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