  Let $\mathcal{L}\{f(t)\} = \int_{0}^{\infty}e^{-st}f(t)dt$ denote the Laplace
transform of $f$. It is well-known that if $f(t)$ is a piecewise continuous
function on the interval $t:[0,\infty)$ and of exponential order for $t > N$;
then $\lim_{s\to\infty}F(s) = 0$, where $F(s) = \mathcal{L}\{f(t)\}$. In this
paper we prove that the lesser known converse does not hold true; namely, if
$F(s)$ is a continuous function in terms of $s$ for which
$\lim_{s\to\infty}F(s) = 0$, then it does not follow that $F(s)$ is the Laplace
transform of a piecewise continuous function of exponential order.
