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Re: Re:"Limit involving square root"

  • To: mathgroup at smc.vnet.net
  • Subject: [mg30253] Re: [mg30235] Re:"Limit involving square root"
  • From: George Woodrow III <georgevw3 at mac.com>
  • Date: Sat, 4 Aug 2001 01:14:22 -0400 (EDT)
  • Sender: owner-wri-mathgroup at wolfram.com

I tried plotting the original function from 0 to 10^16, and got a lot of 
artifacts. There is a strong horizontal line at 5, and spikes. What you 
got were probably artifacts.

If you rewrite the function as x - Sqrt[(x - 9) (x - 1)], it should be 
clear that as x approaches infinity, this approximates x - Sqrt[x^2] = 0.

Don't know why Mathematica is confused. Nlimit[] reports 5 for as the limit and 
6 for the function + 1.

george

On Friday, August 3, 2001, at 12:56  am, Zakir F. Seidov wrote:

> In my case also
> Series is OK, but Limit is NOT:
>
> In[1]:=$Version
> Out[1]="4.0 for Microsoft Windows (December 5, 1999)"
>
> In[2]:=Limit[x - Sqrt[9 - 10 x + x^2], x -> Infinity]
> Out[2]=5
>
> In[3]:=Limit[a + x - Sqrt[9 - 10 x + x^2], x -> Infinity]
> Out[3]=a
>
> In[4]:=
> Normal[Series[a + x - Sqrt[9 - 10 x + x^2], {x , Infinity, 4}] ]
>
> Out[4]=
> 5 + a + 232/x^3 + 40/x^2 + 8/x
>
> What a suprise for us users?!
> Zakir
> """""""""""""""""""""""""
> """""""""""""""""""""""""
> Subject: [mg30253] [mg30235]      Limit involving square root
> Author:       Hugh Goyder <goyder at rmcs.cranfield.ac.uk>
> Organization: Steven M. Christensen and Associates, Inc and
> MathTensor, Inc.
>
> Dear Mathgroup,
>
> Below I take the limit of a function and then the limit of 1 plus the
> same
> function. A plot of the function shows that the first result, (5), is
> correct but the second, (1), is wrong. (Should be 6.) What's
> happening?
>
> In[1]:=$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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