 My Math Forum Equation approximation as variable tends to 0

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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 Tags approximation, equation, variable Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post benmuskler Differential Equations 1 October 17th, 2012 08:06 AM dpwheelwright Algebra 5 August 19th, 2012 12:48 AM Phatossi Algebra 2 March 19th, 2012 08:44 PM superbigio Complex Analysis 1 July 17th, 2007 03:44 PM benmuskler Differential Equations 1 December 31st, 1969 04:00 PM

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