Re: l'Hopital's Rule

*To*: mathgroup at smc.vnet.net*Subject*: [mg24314] Re: [mg24309] l'Hopital's Rule*From*: Andrzej Kozlowski <andrzej at tuins.ac.jp>*Date*: Sun, 9 Jul 2000 04:52:27 -0400 (EDT)*Sender*: owner-wri-mathgroup at wolfram.com

on 00.7.7 1:11 PM, heathw at in-tch.com at 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 > > Actually, you are at least as wrong as Mathematica here. Try, for example, a=-1. In[1]:= f[x_] = (a^2 - a*x)/(a - Sqrt[a*x]) /. {a -> -1} Out[1]= 1 + x ------------- -1 - Sqrt[-x] You can easily check (e.g. by using Table) that Mathematica's answer 0 is now correct, and your answer -2 is wrong. Of course if you put a=1 the correct answe will be indeed your answer as you can check will Mathematica: In[7]:= Limit[(1 - x)/(1 - Sqrt[x]), x -> 1] Out[7]= 2 In fact Mathematica can do the general case, provided you first tell it to simplify the input: Limit[FullSimplify[(a^2 - a*x)/(a - Sqrt[a*x])], x -> a] Out[39]= 2 a + Sqrt[a ] In[10]:= Simplify[%, a > 0] Out[10]= 2 a In[11]:= Simplify[%%, a < 0] Out[11]= 0 -- Andrzej Kozlowski Toyama International University, JAPAN For Mathematica related links and resources try: <http://www.sstreams.com/Mathematica/>