The question looks maybe simple, but it is interesting.
Let us suppose that we have the following function with the stated features:
--$\,\,\,$ $F(x,a)$ was defined as below:
$$ \frac{{ } \int_0^\infty g(x,a) \, dx}{\int_0^\infty h(x,a) \, dx^*} =^* F(x,a) $$ .
--$\,\,\,$ $F(x,a)$ is a finite and continuous. Beside$\,\,\,$ $\int_0^\infty g(x,a_0)dx=\int_0^\infty h(x,a_0)dx=0$ $\,\,\,$ at$\,\,\,$ $a_0$ .
--$\,\,\,$ And we know for all $\,\,\,$ $u$ $\,\,\,$ by including of the neighborhood of the infinity that: $$\int_0^u h(x,a_0)dx≠0$$
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THE QUESTION.Could we obtain the following limit appearance (because we know that$\,\,\,$ "for all $\,\,\,$ $u$ $\,\,\,$ by including of the neighborhood of the infinity that:$\,\,\,$ $\int_0^u h(x,a_0)dx≠0$) $$.$$
$$ \frac{\lim_{u \to \infty} \int_0^u g(x,a) \, dx}{\lim_{u \to \infty}\int_0^u h(x,a) \, dx^*}=^* F(x,a) \,\,\ ⇒ \,\,\ \lim_{u \to \infty} \frac{\int_0^u g(x,a_0) \, dx}{\int_0^u h(x,a_0) \, dx}=F(x,a_0) $$