Re: Re: Strange result of Limit function in Mathematica3.0

*To*: mathgroup at smc.vnet.net*Subject*: [mg16040] Re: [mg16034] Re: [mg15987] Strange result of Limit function in Mathematica3.0*From*: "Kevin J. McCann" <kevinmccann at Home.com>*Date*: Sun, 21 Feb 1999 00:15:17 -0500*Sender*: owner-wri-mathgroup at wolfram.com

I guess I disagree with this philosophy. If Mathematica returned no answer then help it along, but your suggestion implies that one already knows the correct answer, in which case we don't need Mathematica at all. This is a bug. And a bad one at that. Kevin -----Original Message----- From: BobHanlon at aol.com <BobHanlon at aol.com> To: mathgroup at smc.vnet.net Subject: [mg16040] [mg16034] Re: [mg15987] Strange result of Limit function in Mathematica3.0 > >In a message dated 2/19/99 1:45:56 PM, krupa at alpha.sggw.waw.pl writes: > >>It is well known that: >>(2^n+8^n)^(1/n)->8 as n->Infinity >>and >>(2^n+3^n)^(1/n)->3 as n->Infinity >> >>but mathematica3.0 gives very strange results: >> >> >>In[1]:=<<Calculus`Limit` >> >>In[2]:= Limit[(2^n+8^n)^(1/n),n->Infinity] >> >>Out[2]:= 1/E >> >> >>In[3]:=Limit[(2^n+3^n)^(1/n),n->Infinity] >> >>Out[3]:= E >> >>Does Mathematica3.0 give such strange results only >>on my machine or may be I have made some mistake? >> > >Jan, > >Needs["Calculus`Limit`"] > >Limit sometimes needs help. If you simplify the form of the expression >prior to using Limit you will get the results which you expected. > >8*Limit[((1/4)^n + 1)^(1/n), n -> Infinity] > >8 > >3*Limit[((2/3)^n + 1)^(1/n), n -> Infinity] > >3 > >In any case, always do a sanity check of the results from Limit. > >Table[(2^n + 8^n)^(1/n), {n, 6, 10}]//N > >{8.00033,8.00007,8.00002,8.,8.} > >Table[(2^n + 3^n)^(1/n), {n, 22, 26}]//N > >{3.00002,3.00001,3.00001,3.,3.} > > >Bob Hanlon >