My Math Forum Function tends to

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 April 29th, 2017, 09:39 AM #1 Senior Member   Joined: Feb 2015 From: london Posts: 121 Thanks: 0 Function tends to If I have the function: $\displaystyle w = \frac{y}{u^2 (1-e^{-y/u})}$ If y is MUCH smaller than u, could you argue that w = 1/u. ***Disclaimer*** I may well have calculated w wrong which is why I cant seem to argue that w tends to 1/u
 April 29th, 2017, 10:57 AM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,638 Thanks: 959 Math Focus: Elementary mathematics and beyond $$\lim_{y\to0}\frac{y}{u^2(1-e^{-y/u})}=\frac1u$$ Thanks from calypso
 April 29th, 2017, 02:01 PM #3 Senior Member   Joined: Feb 2015 From: london Posts: 121 Thanks: 0 thanks for the quick reply. If its easy to explain, do you mind expanding a little on why that is the case. I can see that when u >> y, $\displaystyle w -> y/u^2$. But I dont understand how u^2 -> u
 April 29th, 2017, 02:55 PM #4 Math Team   Joined: Jul 2011 From: Texas Posts: 2,656 Thanks: 1327 $\displaystyle \dfrac{1}{u^2} \lim_{y \to 0} \dfrac{y}{1 - e^{-y/u}}$ L'Hopital ... $\displaystyle \dfrac{1}{u^2} \lim_{y \to 0} \dfrac{1}{\left(-\frac{1}{u}\right)\left(-e^{-y/u}\right)}$ $\displaystyle \dfrac{1}{u^2} \lim_{y \to 0} \dfrac{u}{e^{-y/u}} = \dfrac{1}{u^2} \cdot \dfrac{u}{1} = \dfrac{1}{u}$ Thanks from calypso
 April 29th, 2017, 03:07 PM #5 Senior Member   Joined: Feb 2015 From: london Posts: 121 Thanks: 0 thanks again, very clear explanation

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