General Problem: Inspect function $L(a)$ for $a\in \mathbb{R}^{n}$, given that:
$$L(a)=\lim_{x\to 0_{+}}\frac{x^{a}}{F(x)}$$
Notation legends:
$x=(x_1,\ldots,x_n),\ a=(a_1,...,a_n),\ x^a=x_{1}^{a_1}\ldots x_{n}^{a_n}$
$F(x)=\sum_{k\in S}F_{k}x^{k},\ S\in 2^{\mathbb{R}^{n}},\ F_k>0, \ \lim_{x\to 0_{+}}$ means $\lim_{x\to (0^{+},\ldots,0^{+})}$
Background: This problem has been solved partially by Cactus on MSE (also crossposted by me). This is what has been proved:
$$P = \{u \in \mathbb{R}^n|\exists p \in \mathrm{Conv}(S), \forall i:u_{i}>p_{i}\}$$ $$ a \in P \Longrightarrow L(a) = 0 \Longrightarrow \frac{x^a}{F(x)} \textrm{ is bounded around } (0^+,\ldots,0^+) \Longrightarrow a \in \overline{P}. $$
Consequently, it is possible to determine how $L(a)$ behaves if $a\in \mathbb{R}^{n}\backslash \overline{P}$ (in fact $L(a)=+\infty$ or $L(a)$ DNE, I did this part myself).
True problem: Inspect $L(a)$ for $a\in\partial P$.
Comment: From my observation, there is a finite closed subset $H$ (polytope like) of $\partial P$ such that: $a\in H \implies L(a)$ DNE (or rarely a positive number) , and $a\in (\partial P)\backslash H \implies L(a)=0$. I attempted to prove this by standard methods (choosing different paths and squeeze theorem) but I couldn't even describe $H$ explicitly, generally, I feel like I'm lacking properties of $\partial P$ to produce such proof.
