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 November 29th, 2016, 10:00 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. $\displaystyle h(s) = \frac{1-e^{-as}}{as -1+e^{-as}}$ (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 10:29 PM.
 November 29th, 2016, 10:24 AM #2 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 125 2) is from Maclaurin series. Keeping only terms in s, h(s)=h(0)+h'(0)s Thanks from topsquark
 November 29th, 2016, 01:45 PM #3 Global Moderator   Joined: May 2007 Posts: 6,770 Thanks: 700 $\displaystyle 1-e^{-as}=1-1+as+o(as)$ $\displaystyle as-1+e^{-as}=as-1+1-as+\frac{(as)^2}{2}+o((as)^2)$ Thanks from topsquark
 November 30th, 2016, 12:00 AM #4 Global Moderator   Joined: Dec 2006 Posts: 20,757 Thanks: 2138 W|A gives $\displaystyle h(s) = \frac{2}{as} - \frac13 + \frac{as}{18} - \frac{a^2s^2}{270} - \,.\,.\,.$
 December 1st, 2016, 11:37 AM #5 Senior Member   Joined: Feb 2015 From: london Posts: 121 Thanks: 0 Thanks all, that clears it up for me

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