In whatever little I have learnt about calculus of two variable functions I have been having some serious problems in the way continuity of a function is defined.(adsbygoogle = window.adsbygoogle || []).push({});

We say that a function is not continuous if we can find two paths of approach along which the value of the independent variable is different.Now if I have to find such paths leading to a say a point,say the origin(0,0,0) the simplest way I can think of is to put the function equal to some constant c and obtain a contour curve for that value.Now in the neighborhood of the point under consideration(origin here),if I can limit the z-value of the contour lines crossing the neighborhood :

f(0,0)-ε< c_{ε }<f(0,0)+ε

where ε>0 is an arbitrary number,I can say that the function is continuous at that point.This follows from the formal definition of continuity and is easily understood.

But I will put forward two examples:

1.f(x,y)=x^{3}+y^{3}/x-y

This function is not continuous at (0,0).A few contour lines for z={1,5,3} are here:

http://rechneronline.de/function-graphs/" [Broken]

By this I can conclude that the function is non-continuous at (0,0).

2.f(x,y)=xy/√ (x^{2}+y^{2})

This function comes out to be continuous at (0,0) by the formal δ-ε method.However the corresponding contour lines look like this:

http://rechneronline.de/function-graphs/" [Broken]

Now I have paths to (0,0) that have different z-values even though the function is continuous at

(0,0).Can somebody help me?

Edit:The graphs linked are not working.Please use these functions:

Function 1:

1.x^3+y^3-x+y

2.x^3+y^3-3*x+3*y

3.x^3+y^3-5*+5*y

Function 2:

4.(x*y)-sqr(x^2+y^2)

5.(x*y)-3*sqr(x^2+y^2)

6.(x*y)-3*sqr(x^2+y^2)

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# Continuity for Two Variable Function

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