Does $\lim_{s\to \infty}F(s)=0$ for all Laplace transforms?

Question

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 $s\geq\delta$.

Answer 1

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:

$$\lim_{s\to \infty}F(s)=\lim_{s\to \infty} \int_0^\infty |f(t)|e^{-st} \, dt<\lim_{s\to \infty} \int_0^\infty M e^{-st+\alpha t} \, dt=0$$

Answer 2

Choose finite $M>0$ with $|f|<Me^{\alpha t}$ so $s>\alpha\implies |F|\le\frac{M}{s-\alpha}$. Choose $\delta>\alpha+\frac{M}{\epsilon}$.