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Calculate this limit with Wolfram|Alpha?

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POSTED BY: Dvora Peretz
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The limit is 0 if you work on the real numbers, but what about the complex numbers? Try

Plot[Abs[(y^2 Sin[x])/(
   x^2 + y^2) /. {x -> t + t^3, y -> I t}], {t, -1, 1}]

I don't know if Wolfram|Alpha had this in mind, though.
POSTED BY: Gianluca Gorni

What about this?

f = (y^2 Sin[x])/(x^2 + y^2);

Limit[Limit[f, x -> 0, Assumptions -> x \[Element] Reals], y -> 0,  Assumptions -> y \[Element] Reals]

Limit[Limit[f, y -> 0, Assumptions -> y \[Element] Reals], x -> 0,  Assumptions -> x \[Element] Reals]

Plot3D[f, {x, -1, 1}, {y, -1, 1}, AxesLabel -> {x, y, "f"}]

The "approach" from "in between" is contained in the answer of Sander
POSTED BY: Hans Dolhaine

If you do this in Mathematica you DO get the right answer:

(y^2 Sin[x])/(x^2+y^2)
% /. x-> c y
Limit[%,y->0]
POSTED BY: Sander Huisman

This limit is very simple to calculate: (y^2 )/(x^2+y^2) is bounded by 1, and Sin x goes to 0. My students use Alpha to check their work and in this case Alpha got it wrong... ):
POSTED BY: Dvora Peretz

Tnks. u r probably right. It was nice if Alpha would understand it "as is" and give the option of complex numbers. Most engineering students don't even know of this option.
POSTED BY: Dvora Peretz

With complex numbers it is not bounded by 1... but I think by default Mathematica (Wolfram Language) does the limits with Real numbers... I could be wrong though...
POSTED BY: Sander Huisman

Limit[Limit[......]] also works indeed. The assumptions are not even necessary actually.

From the plot it is quite clear that the value is not direction-dependent. I would be surprised that WolframAlpha by default would include complex numbers in the limit while Mathematica doesn't...
POSTED BY: Sander Huisman

At present it seems that Mathematica only makes limits in one variable. Wolfram|Alpha apparently can handle more general limits, perhaps assuming complex variables. Iterated limits or directional limits are not the same as a full two-variable limit. Wolfram|Alpha's answer is so vague that it may simply mean that it can't do it.
POSTED BY: Gianluca Gorni

Yes, u r probably right about Alpha. tnks
POSTED BY: Dvora Peretz

When u r dealing with limits of multi variable function u need to simultaneously approach the point with all variables, which means Sqrt[(x-x0)^2+(y-y0)^2]->0, when u r using Limit[Limit[,x->0],y->0] u get a repeated limit which actually has nothing to do with the existance of the limit at the discussed point, for example check the function

f(x,y)=xSin[1/x] for (x,y) not equal to (0,0) and f(0,0)=0.
POSTED BY: Dvora Peretz

Yes, me too, especially as many students use Alpha and they know next to nothing about analytic functions
POSTED BY: Dvora Peretz

u need to simultaneously approach the point with all variables

hmmm, good point. Isn't it so, that the limit must be the same indepent of the arbritrary path on which you approach the point of interest? How can we be sure that there doesn't exist just one way giving another result, even when we have tested lots of other ways?
POSTED BY: Hans Dolhaine

U need to use different strategies, such as converting into polar coordinates, or the Sandwich theorem . For Example f(x,y)=y^2*Sin(y/x) : The Sin is bounded by 1, and y^2 goes to 0, etc. In polar form, if the limit is independent of the angle u can prove existence of limits.
POSTED BY: Dvora Peretz

It is not enough that the limit is independent of the angle. You also need that the limit is uniform in the angle. A classic couterexample was found by Peano:

(x^2 - y) (x^2 - y/3)

It has limit 0 along all radiuses, but not along parables.
POSTED BY: Gianluca Gorni

I'm not sure I follow u, this product goes to 0 when {x,y}->{0,0}, or did u mean it should be divided? In this case :

Sure, That what I meant when I said it should be independent - i.e. if the limit depend on the angle there is no limit. In yr example, after converting to polar u get: {rcos(a)-sin(a)}/{rcos(a)-sin(a)/3} -> 3sin(a)/sin(a)=3 for all a not equal to 0,Pi

and it does depend on the angle, if a=0,Pi u can't divide so it doesn't go to 0 on all radiuses.
POSTED BY: Dvora Peretz

Sorry, i got confused. That example was for a different purpose: it has a minimum at the origin along all straight lines, but not along parabolas. Try this:

Plot3D[(y/(x^2) )/(1 + (y/x^2)^2), {x, -1, 1}, {y, -1, 1}]

This has no limit at the origin, but

(y/(x^2) )/(1 + (y/x^2)^2) == (x^2 y)/(x^4 + y^2)

so that it has limit 0 along all radiuses. I am in a hurry, I hope I didn't make mistakes...
POSTED BY: Gianluca Gorni

Yes. I already figured it out while trying yr example. But u right about using polar coordinates to calculate a limit (usually at (0,0)). I usually use it to show there is no limit - if we r left with expressions that depends on the angle. Ofcourse, if u try to show that there is a limit we need to asume that the angle is not a constatnt which makes it a linear path (on the radiuses), and therefore not enough to prove the existence of the limit. Obviously converting a 2 variable problem in to only one variable problem (when we "ignore" the angle) does not make much sense. Tnks for yr example, Shabat Shalom
POSTED BY: Dvora Peretz
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