Group Abstract Group Abstract

Message Boards Message Boards

1
|
13.1K Views
|
17 Replies
|
3 Total Likes
View groups...
Share
Share this post:

Calculate this limit with Wolfram|Alpha?

enter image description here
POSTED BY: Dvora Peretz
17 Replies
Sort By:

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
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

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

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
POSTED BY: Hans Dolhaine

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

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

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

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

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

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

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
Reply to this discussion
Community posts can be styled and formatted using the Markdown syntax.
Tag limit exceeded
Note: Only the first five people you tag will receive an email notification; the other tagged names will appear as links to their profiles.
Reply Preview
Attachments
Remove
or Discard