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MathGroup Archive 2000

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Re: More about l`Hopital`s rule

  • To: mathgroup at
  • Subject: [mg24586] Re: [mg24560] More about l`Hopital`s rule
  • From: Otto Linsuain <linsuain+ at>
  • Date: Tue, 25 Jul 2000 00:56:17 -0400 (EDT)
  • References: <>
  • Sender: owner-wri-mathgroup at

Excerpts from mail: 24-Jul-100 [mg24560] More about l`Hopi.. by Joaqu~n
Echavarri at client 
> 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?

 Joaquin, the right answer is what Mathematica is giving you ( actually
that could depend on what branch of Sqrt[z] one considers, but let's
stick to ( -Pi< Arg[z] <= +Pi ) => ( Arg[Sqrt[z]] = Arg[z]/2 ) which is
the branch we are all used to ).

 Anyway, As x->0, { Sqrt[x+a^2], Sqrt[x+b^2] } -> { Sqrt[a^2], Sqrt[b^2] }
this is just saying that Sqrt is continuous, which is true away from the
branch cut.

 So in principle, Mathematica's answer is correct, unless it happens
that this ratio is indeterminate.

 Let's then ask whether b - Sqrt[b^2] is zero.

 If -Pi/2 < Arg[b] <= +Pi/2 then the answer is yes, the denominator is zero.

 If, however, -Pi< Arg[b] <= -Pi/2 or +Pi/2 < Arg[b] <= Pi (i.e. Re[b]<0)
 then b-Sqrt[b^2] = 2 b.

 Mathematica's output can then be written as:

 0/0, 0/(2 b), (2 a)/0, (2 a)/(2 b), depending on where a and b are.

 You are right in that the 0/0 case yields b/a upon application of
l'Hopital's rule, but as we have seen that is not the whole story,
Mathematica does not know whether it should apply the rule.

 I have created a short notebook that illustrates this. Evaluate it and
see what happens. The [##]& just forces Plot to evaluate, otherwise you
get an error message:

f[a_,b_]:=Plot[##]&[{Re[#],Im[#}}& @
on -> $DisplayFunction];

 Although the imaginary part plays no role for {2,1}, I have included it
so you can play around with imaginary values of a0 and b0. Hope it
helps. Otto Linsuain. 

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