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: <email@example.com>
- Sender: owner-wri-mathgroup at wolfram.com
You note that
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]
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
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"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?
> 4.1 for Microsoft Windows (November 2, 2000)
> Limit[x - Sqrt[9 - 10 x + x^2],x -> Infinity]
> Limit[1 + x - Sqrt[9 - 10 x + x^2],x -> Infinity]
> I also note that using Series to expand about infinity does give the
> correct answers.
> Hugh Goyder
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