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November 29th, 2016, 11:00 AM  #1 
Senior Member Joined: Feb 2015 From: london Posts: 121 Thanks: 0 
The solution to a question states the following as s tends to 0. $\displaystyle h(s) = \frac{1e^{as}}{as 1+e^{as}}$ (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. 
November 29th, 2016, 11:24 AM  #2 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,603 Thanks: 115 
2) is from Maclaurin series. Keeping only terms in s, h(s)=h(0)+h'(0)s 
November 29th, 2016, 02:45 PM  #3 
Global Moderator Joined: May 2007 Posts: 6,626 Thanks: 622 
$\displaystyle 1e^{as}=11+as+o(as)$ $\displaystyle as1+e^{as}=as1+1as+\frac{(as)^2}{2}+o((as)^2)$ 
November 30th, 2016, 01:00 AM  #4 
Global Moderator Joined: Dec 2006 Posts: 19,865 Thanks: 1833 
WA gives $\displaystyle h(s) = \frac{2}{as}  \frac13 + \frac{as}{18}  \frac{a^2s^2}{270}  \,.\,.\,. $

December 1st, 2016, 12:37 PM  #5 
Senior Member Joined: Feb 2015 From: london Posts: 121 Thanks: 0 
Thanks all, that clears it up for me


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approximation, equation, variable 
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