I wanted to ask, if I have a function of two variables, $$f(x,y)$$ and I want to find its continuity in a certain point $$(0,0)$$ and the function is defined for R except for $$y=-x^2$$ and then I limit the function(approach) to the point $(a,b)$ with the line $$y=kx^2$$ and I find out that the limit does not exist when we approach the point with $y=kx^2$ because it equals $$k^2/(1+k^4)$$ (so something dependant on the way)

can I even use $y=kx^2$ since the function does no exist on $y=-x^2$? and if I can, does it mean that it does not continue at point (a,b) because the limit does not exist there? I feel like I can't use it but I don't know why I can't use it for k>0.

  • $\begingroup$ yes it is (0,0) $\endgroup$ – Zuzana Mitterová May 17 '18 at 16:40

Since the function is not defined for $y=-x^2$, to discuss about continuity, we firstly need to assign a value to the function at $(a,b)$ with $b=-a^2$ that is $f(a,b)=L$.

In this way, if the limit $L$ at the point $(a,b)$ exist we can make tha function continuos.

But in this case, since the limit depends upon $k$ for the trajectories $y=kx^2$, the function never can be continuos at that point.


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