consider $f\in L^1(R_+)$ and define Laplace transform $\mathcal{L}f(z):=\int_0^{\infty} f(s)e^{-zs}\mathbb{d}s. $
How can I prove $\lim_{\mathbf{Re}z\rightarrow\infty}\mathcal{L}f(z) = 0?$
Intuitively it is obvious, but without exponential property I can't figure out how to tackle it.