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MathGroup Archive 2001

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Re: Limit involving square root

  • To: mathgroup at smc.vnet.net
  • Subject: [mg30229] Re: Limit involving square root
  • From: "Allan Hayes" <hay at haystack.demon.co.uk>
  • Date: Fri, 3 Aug 2001 00:56:00 -0400 (EDT)
  • References: <9k880u$4vh$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Hugh,
You note that

expr=1+x-Sqrt[x^2-10x+9];

Limit[expr,x\[Rule]Infinity]

        1

is incorrect.

I don't know why, but  here are a couple of observations.

But using the transformation  y= x-5 we get

Limit[y+6 - Sqrt[y^2-16],y\[Rule]Infinity]

        6

Also, by hand, we get

6  +y- Sqrt[y^2-16] =

6 +( y^2-(y^2-16))/(y+ Sqrt[y^2-16] ) =

6 +16/(y+ Sqrt[y^2-16] )

Which evidently tends to 6 (this technique could, of course, be used without
the substitution).

--
Allan
---------------------
Allan Hayes
Mathematica Training and Consulting
Leicester UK
www.haystack.demon.co.uk
hay at haystack.demon.co.uk
Voice: +44 (0)116 271 4198
Fax: +44 (0)870 164 0565

"Hugh Goyder" <goyder at rmcs.cranfield.ac.uk> wrote in message
news:9k880u$4vh$1 at smc.vnet.net...
> 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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