# 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}

1. 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?

• $f(x)=e^{2x}$.${}$ – David Mitra Feb 2 '15 at 22:20
• @DavidMitra why not make that an answer? – Steven Gubkin Feb 2 '15 at 22:21