Re: l'Hopital's Rule

*To*: mathgroup at smc.vnet.net*Subject*: [mg24418] Re: l'Hopital's Rule*From*: "John L. Jechura Jr." <JLJechura at MarathonOil.com>*Date*: Tue, 18 Jul 2000 00:58:15 -0400 (EDT)*Organization*: Marathon Oil Company*References*: <8k3oi3$42k@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

There are actually two answers depending upon the restrictions on a. If a>0 (as you have assumed), then the limit is zero. However, if a<0, then a-Sqrt[a^2] is not zero but rather 2a; in this case, the limit would be zero. John Jechura Sr. Engineer Marathon Oil Company <heathw at in-tch.com> wrote in message news:8k3oi3$42k at smc.vnet.net... > 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 > >