Tauberian theorem when limit is zero Let $h \geq 0$ be a non-negative increasing function with Laplace transform $H$. Let $\rho \geq 0$ be a constant.
A simple Tauberian theorem says that the following two statements are equivalent:
I. $\quad$ $\beta^\rho H(\beta) \rightarrow \gamma \in (0,\infty) \quad \textrm{as} \quad \beta \rightarrow 0 $
II. $\quad t^{1-\rho} h(t) \rightarrow \frac{\gamma}{\Gamma(\rho)} \quad \textrm{as} \quad t \rightarrow \infty.$
(I have taken the slowly varying function in the normal statement simply to be constant). 
I am wondering what conclusions (if any) can be drawn about the asymptotic behaviour of $h$ at infinity in the cases when:
$$\beta^\rho H(\beta) \rightarrow 0 \quad \textrm{as} \quad \beta \rightarrow 0 $$
and/or
$$\beta^\rho H(\beta) \rightarrow \infty \quad \textrm{as} \quad \beta \rightarrow 0 .$$
I cannot find such a statement anywhere online, so I guess either the question is silly or the answer is that nothing can be said.
Many thanks for your help.
 A: I and II give precise asymptotics for $H$ and $h$. However, the statement $I^{\prime}$ below is imprecise. With some additional assumptions, you may be able to make some statement with inequalities, but I doubt it will be as nice as the original equivalence.
I$^{\prime}$: $\beta^{\rho}H(\beta)\rightarrow 0$ as $\beta\rightarrow 0$
Note that $H(\beta)\sim \beta^{-r}$ will satisfy I$^{\prime}$ for any $r<\rho$.  
Also, I$^{\prime}$ allows for things like $H(\beta)\sim \beta^{-1}\sin(|\beta|^{-1})$, which are not considered in I.  
A: Oops. Some of what's below is wrong. The part about $\lim=0$ is fine, and the part about $\lim=\infty$ in the case $0\le\rho\le 1$ is fine. But the counterexample in the case $\lim=\infty$, $\rho>1$ has big problems. The construction is "soft" enough that for all we know we got $H(s)=\infty$. A lawyer might argue that there's nothing about $H(s)$ being finite in the OP, but that's clearly not what was intended.
Right now it's not clear to me whether this can be fixed, or even whether the answer is yes or no in that case.
The part that's still ok:
Apologies for writing $H(s)$ instead of $H(\beta)$; it's $s$ in my notes, if I changed it to $\beta$ I'd miss one.
First Question Yes, $s^\rho H(s)\to0$ implies $t^{1-\rho}h(t)\to0$.
Suppose not; there exist $t_n\to\infty$ with wlog $$t_n^{1-\rho}h(t_n)\ge1.$$Let $s_n=1/t_n$. Now, $$h(t)\ge t_n^{\rho-1}
\quad(t>t_n),$$so $$H(s_n)\ge t_n^{\rho-1}\int_{t_n}^\infty e^{-ts_n}\,dt
=t_n^\rho e^{-1};$$hence $s_n^\rho H(s_n)\ge e^{-1}$.
Second Question First, if $0\le\rho\le 1$ then $s^\rho H(s)\to\infty$ implies $t^{1-\rho}h(t)\to\infty$.
If $0\le\rho<1$ then $t^{1-\rho}h(t)\to\infty$ regardless.
Suppose $\rho=1$. If $h(t)$ does not tend to infinity then $h$ is bounded; hence $H(s)\le c/s$, so $sH(s)$ does not tend to infinity.
The bogus part:
But if $\rho>1$ then $s^\rho H(s)\to\infty$ does not imply $t^{1-\rho}h(t)\to\infty$:
Let $s_n$ be any sequence decreasing to $0$. The example is going to be $$h=\sum_{n=1}^\infty c_n\chi_{[A_n,A_{n+1})},$$where $c_{n+1}>c_n$ and $A_{n+1}\ge A_n+1$. The $A_n$ and $c_n$ will be chosen in the following order: $A_1,c_1,A_2,c_2,\dots$.
Let $A_1=1$.
Suppose $A_n$ has been chosen. We will have $$H(s)\ge c_n\int_{A_n}^{A_n+1}e^{-st}\,dt.$$It's clear that if $c_n$ is large enough this will imply $$s^\rho H(s)\ge n\quad(s_{n+1}\le s\le s_n).$$ 
Having chosen $c_n$ large enough to make that happen, if we now choose $A_{n+1}$ large enough we will  have $$(A_{n+1}-1)^{1-\rho}c_n\le 1.$$
