Proving that the derivative always diverges faster than the original function

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.)

• What do you mean by "smooth" here? Clearly this condition precludes even continuity at $x = a$. Unless $a$ can be $\pm \infty$, in which case I'd point to the example $f(x) = x$. Commented Jul 29, 2018 at 7:04
• If the function has a divergence in the form of a pole, then it could be expanded around $x=a$, writing $f(x) = (x - a)^{-\beta} + ...$. For such functions, at least, one can then differentiate and examine your ratio. For other functions, some more detailed analysis would be required
– lux
Commented Jul 29, 2018 at 7:06
• @TheoBendit By smooth I mean differentiable. And $a$ is supposed to be a finite number (added it to the post for clarity). I'm not following the comment about even continuity. Commented Jul 29, 2018 at 7:09
• $f$ is "a differentiable real function", but how can it be differentiable at $a$? Commented Jul 29, 2018 at 7:50
• The claim is not true in general; one counterexample is $f(x)=1/x^2+\sin(1/x^3)$, with $a=0$. You can weaken it to $\limsup_{x\to a}[f'(x)/f(x)]=+\infty$, or you can assume monotonicity of $f$. Commented Jul 29, 2018 at 8:49

3 Answers

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 $f(x) = \frac1x + sin\Big(\frac1x\Big)$

• Thanks! It was hard to choose between accepting this answer or RRL's. In the end, I went with RRL's, since his is a bit more constructive. Either way, thanks again, and +1 Commented Jul 29, 2018 at 18:10
• How did you search such function? Using intuition? Commented Apr 3 at 2:13

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.

• Thanks! I might use this fact as a small intermediate step in a physics paper which I am in the process of writing. I would like to thank you in the acknowledgements, would you be okay with that? If so, would you like me to acknowledge you by your real name, or as anonymous? (In case of the former feel free to send it to my email which you can find on google.) Commented Jul 29, 2018 at 18:16
• @RubenVerresen: You're welcome. No need for acknowledgement, otherwise anonymous is fine.
– RRL
Commented Jul 29, 2018 at 21:00

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 $$\lim_{x_2\to a} \frac{f'(x) }{f(x_2)} = \infty$$

• how did you remove the $f(x_1)$ from the denominator? Commented Jul 29, 2018 at 9:59
• how does $f'(x)$ tend to infinity? Commented Jul 29, 2018 at 10:01
• @CalvinKhor Ah, nice. This should be where it goes wrong. It's actually only unbounded; If it were bounded, take the sup of $f'(x)$ in $(a-\epsilon, a]$, and we'd have contradiction on the assumption that $\lim_{x\to a} f(x) = \infty$, as $f(a) < f(a-\epsilon) + sup \cdot \epsilon$ Commented Jul 29, 2018 at 10:06
• yeah. If you assume $f$ is monotone then it should work Commented Jul 29, 2018 at 10:09
• @RubenVerresen Thank you! I've reformulated my answer based on the given input, hopefully without also reinstating new mistakes Commented Jul 29, 2018 at 22:54