Does $\lim_{s\to \infty}F(s)=0$ for all Laplace transforms? Let $f(t)$ be a piece-wise continous function of exponential order $\alpha$.
Then $F(s)$ exists.
I must prove then that $\lim_{s\to\infty} F(s)=0$ but i have no idea on how to do it.
I tried to prove it by the $\varepsilon,\delta$ definition of limits, using the piece-wise continous and exponential order properties of $f(t)$, but didn't reach the result. Perhaps I'm doing something wrong? 
My definition of  the limit would be that for every $\varepsilon>0\ \exists\ \delta>0$  such that $|F(s)|<\varepsilon$ for every $s\geq\delta$.

If 
$$|f(t)|\lt Me^{\alpha t}$$
then: 
$$F(s)=\int_{0}^{\infty} |f(t)|e^{-st} \, dt < \int_0^\infty Me^{-st+\alpha t} \, dt$$
thus:
$$\lim_{s\to \infty}F(s)=\lim_{s\to \infty} \int_0^\infty |f(t)|e^{-st} \, dt<\lim_{s\to \infty} \int_0^\infty M e^{-st+\alpha t} \, dt=0$$
