Monotonicity of quotient function from the monotonicity of original functions Let,
$h,g:\mathbb{R}^+ \rightarrow \mathbb{R}^+$ be two continuous and increasing functions such that,
$\frac{h(t)}{g(t)} \rightarrow 0$ as $t\rightarrow 0$.
From this can we show that near 0,  $\frac{h(t)}{g(t)}$ is increasing?
I need this because while reading about Hausdorff dimension, there is a generalized definition where Dimension function is used to define Hausdorff dimension. Above mentioned $h,g$ are such functions. There it is written that if $h,g$ satisfies the above mentioned conditions and $\mathcal{H}^g(F)<\infty$ then $\mathcal{H}^h(F)=0$, where $F \subset \mathbb{R}^n$. Now if ${U_i}$ is a $\delta$ cover of $F$ then, if it can be shown that $h/g$ is increasing near zero then
$h(|U_i|)=\frac{h(|U_i|)}{g(|U_i|)}g(|U_i|)\leq \frac{h(\delta)}{g(\delta)}g(|U_i|)$
And now raising both sides to a power $s\geq0$ and summing and then taking limit as $\delta \rightarrow 0$, We can get the desired result. This was my plan. But without that monotonicity of $h/g$ the proof will be incomplete.
Please help.

As mentioned, you don't get that quotient to be increasing, however that's not an issue for what you want. In order to bound the quantities $h(|U_i|)/g(|U_i|)$, use the following fact:
If $f:(0,\infty)\to \mathbb{R}$ is such that $\lim_{t\to0} f(t)=0$, then the function $F(t):= \sup_{s\in (0,t)} |f(s)|$ satisfies $\lim_{t\to 0}F(t)=0$.
This sort of "self-improvement" of the limit condition is immediate once you unpack the $\varepsilon$, $\delta$ definition of limit.
Now just apply this with $f(t)=h(t)/g(t)$ to get the desired conclusion.
