       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.

-----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:=
\$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