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MathGroup Archive 2000

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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



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