A problem on $(a, b)$-homogeneous or simply homogeneous function Let $f$ be a real $(a, b)$-homogeneous or simply homogeneous function that means $f(ax)=bf(x)$ for $a,b>0$ and $0 \leq x  \leq \frac{1}{a}$. I want find this limit: $$\lim_{x\to0^+}f(x)$$
 A: We'll need to impose some restriction on $f$ in order to have any chance of the limit existing. So in what follows we will assume $f$ is continuous. (Less restrictive assumptions might be made.)
First we rewrite the homogeneity relation in a more convenient form. Since the relation holds for $0 \le x \le 1/a$ we write $x=t/a$ and note that for $0 \le t \le 1$ we have the required inequality on $x$, and since $ax=t$ the relation now reads
$$[1]\ \ \ \ f(t)=bf(t/a),\ \ 0 \le t \le 1.$$
Now suppose $a<1$. Then the interval $[0,1]$ is partitioned by the powers of $a$ into subintervals $$...[a^3,a^2],\ [a^2,a],\ [a,1]$$ and the function $f$ is completely determined by its values on $[a,1]$, and may be arbitrary in that interval with the only restriction from $[1]$ being $f(a)=bf(1).$ Since $f$ is continuous we can let $m,M$ respectively be the minimum and maximum of $f(t)$ on the interval $[a,1].$ Now if $t \in [a^{k+1},a^k]$ the relation [1] gives $f(t)=b^kf(t')$ for some $t' \in [a,1]$. This gives 
$$b^km \le f(t) \le b^kM.$$ So provided $b<1$ we may conclude that $\lim_{t \to 0+}f(t)=0,$ which implies $\lim_{x \to 0+}f(x)=0$ since $x=t/a.$
On the other hand, if $b>1$ and $m>0$ we arrive at $\lim_{t \to 0+}f(t)=\infty.$ And if all we assume is that the maximum $M>0$ we see that as $t \to 0^+$ the values of $f$ are unbounded, so again the limit fails to exist.
The case of $a>1$ has a similar way to work things out, with again the limit being $0$, or nonexistent depending on $m,M$, according to on which side of $1$ the number $b$ lies. There remain the cases of one of $a,b$ being $1$; these are not difficult to analyze. Note that if $a=b=1$ the relation says nothing about $f$.
