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RE: Limit involving square root

• To: mathgroup at smc.vnet.net
• Subject: [mg30204] RE: [mg30167] Limit involving square root
• From: Bradley Stoll <BradleyS at Harker.org>
• Date: Thu, 2 Aug 2001 03:15:59 -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: [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