Does $f'(x)$ always remain close to $f(x)/x$ as $x \rightarrow 0$? Consider a function $f: (0,1): \to  (0,1)$ such that


*

*$\lim_{x \rightarrow 0+} f(x) = 0$, $\lim_{x \rightarrow 1-} f(x) = 1$;

*$f$ has a power series expansion around $x=1$ that converges in $(0,1)$:
$$
f(x) = 1 - \sum_{k=1}^\infty c_k(1-x)^k.
$$
with $c_k \geq 0$ for all $k$. Note that the condition $\lim_{x \rightarrow 0+} f(x) = 0$ implies $\sum_{k=1}^\infty c_k = 1$.


A typical example may be $f(x) = x^{2/3}$. Note that the derivative of $f$ may tend to $\infty$ as $x$ approaches $0$ from the right, as in this example.
I would like to prove (or find a counter-example) that the derivative $f'(x)$ remains asymptotically close to $f(x)/x$ as $x \rightarrow 0$. By that I mean any of the following:


*

*the limit
$$
\lim_{x \rightarrow 0} f'(x)x/f(x)
$$
exists and is finite and nonzero;

*or more generally that $f'(x) = \Theta(f(x)/x)$ as $p \rightarrow 0$, where $\Theta$ means "bounded below and above asymptically" (see definition here);

*or any similar result that asserts that $f'(x)$ does not "deviate too much" from $f(x)/x$ as $x \rightarrow 0$.


For the example function this is true. In fact the limit exists and equals $2/3$. This seems to be the case (with other values) for all functions I am testing, but how to prove it/disprove it?
 A: Here is a counterexample: Define
$$\tag 1 f(x) = \sum_{k=1}^{\infty}\frac{x^{1/k}}{2^k},$$
the series converging uniformly on $[0,1].$ Each $x^{1/k}$ can be written
$$x^{1/k} = 1-\sum_{n=1}^{\infty}c_k(n)(1-x)^n,$$
where $0\le c_k(n)\le 1$ for all $k,n.$ Thus
$$f(x) = 1 - \sum_{k=1}^{\infty}\frac{1}{2^k}\sum_{n=1}^{\infty}c_k(n)(1-x)^n =  1 - \sum_{n=1}^{\infty}\left (\sum_{k=1}^{\infty}\frac{c_k(n)}{2^k}\right )(1-x)^n.$$
So we see $f$ has the desired form.
Claim: $\lim_{x\to 0^+}\dfrac{xf'(x)}{f(x)} = 0.$
Proof: Note that differenitaing termwise in $(1)$ gives a  unifomly convergent series on $[a,1]$ for any $a\in (0,1].$ Thus
$$f'(x) = \sum_{k=1}^{\infty}\frac{x^{1/k-1}}{k2^k}$$
for $x\in (0,1].$ Therefore
$$\frac{xf'(x)}{f(x)} = \frac{ \sum_{k=1}^{\infty}x^{1/k}/(k2^k)}{ \sum_{k=1}^{\infty}x^{1/k}/2^k}.$$
Let $N\in \mathbb N.$ Define $S_N(x) = \sum_{k=1}^{N}\dfrac{x^{1/k}}{2^k},$ $T_N(x) = \sum_{k=N+1}^{\infty}\dfrac{x^{1/k}}{2^k}.$ Then
$$\tag 2\frac{xf'(x)}{f(x)} \le \frac{S_N(x) +T_N(x)/N}{S_N(x) + T_N(x)} \le \frac{S_N(x)}{x^{1/(N+1)}/2^{N+1}} + \frac{1}{N}.$$
As $x\to 0^+,$ the right side of $(2) \to 0+1/N = 1/N.$ Because $N$ was arbitrary, the $\limsup_{x\to 0^+}$ of the left side of $(2)$ is $0,$ proving the claim.
A: What follows does not answer the original question. However, it does show that even if $f'(x)$ can diverge to infinity for $x\to 0$, with the constraints above $xf'(x)$ does not; so to prove any counterexample, one must exploit $f(x)\to 0$.
$|xf'(x)|\leq 1$ for any $f(x)$ satisfying the constraints above and any $x\in(0,1)$. Note that $\forall x\in(0,1)$:
$|xf'(x)|=\left|x\sum_{k=1}^{\infty}\left(c_k k(1-x)^{k-1}\right)\right|\leq \sum_{k=1}^{\infty}\left| c_k k \max_{x\in(0,1)}\left(x(1-x)^{k-1}\right)\right|$.
Since $c_k\geq 0$ and $\sum_{k=1}^\infty c_k = 1$, observing that $x(1-x)^{k-1}$ is maximized by $x=1/k$ as its derivative equals $(1-x)^{k-1}+x(k-1)(1-x)^{k-2}(-1)=(1-x)^{k-2}(1-kx)$, one immediately obtains:
$|xf'(x)|
\leq c_1 + \sum_{k=2}^{\infty} c_k \left(1-\frac{1}{k}\right)^{k-1}
= c_1+\sum_{k=2}^{\infty} c_k \frac{k}{k-1} \left(1-\frac{1}{k}\right)^{k}
\leq c_1 + \sum_{k=2}^{\infty} (c_k 2e^{-1})
\leq 1$.
A: Not quite a counterexample (though I initially thought it was, and why it fails might be of interest):  Let 
$$
f(x) = \frac{x}{\sin 1} \sin \left( \frac{1}{x} \right).
$$
We obviously have $\lim_{x \to 1} f(x) = 1$;  and in the limit $x \to 0$, we note that 
$$-\frac{x}{\sin 1} \leq f(x) \leq \frac{x}{\sin 1},$$ so $\lim_{x \to 0} f(x) = 0$ by the squeeze theorem.
This is a holomorphic function in the complex plane, and so its radius of convergence in a series about $x = 1$ reaches to the pole nearest to $x = 1$.  But the nearest (and only) pole of this function is at $x = 0$.  So the power series converges on the ball $(0,2)$.
However, we also have
$$
f'(x) = \frac{1}{\sin 1} \left[\sin \left( \frac{1}{x} \right) - \frac{1}{x} \cos \left( \frac{1}{x} \right) \right],
$$
and 
$$
\frac{f'(x) x}{f(x)} = \frac{x \sin (1/x) - \cos (1/x)}{x \sin(1/x)} = 1 - \frac{1}{x \tan (1/x)}.
$$
This function does not have a well-behaved limit as $x \to 0$.
However, this is not quite a counterexample since the power series coefficients are not all positive.  In fact, the signs of the $c_k$ for this series seem to generally be alternating.
