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**chisigma** If the Laplace Tranform $\displaystyle \mathcal {L} \{\varphi(t)\}= f(s)$ exists , then the 'Final value theorem' extablishes that...

$\displaystyle \displaystyle \lim_{t \rightarrow \infty} \varphi(t) = \lim_{s \rightarrow 0} s\ f(s)$ (1)

Now is...

$\displaystyle \displaystyle \mathcal{L} \{\varphi^{'} (t)\} = s\ f(s) - \varphi(0)$ (2)

... and taking into account (1) and the fact that is $\displaystyle \displaystyle \lim_{t \rightarrow \infty} \varphi(t)=0$ is...

$\displaystyle \displaystyle \lim_{t \rightarrow \infty} \varphi^{'}(t) = \lim_{s \rightarrow 0} \{ s^{2}\ f(s) - s\ \varphi(0) \} =0$ (3)

Kind regards

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