Re: l'Hopital's Rule

*To*: mathgroup at smc.vnet.net*Subject*: [mg24336] Re: [mg24309] l'Hopital's Rule*From*: Rob Pratt <rpratt at email.unc.edu>*Date*: Sun, 9 Jul 2000 04:52:43 -0400 (EDT)*Sender*: owner-wri-mathgroup at wolfram.com

First, we must assume that a != 0 since otherwise the function of x is undefined. Also, we need not employ L'Hopital's Rule if instead we multiply top and bottom by the conjugate of the denominator, namely a + Sqrt[a*x]. Limit[(a^2 - a*x)/(a - Sqrt[a*x]), x -> a] Limit[(a^2 - a*x)/(a - Sqrt[a*x]) (a + Sqrt[a*x])/(a + Sqrt[a*x]), x -> a] Limit[(a^2 - a*x)(a + Sqrt[a*x])/(a^2 - a*x), x -> a] Limit[a + Sqrt[a*x], x -> a] (we used a != 0 here too) a + Sqrt[a^2] = a + Abs[a] If a > 0, we get 2a. If a < 0, we get 0. Rob Pratt Department of Operations Research The University of North Carolina at Chapel Hill rpratt at email.unc.edu http://www.unc.edu/~rpratt/ On Fri, 7 Jul 2000 heathw at in-tch.com wrote: > Hi, > When I input this: > Limit[(a^2 - a*x)/(a - Sqrt[a*x]), x -> a] > The output is: > 0 > The output should be 2*a. > Can't Mathematica 4 use l'Hopital's Rule? > Thanks, > Heath