Limit of a Two-Variable Function

[ronthepon]

Okay, studies have started, and the first thing in my math 101 course if funtions in two variables.

 

Now, limits for single variable functions were pretty simple, all you had to do was to approach the point in question from two directions.

 

Obviously there are infinite 'directions' from where we can approach the a point defined by two variables. I mean if the variables are 'x' and 'y', there are infinite ways we can move towards, say (0,0) in the x-y plane.

 

I tried using a generalised approach to the point by relating x and y as [math]y = mx + c[/math], obviously we can simply substitute the point in question and get a representation of 'c' in 'm'. Like for (0,0), c becomes zero, and for (1,1), c becomes -m.

 

Hence, we have a general line serving as a direction of approach. We can easily use this general approach line to substitute one of the variables in the function in the terms of the other. Having done that, it shall be easy to find the limit of the function from the 'general direction'. (The limit will contain the 'm' term, and that will give every possible limit from every possible direction.)

 

In the end, all that remains for me to do is vary m from minus infinity to plus infinity, and every direction of approach has been covered.

 

Right?

 

But no, apparently, while this method suggests a limit of zero for the function [math]f(x,y) = \frac{xy^3}{x^2 + y^6}[/math] at (0,0) the function does not have a limit there. (Somehow a 1/2 is also there as a limit from some direction)

 

So who's helping me?

[ronthepon]

Duhhhh...huhhhh... It's apparently in the direction of approach of [math]x = y^3[/math] that the function remains 1/2. Obviously on this path, the limit to zero will be 1/2.

 

:confused:

 

What the heck? How am I supposed to figure such a thing out in a general case? And how the hell can [math]x = y^3[/math] have a constant valus of 1/2 over the function when I've determined that all around the origin, the function turns towards zero?

[Qfwfq]

In the end, all that remains for me to do is vary m from minus infinity to plus infinity, and every direction of approach has been covered.

 

Right?

Careful! The matter is tricky, it is possible for there to be a limit along curves through the point but not the limit at the point. Can't help you much now for lack of time and foggy memories but it is a tricky thing and full of pitfalls.

[ronthepon]

Now what interests me is how the plot of the function would look like. Would the lone [math]x = y^3[/math] just stand out from the x and y tending to zero reigion as a long, twisted peak?

 

If that were so, somehow the lines [math](y = mx)[/math] interesting [math]x = y^3[/math] as close to the latter function would tend to zero, simultaneously remaining = 0.5 right on the [math]x = y^3[/math] line.

 

Nuts really. Anyone know a computer handled way to plot this function in a 3-D plane?

[Erasmus00]

Maple or mathematica can plot this for you. They can be a bit pricey though.

 

As to the question of limit, the only way I know to avoid testing many curves through a given point to get a limit is to use the epsilon-delta type of limit.

-Will

 

edit:Also, consider this rule of thumb method. imagine taking y and x small by plugging in [math]\epsilon[/math] for x and [math]\epsilon '[/math] for y. Keeping leading order terms (dropping the y^6, as we assume both epsilon parameters are small)

 

[math]f=\frac{{\epsilon '}^3}{\epsilon}[/math]

 

We see that if we take epsilon to 0 independent of epsilon' our function blows up. Hence, no defined limit. This method isn't perfect, but often helps diagnose pathology in functions quickly.

[ronthepon]

As to the question of limit, the only way I know to avoid testing many curves through a given point to get a limit is to use the epsilon-delta type of limit.

Quite so. Just learn't about them today.