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November 29th, 2016, 11:00 AM   #1
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The solution to a question states the following as s tends to 0.

$\displaystyle h(s) = \frac{1-e^{-as}}{as -1+e^{-as}}$ (1)

$\displaystyle h(s) = \frac{as +o(s)}{0.5a^2s^2 + o(s^2)} $ (2)

$\displaystyle h(s) approx = \frac{2}{as}$

I don't really understand how they go from step 1 to step 2. To me it seems that as s tends to 0, the top part of the equation just tends to 0?

Last edited by skipjack; November 29th, 2016 at 11:29 PM.
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November 29th, 2016, 11:24 AM   #2
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2) is from Maclaurin series. Keeping only terms in s,

h(s)=h(0)+h'(0)s
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November 29th, 2016, 02:45 PM   #3
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$\displaystyle 1-e^{-as}=1-1+as+o(as)$
$\displaystyle as-1+e^{-as}=as-1+1-as+\frac{(as)^2}{2}+o((as)^2)$
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November 30th, 2016, 01:00 AM   #4
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W|A gives $\displaystyle h(s) = \frac{2}{as} - \frac13 + \frac{as}{18} - \frac{a^2s^2}{270} - \,.\,.\,. $
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December 1st, 2016, 12:37 PM   #5
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Thanks all, that clears it up for me
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