TITLE: a limit involving derivative
QUESTION [10 upvotes]: Find the following limit for a smooth convex function $f:[0,\infty)\to\mathbb{R}^{\geq0}$ such that $f(1)=1$ and
$f'(0)=f(0)=0$
$$\lim_{x\to0^{+}}\frac{f'(f^{-1}(x)a)}{f'(f^{-1}(x))}=L.$$
Is $L$ equal to $\frac{f(a)}{a}?$

REPLY [2 votes]: Counterexample. For $p\in (0,1)$ the function 
$$
f_p:[0,\infty)\to [0,\infty),\ f_p(x)=(1-p)x^2+px^3
$$ 
satisfies all the conditions, however for every $a>0$ we have
$$
\lim_{x \to0}\frac{f_p'(ax)}{f_p'(x)}=\lim_{x \to 0}\frac{2a(1-p)+3a^2px^2}{2(1-p)x+3px^2}=a\ne (1-p)a+pa^2=\frac{f_p(a)}{a}.
$$
Claim: Let $f:[0,\infty) \to [0,\infty)$ be a non-zero smooth function such that 
$$
f(0)=f'(0)=\ldots=f^{(n-1)}(0)=0\ne f^{(n)}(0).
$$
Then
$$
\lim_{x \to 0+}\frac{f'(ax)}{f'(x)}=a^{n-1},
$$
Proof. For $\varepsilon>0$ small enough, we have
$$
f(x)=\frac{f^{(n)}(0)}{n!}x^n+o(x^n)=:c_nx^n+o(x^n) \quad \forall x \in (0,\varepsilon].
$$
Therefore, for every $x \in (0,\varepsilon]$ we have
$$
\frac{f'(ax)}{f'(x)}=\frac{nc_na^{n-1}x^{n-1}+o(x^{n-1})}{nc_nx^{n-1}+o(x^{n-1})}=\frac{a^{n-1}+o(1)}{1+o(1)}.
$$
Hence
$$
\lim_{x \to 0+}\frac{f'(ax)}{f'(x)}=a^{n-1}.
$$
Remark: If $f(x)=x^n$, with $n\ge 2$ an integer, then in fact
$$
\frac{f(a)}{a}=a^{n-1} \quad \forall a>0.
$$