The initial value theorem states that if $$F(s)=\int_0^{\infty} f(t)\exp(-s t) dt$$ then $$f(0)=\lim_{s\rightarrow \infty} s F(s)$$ under some well-behavedness conditions on $f$ (including, for the purposes of this question, continuity).
I am interested in functions with a second parameter $f(t, a)$, where $f(0,a)=f(0)$ is independent of $a$. $$F(s,a)=\int_0^{\infty} f(t, a)\exp(-s t) dt$$ Is it the case that $$f(0)=\lim_{a\rightarrow \infty} a F(a,a) \tag{1}$$ Clearly not: $f(t,a)=\exp(-a t)$ is a counterexample.
My hypothesis is that if both $f(0,a)$ is independent of $a$, and the first derivative $\left. \frac{\partial}{\partial x}f(x,a)\right|_{x=0}$ is independent of $a$, then (1) holds. Question: Is this true?
Examples for which (1) turns out to be true are $f(x,a)=\exp(-a x)x$ and $f(x,a)=\exp(-a x^2)$