Suppose I have a function $f(x)$ defined for $x>0$, such that $f(x)\to\infty$ as $x\to\infty$. Let $F(s)$ be the Laplace transform of $f(x)$. Then can we say in general that $F(s)\to 0$ as $s\to\infty$?
P.S. By solving a few examples such as $f(x)=e^x$ or $f(x)=x^n, n>0$, I know that it tends to zero as $s\to\infty$ in these particular cases: $\mathbb{L}(e^x)=\frac{1}{1-s},~\mathbb{L}(x^n)=s^{-(n+1)}\Gamma(n+1)$.
According to the Final Value Theorem:
if $\displaystyle\lim_{x\to \infty}f(x)$ is finite, then $$\displaystyle\lim_{x\to \infty}f(x)=\displaystyle\lim_{s\to 0}sF(s)$$
But this cannot be applied when the final value does not exist. In your first example, for instance, the final value is infinity while $\displaystyle\lim_{s\to 0}sF(s)=0$.