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$\newcommand\R{\Bbb R}$For $a=(a_1,\dots,a_n)\in\R^n$, let $$L(a):=\lim_{x\downarrow0}R_a(x)$$ if the limit exists, where $$R_a(x):=\frac{x^a}{F(x)},$$ $x=(x_1,\dots,x_n)\in\R^n$, $x\downarrow0$ means that $x_i\downarrow0$ for each $i\in[n]:=\{1,\dots,n\}$, $x^a:=x_1^{a_1}\dotsm x_n^{a_n}$, $F(x):=\sum_{k\in S}x^k$, and $S$ is a finite nonempty subset of $\R^n$.

Is it then always true that $L(a)=0$ iff $$l_p(a):=\lim_{t\downarrow0}R_a((t^{p_1},\dots,t^{p_n}))=0\text{ for all $p=(p_1,\dots,p_n)\in(0,\infty)^n$?}$$

This question is motivated by this previous one.

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Yes. $L(a)=0$ is equivalent to the following condition: the polytope $P:={\rm conv} \, (S)$ contains a point $b=(b_1,\ldots,b_n)$ for which $a_i\geqslant b_i$ for all $i$ and $a_i>b_i$ for at least one index $i$.

If such a point exists, then $F(x)\geqslant x^b$ (indeed, if $b=\sum t_k k$ for $k\in S$ and $t_k\geqslant 0$ and $\sum t_k=1$, then $x^b=\prod_k (x^{k})^{t_k}\leqslant \prod (F(x))^{t_k}=F(x)$), and $x^a/x^b=\prod x_i^{a_i-b_i}$ goes to 0 when $x_i\downarrow 0$.

If such $b$ does not exist, then $P$ and the closed polyhedral cone $C:=a-\mathbb{R}_+^n$ are also disjoint or have only one common point $a$. Then the polyhedral cones $a+\mathbb{R}_+\cdot (P-a)$ and $C$ can be strictly separated by a hyperplane passing through $a$ (it is important here that they are polyhedral). Its equation is $\ell(v)=0$, where $\ell(v)=\sum_{i=1}^n y_i(a_i-v_i)$ for some coefficients $y_i>0$. Then we have $\ell(k)\leqslant 0$ for all $k\in S$, that yields $\prod t^{y_i k_i-y_ia_i}\leqslant 1$ for $0<t<1$, and $F(t^{y_1},\ldots,t^{y_n})\leqslant |S|t^a$.

So, either the limit is 0, or a limit is not 0 along some curve $(t^{y_1},\ldots,t^{y_n})$.

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