Re: How would you evaluate this limit in Mathmatica

*To*: mathgroup at smc.vnet.net*Subject*: [mg62780] Re: How would you evaluate this limit in Mathmatica*From*: "Dana" <Dana.OnTheBeach at comcastl.com>*Date*: Mon, 5 Dec 2005 03:37:42 -0500 (EST)*Sender*: owner-wri-mathgroup at wolfram.com

> ..I am interested in getting comments about doing this in > Mathematica. > I'm also interested in how one would teach solving limits using Mathematica. Hi. Just something different.. equ = ((1 + 4*x^2)^(1/4) - (1 + 5*x^2)^(1/5))/(a^(-x^2/2) - Cos[x]) ContourPlot[equ, {x, -1, 1}, {a, 2, 4}, Contours -> 20, PlotPoints -> 300]; This picture suggest a limit at x=0, with a special case of 'a' being near 2.7 (ie E) The White/Black band "suggests" a discontinuity jump from high to low. Interchanging the option to the following "suggests" that the limit is zero. (Adjust as desired..) Contours -> {0, .01} Contours -> {0, -.01} Contours -> {-.01,0, .01} Other options w/ Mathematica... a = E; Needs["NumericalMath`NLimit`"] Numerically, NLimit isn't so hot on the answer being 6 6 - NLimit[equ, x -> 0, Terms -> 13, WorkingPrecision -> 50] 0. * 10^-4 But the option "SequenceLimit" worked harder in convincing us the answer is 6. 6 - NLimit[equ, x -> 0, Terms -> 13, WorkingPrecision -> 50, Method -> SequenceLimit] 3. * 10^-23 Again, just some more options. (I Think David's comments are excellent !!) -- Dana DeLouis