TITLE: Is there any function which grows 'slower' than its derivative?
QUESTION [27 upvotes]: 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?

REPLY [3 votes]: 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 $C$.