Re: Limit involving square root

*To*: mathgroup at smc.vnet.net*Subject*: [mg30229] Re: Limit involving square root*From*: "Allan Hayes" <hay at haystack.demon.co.uk>*Date*: Fri, 3 Aug 2001 00:56:00 -0400 (EDT)*References*: <9k880u$4vh$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

Hugh, You note that expr=1+x-Sqrt[x^2-10x+9]; Limit[expr,x\[Rule]Infinity] 1 is incorrect. I don't know why, but here are a couple of observations. But using the transformation y= x-5 we get Limit[y+6 - Sqrt[y^2-16],y\[Rule]Infinity] 6 Also, by hand, we get 6 +y- Sqrt[y^2-16] = 6 +( y^2-(y^2-16))/(y+ Sqrt[y^2-16] ) = 6 +16/(y+ Sqrt[y^2-16] ) Which evidently tends to 6 (this technique could, of course, be used without the substitution). -- Allan --------------------- Allan Hayes Mathematica Training and Consulting Leicester UK www.haystack.demon.co.uk hay at haystack.demon.co.uk Voice: +44 (0)116 271 4198 Fax: +44 (0)870 164 0565 "Hugh Goyder" <goyder at rmcs.cranfield.ac.uk> wrote in message news:9k880u$4vh$1 at smc.vnet.net... > 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 >