Final value theorem

Remarks

Since $f$ has a finite limit, it is bounded for large $t$, so $F(s)$ converges for $s > 0$. Since $s \int_0^\infty e^{-s t} \, dt = 1$, the difference $s F(s) - f(\infty)$ equals $s \int_0^\infty e^{-s t} (f(t) - f(\infty)) \, dt$. As $s \to 0^+$ the weight $s e^{-s t}$ spreads its mass over large $t$, where $f(t) - f(\infty)$ is small, and the part over a fixed interval $[0, T]$ tends to $0$. Example: for $f(t) = 1 - e^{-t}$, $s F(s) = 1 - s/(s + 1) \to 1 = f(\infty)$. The hypothesis that $f(\infty)$ exists is needed: for $f(t) = \sin t$, $s F(s) = s / (s^2 + 1) \to 0$, but $\sin t$ has no limit.

Going deeper

For a student

In control theory one uses it to find the steady-state error of a stable system. For a rational $F$, the limit $f(\infty)$ exists when all poles of $s F(s)$ lie in the open left half-plane; then the theorem applies. If $s F(s)$ has a pole on the imaginary axis or to its right, the formula can give a wrong answer, as the example $\sin t$ shows. The converse direction, from $F$ to $f$, needs extra hypotheses and belongs to the Tauberian theorems. Its companion is the initial value theorem.