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2 vars limit with WolframAlpha

Posted 10 years ago

Hello everybody!

Does anybody know why WolframAlpha gives the answer "0" to

    limit[(x*y^2)/(x^2 + y^4),{x,y}->{0,0}] 

against the fact that, although on many paths the limit is 0, in each neighborhood of the origin there are:

  • an infinite number of points (of the kind x = y^2) on which the function simplifies to the constant 1/2

  • an infinite number of points (of the kind x = - y^2) on which the function simplifies to the constant - 1/2.

In my opinion the right answer should then be "limit does not exist, is path dependent or cannot be determined", (the same WolframAlpha gives, for example, to limit[(x*y)/(x^2 + y^2),{x,y}->{0,0}], which can't be "0" because the function simplifies to the constant 1/2 if you set y = x)

Infact both functions have a discontinuity in the origin, and this can also be shown by Plot3D (using more PlotPoints makes it clearer)

Plot3D[(x y^2)/(x^2 + y^4), {x, -.5, .5}, {y, -.5, .5}, 
 PlotRange -> {-.5, .5}, PlotPoints -> 150]

Plot3D[(x y)/(x^2 + y^2), {x, -.5, .5}, {y, -.5, .5}, 
 PlotRange -> {-.5, .5}]

Am I somehow wrong or is it a bug? Does Mathematica 8, 9 or 10 calculate this kind of limits? (I'm only up to version 7...)

To understand better, I also asked WolframAlpha for

    limit[(x^a*y^(a*b))/(x^(2*a) + y^(2*a*b)), {x, y} -> {0, 0}]

with a = 1,2,3,4,5 and b = 1,2,3,4 and still got the answer "0" (except for b = 1) against the fact that the function simplifies to 1/2 on the points where x = y^b (and to the constant -1/2 for x = -y^b if a is odd).

It looks as if the algorithm "checked" only the paths like x = y and not those of the kind x = y^b.

The same fact occurs if you change x and y and ask for

limit[(x^(a*b) y^a)/(x^(2*a*b) + y^(2*a)), {x, y} -> {0, 0}]

Thank you very much, bye!

Mario Gianini

POSTED BY: Mario Gianini

It's a weakness in that Alpha limit code.

Mathematica does not at this time have code dedicated to multivariate limits.

POSTED BY: Daniel Lichtblau
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