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RE: Limit involving square root
*To*: mathgroup at smc.vnet.net
*Subject*: [mg30222] [mg30204] RE: [mg30167] Limit involving square root
*From*: Bradley Stoll <BradleyS at Harker.org>
*Date*: Fri, 3 Aug 2001 00:55:54 -0400 (EDT)
*Sender*: owner-wri-mathgroup at wolfram.com
Your eyes (and domain) are deceiving you. At first glance, I thought the
result to the first function should've been zero. Then after evaluating, at
what I thought were very large numbers, I thought I was mistaken and indeed
the answer was 5. But further investigations reveal that the first limit is
in fact 0. Try this to see something interesting. Evaluate at 10^15, 10^16
and 10^17. I think you will be surprised. Also, I tried using Mathematica
4.1 to evaluate the first limit directly and it just gave me back my input.
Bradley Stoll
-----Original Message-----
From: Hugh Goyder [mailto:goyder at rmcs.cranfield.ac.uk]
To: mathgroup at smc.vnet.net
Subject: [mg30222] [mg30204] [mg30167] Limit involving square root
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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