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

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.

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.

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

Yes, me too, especially as many students use Alpha and they know next to nothing about analytic functions

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?

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...

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.