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.
 A: 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.
A: 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. 
