I suppose is well known to you the so called 'Final value theorem' , connected to the Laplace Trasform:

If...

$\displaystyle f(s) = \mathcal{L} \{\varphi (t)\}$

... then is...

$\displaystyle \lim _{t \rightarrow \infty} \varphi (t) = \lim_{s \rightarrow 0} s\cdot f(s) $

... provided that both limits exist.

Let's take an example and consider the function...

$\displaystyle \varphi (t) = \sum_{n=0}^{\infty} \mathcal {U} (t-n)$ (1)

... where $\displaystyle \mathcal {U}$ is the so called 'Haeviside step function'. This function is represented here...

In this case...

$\displaystyle f(s) = \frac{1}{s\cdot (1-e^{-s})}$ (2)

... so that is...

$\displaystyle \lim_{t \rightarrow \infty} \varphi(t) = \lim_{s \rightarrow 0} \frac{1}{1-e^{-s}} = + \infty$ (3)

... and it seems to be all right. Let' try now with another function...

$\displaystyle \varphi (t) = \sum_{n=0}^{\infty} \mathcal {U} (t-n) - t$ (4)

... that is represented here...

It is quite obvious that such type of function doesn't have limit for $\displaystyle t \rightarrow \infty$. In this case is...

$\displaystyle f(s)= \frac{s -1 + e^{-s}}{s^{2}\cdot (1-e^{-s})}$ (5)

... and if we try to apply the 'Final value theorem', using three times the L'Hopital rule, we obtain...

$\displaystyle \lim_{s \rightarrow 0} s\cdot f(s) = \frac{1}{2}$ (6)

... and that's is a little surprising! ...

At this point the questions for You are...

a) did I make some mistrake?...

b) if not how to interpret all that?...

Any help will be gratefully accepted!...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$