I would like to expect that, if $h(t)=f(t)\ast g(t)$, ie,
$$h(t)=\int_0^t f(t-t')g(t')dt',$$
and $\lim_{t\to \infty} f(t)=\lim_{t\to \infty} g(t)=0,$ so $\lim_{t\to \infty} h(t)=0$ (or maybe I have at least some other information).
However, I have uncertanties for proceed.
Thank you so much.
As written in the comments it is not true. A counterexample would be with $f(s)=1/\sqrt{s}=g(s)$. Then we get $$ h(t)=\int_0^1 \frac{1}{\sqrt{1-x}\sqrt{x}} dx = \pi.$$ Thus, $\lim_{t\rightarrow\infty} h(t)=\pi \neq 0$. With a different power, $f(s)=s^{-1/4}=g(s)$ we get that $\lim_{t\rightarrow\infty} h(t)=\infty$. Indeed, we have in this case $$ h(t) = \int_0^t \frac{1}{(t-s)^{1/4}} \frac{1}{s^{1/4}} ds = \int_0^t \frac{1}{t^{1/4}}\frac{1}{(1-s/t)^{1/4}} \frac{1}{s^{1/4}} ds \stackrel{x=s/t, ds=tdx}{=} t^{3/4} \int_0^1 \frac{1}{(1-x)^{1/4} x^{1/4}} dx. $$