lim_{h\rightarrow 0^+}1>0$$
which is true, you can't necessarily, say, multiply through by $h$ to get
$$\lim_{h\rightarrow 0^+}h>0\cdot h$$
which is false. This is essentially what you do, and why your conclusion is wrong. You have to regard the $h$ as belonging to the limit - that is $\lim_{h\rightarrow 0^{+}}f(h)$ is a constant - it does not depend on $h$, because there is no notion of "$h$" outside of the limit. 
For an example of a function that does not satisfy this, you can take $e^{\alpha x}$ for any $\alpha>1$. Another solution is $xe^x$. It's worth noting that since the solution to $f'=f$ is $x\mapsto e^x$ we can prove that any solution grows faster than exponentially.
A: My personal favorite counterexample is $x^x$. When you differentiate it, you get $x^x[ln(x)+1]$, which is greater than $x^x$ over $[0,\infty]$ (which is the largest interval over which the derivative is defined). And, while it's not possible to get a closed form integral, doing some numerical integration, maybe in Mathematica, shows that the integral grows more slowly than the function itself. Of course, we would have to do quite a bit more work to actually prove this.
A: Even more generally than what other answers have brought up (though this implicit in a step of GEdgar's proof), if you let $g$ be a continous function, and $G$ an arbitary primitive of $g$ (that is, G' = g), then if $f(x) = e^{G(x)}$, we have $f' = gf$. So the ratio of a function to it's derivative can grow at essentially arbitary rates. 
Edit: as Marc van Leeuwen points out, an example gotten from this is that $f(x) = e^{x^2}$ satisfies the property that $f'(x)/f(x) = 2x$ so the derivative grows asymptotically faster than the function.
