I'm trying to find the limit of the Laplace transform of a function $f(t)$ of exponential order 2. The function is also piecewise continuous. I believe it is zero using the squeeze theorem from calculus but I'm stuck on one part of the proof.
Of exponential order 2: there exists numbers $K>0$, $M>0$ such that $-Ke^{2t}\leq f(t) \leq Ke^{2t}$ whenever $t \geq M$.
Then
$$-\int_M^{\infty} Ke^{-st}e^{2t} \, dt \leq \int_M^{\infty}e^{-st}f(t) \, dt \leq \int_M^{\infty} Ke^{-st}e^{2t} \, dt$$
The right-hand side evaluates to $\frac{Ke^{(2-s)M}}{s-2}$ and the left-hand side is just the negative of that.
Now as $s$ approaches infinity the right- and left-hand sides approach zero squeezing the middle integral to zero. But the middle integral is not quite the Laplace because it is missing the integral over the interval $[0, M]$.
The integral over $[0,M]$ is
$$\int_0^M e^{-st}f(t)\, dt$$
which is integrable since $f$ is piecewise continuous, but is this necessarily zero as I think it should be?
I thought of using integration by parts, but that assumes that $f$ is differentiable, and furthermore $f^{\prime}$ would need to be at least piecewise continuous I think so that the integrand containing it is integrable. But these assumptions are not in the problem statement.