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MathGroup Archive 2005

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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


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