Proving that the derivative always diverges faster than the original function

Question

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.)

Answer 1

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.

Answer 2

Edit: As the comments and other answers indicated, the original assumption was incorrect.
It can however be salvaged by adding additional requirements.
By requiring $f'$ to be monotone in $(a-\epsilon,a)$ for some $\epsilon>0$ we can proof $\lim_{x\nearrow a} \frac{f'(x)}{f(x)}=\infty$ .
Let further $\to$ stand for $\nearrow$ (I feel it doesn't fit nicely into the equations), so that e.g. $\lim_{x\to a} f(x) $ means the limit of $x$ tending to $a$ from the left.

Using the mean value theorem:

Let $\lim_{x\to a} f(x) = \infty$. Define the points $x_1,x_2$ so that $x_1<x_2<a$.

Then, per mean value theorem there exists an $x\in (x_1,x_2)$ so that $$f'(x) = \frac{f(x_2) - f(x_1)}{x_2-x_1} \\\Leftrightarrow\\ \frac{f'(x) }{f(x_2) - f(x_1)} = \frac 1{x_2-x_1} $$ Now, let $x_2\to a$. Then by the above equation, we have (in doubt, check the comments):

$$\lim_{x_2\to a} \frac{f'(x) }{f(x_2) - f(x_1)} = \lim_{x_2\to a }\frac 1{x_2-x_1} \\\Leftrightarrow\\ \lim_{x_2\to a} \frac{f'(x) }{f(x_2) - f(x_1)} =\frac 1{a-x_1}$$ To further simplify the above equation, we'll now show $\lim_{x\to a}f'(x)=\infty$.

For this, let's assume $\lim_{x\to a}f'(x) = c$ for some $c\in\mathbb{R}$.
Then $f'(x)$ would have a supremum in the interval $(a-\epsilon,a]$:
$$\sup := \sup\{f'(x)\mid x\in (a-\epsilon,a]\}$$ However, this would imply $f(a) < f(a-\epsilon) + sup \cdot \epsilon <\infty$, resulting in a contradiction.

Therefore, $\lim_{x\to a}f'(x) $ has to be unbounded, and as $f'$ is monotone for this consideration, we have $\lim_{x\to a}f'(x) =\pm \infty$.
Obviously, $\lim_{x\to a}f'(x) =- \infty$ can't be the case, so we can conclude $$\lim_{x\to a}f'(x) = \infty$$

With this result, we can now simplify our equation: $$ \lim_{x_2\to a} \frac{f'(x) }{f(x_2) - f(x_1)} =\frac 1{a-x_1} \\\Leftrightarrow\\ \lim_{x_2\to a} \frac{f'(x) }{f(x_2)} = \frac 1{a-x_1} $$ (as both $f'(x)$ and $f(x_2)$ tend to $\infty$, while $f(x_1)$ is finite)

Therefore, if we let $x_1\to a$, we can conclude $$\lim_{x_2\to a} \frac{f'(x) }{f(x_2)} = \infty$$

Answer 3

This is actually wrong. For this function: there are values of x arbitrarily close to $0$ such that $\frac{f'(x)}{f(x)} = 0$. If you want an explicit expression, take $f(x) = \frac1x + sin\Big(\frac1x\Big)$