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MathGroup Archive 2001

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Re:"Limit involving square root"

  • To: mathgroup at smc.vnet.net
  • Subject: [mg30235] Re:"Limit involving square root"
  • From: seidovzf at yahoo.com (Zakir F. Seidov)
  • Date: Fri, 3 Aug 2001 00:56:04 -0400 (EDT)
  • Organization: The Math Forum
  • References: <9k880u$4vh$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

In my case also
Series is OK, but Limit is NOT:

In[1]:=$Version
Out[1]="4.0 for Microsoft Windows (December 5, 1999)"

In[2]:=Limit[x - Sqrt[9 - 10 x + x^2], x -> Infinity] 
Out[2]=5

In[3]:=Limit[a + x - Sqrt[9 - 10 x + x^2], x -> Infinity] 
Out[3]=a

In[4]:=
Normal[Series[a + x - Sqrt[9 - 10 x + x^2], {x , Infinity, 4}] ]

Out[4]=
5 + a + 232/x^3 + 40/x^2 + 8/x

What a suprise for us users?!
Zakir
"""""""""""""""""""""""""
"""""""""""""""""""""""""
Subject: [mg30235]      Limit involving square root
Author:       Hugh Goyder <goyder at rmcs.cranfield.ac.uk>
Organization: Steven M. Christensen and Associates, Inc and
MathTensor, Inc.

Dear Mathgroup,

Below I take the limit of a function and then the limit of 1 plus the
same
function. A plot of the function shows that the first result, (5), is
correct but the second, (1), is wrong. (Should be 6.) What's
happening?

In[1]:=$Version

Out[1]=
4.1 for Microsoft Windows (November 2, 2000)

In[2]:=
Limit[x - Sqrt[9 - 10 x + x^2],x -> Infinity]

Out[2]=5

In[3]:=
Limit[1 + x - Sqrt[9 - 10 x + x^2],x -> Infinity]

Out[3]=1

I also note that using Series to expand about infinity does give the
correct answers.

Thanks

Hugh Goyder




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