
Elementary Math Fractions, Percentages, Word Problems, Equations, Inequations, Factorization, Expansion 
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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: $\displaystyle w = \frac{y}{u^2 (1e^{y/u})}$ 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 
$$\lim_{y\to0}\frac{y}{u^2(1e^{y/u})}=\frac1u$$

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 ... $\displaystyle \dfrac{1}{u^2} \lim_{y \to 0} \dfrac{1}{\left(\frac{1}{u}\right)\left(e^{y/u}\right)}$ $\displaystyle \dfrac{1}{u^2} \lim_{y \to 0} \dfrac{u}{e^{y/u}} = \dfrac{1}{u^2} \cdot \dfrac{u}{1} = \dfrac{1}{u}$ 
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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