Re:"Limit involving square root"
- To: mathgroup at smc.vnet.net
- Subject: [mg30235] Re:"Limit involving square root"
- From: seidovzf at yahoo.com (Zakir F. Seidov)
- Date: Fri, 3 Aug 2001 00:56:04 -0400 (EDT)
- Organization: The Math Forum
- References: <9k880u$4vh$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
In my case also Series is OK, but Limit is NOT: In[1]:=$Version Out[1]="4.0 for Microsoft Windows (December 5, 1999)" In[2]:=Limit[x - Sqrt[9 - 10 x + x^2], x -> Infinity] Out[2]=5 In[3]:=Limit[a + x - Sqrt[9 - 10 x + x^2], x -> Infinity] Out[3]=a In[4]:= Normal[Series[a + x - Sqrt[9 - 10 x + x^2], {x , Infinity, 4}] ] Out[4]= 5 + a + 232/x^3 + 40/x^2 + 8/x What a suprise for us users?! Zakir """"""""""""""""""""""""" """"""""""""""""""""""""" Subject: [mg30235] Limit involving square root Author: Hugh Goyder <goyder at rmcs.cranfield.ac.uk> Organization: Steven M. Christensen and Associates, Inc and MathTensor, Inc. 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