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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