Re: More about l`Hopital`s rule

*To*: mathgroup at smc.vnet.net*Subject*: [mg24588] Re: [mg24560] More about l`Hopital`s rule*From*: Rob Pratt <rpratt at email.unc.edu>*Date*: Tue, 25 Jul 2000 00:56:19 -0400 (EDT)*Sender*: owner-wri-mathgroup at wolfram.com

The "right" answer depends on the signs of a and b: If a and b are both positive, L'Hospital's Rule applies, and we get b/a. But if a and b are both negative, L'Hospital's Rule doesn't apply, and the limit is a/b. If a and b are both 0, the limit is 1. If a = 0 and b > 0, then the limit is +Infinity. etc etc (There are numerous special cases to consider.) 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 Mon, 24 Jul 2000, JoaquXn GonzXlez de Echavarri wrote: > Wrong answer: > > In= Limit[(Sqrt[x + a^2] - a)/(Sqrt[x + b^2] - b), x -> 0] > > 2 > a - Sqrt[a ] > Out= ------------ > 2 > b - Sqrt[b ] > > The right answer is b/a. > > Any suggestion? > > Joako