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The equation is quite simple, however it is not very obvious to me to have the following relationship: $$\begin{equation} \frac{1-exp(-\kappa(T-t))}{\kappa}\rightarrow(T-t) \quad \rm{when\space} \kappa \rightarrow 0 \end{equation}$$

Thanks in advance!

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  • $\begingroup$ You should ask this question in other site $\endgroup$ – user16651 Jul 11 '16 at 12:04
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:D Is it a joke? $$\underset{\kappa \to 0 }{\mathop{\lim }}\,\frac{1-e^{-\kappa(T-t)}}{\kappa}=\underset{\kappa \to 0 }{\mathop{\lim }}\,\frac{\frac{d}{d\kappa}\left(1-e^{-\kappa(T-t)}\right)}{\frac{d}{d\kappa}\kappa}=\underset{\kappa \to 0 }{\mathop{\lim }}\,(T-t) e^{-\kappa(T-t)}=T-t$$

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  • $\begingroup$ That is a brilliant way to do it (ofc it has to be defined for both denominator and numerator for kappa)! $\endgroup$ – Donkey_JOHN Jul 11 '16 at 12:14
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    $\begingroup$ Ok it is brilliant way :D :D :D $\endgroup$ – user16651 Jul 11 '16 at 12:16
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    $\begingroup$ :) It's called l'Hospital rule @Donkey_JOHN, see: en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule $\endgroup$ – Quantuple Jul 11 '16 at 13:04

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