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MathGroup Archive 2000

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More about l`Hopital`s rule

  • To: mathgroup at smc.vnet.net
  • Subject: [mg24560] More about l`Hopital`s rule
  • From: "Joaquín González de Echavarri" <jge at clientes.euskaltel.es>
  • Date: Mon, 24 Jul 2000 03:04:42 -0400 (EDT)
  • Organization: Euskaltel
  • Sender: owner-wri-mathgroup at wolfram.com

Different answers:

In= Limit[(a^2 - a*x)/(a - Sqrt[a*x]), x -> a]

Out= 0

In=Limit[Simplify[(a^2 - a*x)/(a - Sqrt[a*x])], x -> a]

Out= 0

In= Limit[FullSimplify[(a^2 - a*x)/(a - Sqrt[a*x])], x -> a]

               2
Out= a + Sqrt[a ]

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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