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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.)

Ruben Verresen
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  • 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 $f(x) = x$. – Theo Bendit Jul 29 '18 at 07:04
  • 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 Jul 29 '18 at 07:06
  • @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. – Ruben Verresen Jul 29 '18 at 07:09
  • $f$ is "a differentiable real function", but how can it be differentiable at $a$? – Robert Israel Jul 29 '18 at 07:50
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    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 $f$. – user254433 Jul 29 '18 at 08:49

3 Answers3

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

Eulerrr
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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.

RRL
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  • 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.) – Ruben Verresen Jul 29 '18 at 18:16
  • @RubenVerresen: You're welcome. No need for acknowledgement, otherwise anonymous is fine. – RRL Jul 29 '18 at 21:00
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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 $$\lim_{x_2\to a} \frac{f'(x) }{f(x_2)} = \infty$$

Sudix
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  • Why is the first equation after the words: "Then by the above equation, we have:" correct? – Eulerrr Jul 29 '18 at 09:23
  • You can't prove in this way as the limit may not exist. – RRL Jul 29 '18 at 09:45
  • @RRL Well, since there are counter-examples, this is now a given, but I don't really find the point where it all goes wrong – Sudix Jul 29 '18 at 09:50
  • @Eulerrr Instead of $$\frac{f'(x) }{f(x_2) - f(x_1)} = \frac 1{x_2-x_1}$$ look at $$\frac{f'(x) }{f(x_2) - f(x_1)} - \frac 1{x_2-x_1} = 0$$ Now put in any sequence $a_n$ with limit $a$. Clearly, for all $n\in\mathbb{N}$ the equation holds. So, for all $\epsilon>0$ there's clearly an $n\in\mathbb{N}$ so that for all $m>n$, the equation $$\frac{f'(x) }{f(x_n) - f(x_1)} - \frac 1{x_n-x_1} <\epsilon$$ holds. This is the definition of the limit, so we have $$\lim_{x\to a} \frac{f'(x) }{f(x_n) - f(x_1)} - \frac 1{x_n-x_1} =0$$ . As $$\lim_{x\to a} \frac 1{x_n-x_1} $$ exists, we get the equation – Sudix Jul 29 '18 at 09:56
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    how did you remove the $f(x_1)$ from the denominator? – Calvin Khor Jul 29 '18 at 09:59
  • @CalvinKhor All other parts tend to infinity, so $f(x_1)$ becomes insignificant – Sudix Jul 29 '18 at 10:00
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    how does $f'(x)$ tend to infinity? – Calvin Khor Jul 29 '18 at 10:01
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    @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 $f(a) < f(a-\epsilon) + sup \cdot \epsilon$ – Sudix Jul 29 '18 at 10:06
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    yeah. If you assume $f$ is monotone then it should work – Calvin Khor Jul 29 '18 at 10:09
  • @Sudix it still doesnt work, you can think of a function that becomes flat then smoothly changes to have huge gradient, repeatedly as you approach $a$. What you end up proving is the limsup is infinite. You need $f'$ to be monotone – Calvin Khor Jul 29 '18 at 15:49
  • +1 since I don't think quality posts should be punished even if they contain a mistake (the score was negative at the time of writing), and the ensuing comments were very insightful. In fact, it might be instructive for future visitors for you to update your post with the correction? – Ruben Verresen Jul 29 '18 at 18:20
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    @RubenVerresen Thank you! I've reformulated my answer based on the given input, hopefully without also reinstating new mistakes – Sudix Jul 29 '18 at 22:54