Re: How would you evaluate this limit in Mathmatica

*To*: mathgroup at smc.vnet.net*Subject*: [mg62870] Re: How would you evaluate this limit in Mathmatica*From*: "Chris Chiasson" <chris.chiasson at gmail.com>*Date*: Tue, 6 Dec 2005 23:11:31 -0500 (EST)*References*: <dmtc0b$foe$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

David Park was kind enough to send me his notebook on this problem and gant permission for me to publish this post. His notebook contains a lot of explanation and visualizations created using the (commercially available) Cardano 3 package. Based off of his work, both in the notebook and in the Cardano 3 package, I have created four plots that (somewhat) illustrate why one must take care in evaluating the limit posed in this thread. Here is the code to create the plots: f[a_][x_]=((1+4*x^2)^(1/4)-(1+5*x^2)^(1/5))/(a^(-(x^2/2))-Cos[x]) SetOptions[DensityPlot,Mesh\[Rule]False,PlotPoints\[Rule]400, ColorFunctionScaling\[Rule]False, ColorFunction\[Rule] Function[RGBColor[Min[Max[0,Re[#]]^(1/1.5),1],Min[Abs[Im[#]]^(1/30),1], Min[Abs[Min[0,Re[#]]]^(1/1.5),1]]],FrameLabel\[Rule]{"x","a"}]; SetOptions[Plot3D,PlotPoints\[Rule]400,ViewPoint\[Rule]{0,0,3}, Mesh\[Rule]False,AxesLabel\[Rule]{"x","a","f"}]; DensityPlot[f[a][x],{x,-10,10},{a,-10,10}]; Plot3D[f[a][x],{x,-10,10},{a,0,10}]; DensityPlot[f[a][x],{x,-.01 E,.01 E},{a,.99 E,1.01 E}]; Plot3D[f[a][x],{x,-.01 E,.01 E},{a,.99 E,1.01 E}]; For the density plots, I have spliced the positive real parts of the function into the red channel, the negative real parts of the function into the blue channel, and the absolute value of the imaginary part of the function into the green channel. Black just indicates the function is close to zero. You can clearly see the discontinuity and the zero David mentioned passing straight "through" {0,E}. Finding the limit at the confluence of these features looks to be a risky venture at best. So, one of the questions still on my mind is: Does Mathematica make an error in the following evaluation: In[74]:= Limit[f[a][x],x\[Rule]0] Out[74]= 0 Regards, P.S. The exponents in the color function allow one to amplify the color scaling, which I have specifically used to make sure the function isn't complex for positive a. -- http://chrischiasson.com/contact/chris_chiasson