Services & Resources / Wolfram Forums
-----
 /
MathGroup Archive
2000
*January
*February
*March
*April
*May
*June
*July
*August
*September
*October
*November
*December
*Archive Index
*Ask about this page
*Print this page
*Give us feedback
*Sign up for the Wolfram Insider

MathGroup Archive 2000

[Date Index] [Thread Index] [Author Index]

Search the Archive

Re: More about l`Hopital`s rule

  • To: mathgroup at smc.vnet.net
  • Subject: [mg24571] Re: [mg24560] More about l`Hopital`s rule
  • From: Andrzej Kozlowski <andrzej at tuins.ac.jp>
  • Date: Tue, 25 Jul 2000 00:56:04 -0400 (EDT)
  • Sender: owner-wri-mathgroup at wolfram.com

on 7/24/00 9:04 AM, Joaquín González de Echavarri at
jge at clientes.euskaltel.es wrote:

> 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
> 


This is basically the same situation as in the original problem in the
l'Hospital rule thread. You really should have read posted answers to that
thread before sending your own.

The problem is that you are quite wrong when you write:

> The right answer is b/a.

You could easily check yourself that this is not always so, e.g.

In[1]:=
f[x_] := (Sqrt[a^2 + x] - a)/(Sqrt[b^2 + x] - b)

In[2]:=
a = -6; b = -3;

In[3]:=
f[0.0001]
Out[3]=
2.

or

In[4]:=
Limit[f[x], x -> 0]
Out[4]=
2

This is a/b not b/a!

On the other hand:

In[5]:=
a = 6; b = 3;

In[6]:=
Limit[f[x], x -> 0]
Out[6]=
1
-
2


The point is that Mathematica simply chose a differnet branch branch of the
multivalued solution set and you are chosing a different one. If
Mathematica's answer is "wrong" then so is yours. In fact, Mathematica's
answer covers more cases than your won.


One can find all the answers using Mathematica if one approaches the problem
a little more carefuly:

In[1]:=
h[x_] := Sqrt[a^2 + x] - a + O[x]^2;
  g[x_] := Sqrt[b^2 + x] - b + O[x]^2;

In[2]:=
conds = Partition[
    Flatten[NestList[RotateLeft,
        Outer[Apply, {Less, Greater}, {{a, 0}, {b, 0}}, 1] // Flatten, 1]],
2]
Out[2]=
{{a < 0, b < 0}, {a > 0, b > 0}, {b < 0, a > 0}, {b > 0, a < 0}}


In[59]:=
Limit[Simplify[Normal[h[x]]/Normal[g[x]], #], x -> 0] & /@ conds
Out[59]=
 a  b
{-, -, 0, -(Infinity Sign[a b])}
 b  a

This last element in this list should be , of course, just Infinity.



-- 
Andrzej Kozlowski
Toyama International University, JAPAN

For Mathematica related links and resources try:
<http://www.sstreams.com/Mathematica/>





  • Prev by Date: Re: More about l`Hopital`s rule
  • Next by Date: Re: Malfunction of Position or where am I wrong?
  • Previous by thread: Re: More about l`Hopital`s rule
  • Next by thread: Re: Re: More about l`Hopital`s rule