Ratio of two positive functions

Let $g(x)$ and $h(x)$ be two nonnegative functions over $[0,a]$ with $g(0)=h(0)=0$, $$ g'(x)\leq h'(x), $$ and $g(x)>0$ and $h(x)>0$ for $x\in(0,a)$. Let $$ f(x) = \frac{g(x)}{h(x)} $$ over $(0,a)$ with $$ \lim_{x\to 0}f(x)=1. $$ (An example is $g(x)=\sin 2x$ and $h(x)=2\sin x$ for $a=\frac{\pi}{2}$.) Is it true that $f(x)$ is non-increasing over $(0,a)$ (since $g(x)$ does not "grow" faster than $h(x)$)?

Solution 1:

Take $g,h\colon(0,1]\to \mathbb{R}$ defined by $$ h(x)=x, \qquad g(x)=(1-\sin(x)\cos(x))) x $$ so that (1) $f,g\geq 0$, (2) $\lim_{x\to 0^+} h(x)=\lim_{x\to 0^+} g(x)=0$, (3) $\lim_{x\to 0}\frac{g(x)}{h(x)} = 1$, (4) $g'(x) \leq h'(x)=1$ for $x\in(0,1]$.

However, $$ \frac{g(x)}{h(x)} = \frac{1}{1-\sin(x)\cos(x)} $$ which is not monotone on $(0,1]$.

Solution 2:

Nope. Take $g(x)=x^2-\exp(-1/x)$, $h(x)=x^2$ and $a=1$. (Note that $g$ can be defined in $x=0$ by putting $g(0):=0$.)