real analysis - 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:

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




2. Thanks to this property, I can write:

$$\begin{align} & f'(x) > f(x) > 0 \\ & \lim_{h \rightarrow 0^+} \frac{f(x + h) - f(x)}h > f(x) > 0 \\ & \lim_{h \rightarrow 0^+} f(x + h) - f(x) > h f(x) > 0 \\ & \lim_{h \rightarrow 0^+} f(x + h) > (h + 1) f(x) > f(x) \\ & \lim_{h \rightarrow 0^+} \frac{f(x + h)}{f(x)} > h + 1 > 1 \end{align}$$

3. 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?

Answer

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.