TITLE: Proving that the derivative always diverges faster than the original function
QUESTION [6 upvotes]: Let $f$ be a differentiable real function. What is the simplest/neatest way of proving that $\lim_{x \to a} f(x) = \infty$ implies that $ \lim_{x\to a} \frac{f'(x)}{f(x)} = \infty$? It seems like such a simple statement that perhaps there is even a proof that avoids equations altogether? (Note that $a$ is a finite number.)
Here is the one I came up with so far, which works by proving that integrating $\frac{f'(x)}{f(x)}$ in the vicinity of $x = a$ blows up. Indeed, $\int \frac{f'(x)}{f(x)} \mathrm d x = \int \frac{\mathrm d}{\mathrm d x} \left( \ln f(x) \right) \mathrm dx = \ln f(x) + c$, and then we can use the fact that $\lim_{x \to a} f(x) = \infty$ implies that $\lim_{x \to a} \ln f(x) = \infty$.
EDIT: the above statement is not true! The catch is that $f'/f$ might not have a well-defined limit. The best one can show is that $f'/f$ is unbounded in any neighbourhood of $a$. (See RRL's answer for the proof, and Euler's answer for a counter-example to the  original statement.)

REPLY [6 votes]: The most you can show is that $f'/f$ is unbounded in any neighborhood of $a$.
With $a < x < y$, it follows by the mean value theorem that there exists a point $\xi_{x,y}$ between $x$ and $y$ such that
$$\log f(x) - \log f(y) = \frac{f'(\xi_{x,y})}{f(\xi_{x,y})}(x - y),$$
and 
$$\lim_{x \to a+}\frac{f'(\xi_{x,y})}{f(\xi_{x,y})} = \lim_{x \to a+} \frac{\log f(x) - \log f(y)}{x-y} = \infty.$$
Hence, on any interval $(a,y]$ no matter how small we can find a sequence of points $(\xi_n)$ such that $f'(\xi_n)/f(\xi_n) \to  \infty.$ The mean value theorem is non-constructive with respect to the intermediate point, so we cannot determine that $\xi_n \to a$ or more generally that $\lim_{x \to a} \xi_{x,y}= a.$  This shows, at least, that $f'/f$ must be unbounded in any neighborhood of $x=a.$
The answer by @Eulerr is an example where $f'/f$ is unbounded but the limit does not exist.
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