       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:=
> \$Version
>
> Out=
> 4.1 for Microsoft Windows (November 2, 2000)
>
> In:=
> Limit[x - Sqrt[9 - 10 x + x^2],x -> Infinity]
>
> Out=
> 5
>
> In:=
> Limit[1 + x - Sqrt[9 - 10 x + x^2],x -> Infinity]
>
> Out=
> 1
>
>
>
> I also note that using Series to expand about infinity does give the
>
>
> Thanks
>
> Hugh Goyder
>

```

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