# Does a strictly increasing, differentiable function that converges to 0 with an unbounded slope necessarily behave as a power function?

Suppose $$f:\mathbb{R}_+\rightarrow \mathbb{R}_+$$ satisfies:

• $$f$$ is continuous,
• $$f(0)=0$$,
• $$f$$ is differentiable on $$\mathbb{R}_{++}$$ with $$f'(x)>0$$, and
• $$f'(0)=\infty$$.

A canonical example of such a function would be $$ax^b$$ for $$a>0$$, $$b\in(0,1)$$.

My question is, do all functions satisfying the above conditions also satisfy $$f\sim ax^b$$ as $$x\rightarrow 0$$ for some $$a>0$$, $$b\in(0,1)$$? Here, I use $$\sim$$ in the sense of $$\lim_{x\rightarrow0}ax^b/f(x) = 1$$.

My gut says the answer should be no, but I've been unable to prove it/find any counter-examples.

Your intuition is correct, here is an example.

Let $$f(x) = -\frac 1 {\ln x}$$ over $$(0,0.5]$$ prolongated at $$x=0$$ by $$0$$ and for $$x>0.5$$ differentially with an increasing function (which is possible, I can give details if you need).

Consider $$h_b(x) = \frac{x^b} {f(x)} = -x^b \ln x$$. We know that, for all $$b>0$$ (standard result) : $$\lim_{x \to 0^+} h_b(x) =0.$$

Thus, $$f \not \sim ax^b$$ when $$x \to 0^+$$ for all choices $$a>0, b \in (0,1)$$.

Note : it is always a good reflex, when searching for something that will grow/decrease/vary faster than a range of power of $$x$$, to look at some function involving the exponential or logarithmic functions.

• Thank you! This is perfect. May 20 '20 at 12:46
• You are welcome. @CornerSolution May 20 '20 at 14:33

Let's consider the function $$f(t)$$ defined by

• $$f(t) = (-1) / \log(t)$$ for $$0 < t < 1/e$$,
• $$f(0) = 0$$ ,
• $$f(t) = e t$$ for $$t \ge 1/e$$

This function 'grows faster' than any power, which you can check by L'Hopital. This solution is inspired by the fact that the taylor series of the function $$e^{-1/t}$$ is zero, so analogously it 'grows slower' than any polynomial.

• Great, thank you so much! May 20 '20 at 12:47