Continuity of function with 2 variables

Question

Could someone please explain how to determine whether a two valued function is continuous at a certain point using straight paths $y=mx$? For example, say I had the function $$f(x,y)=\frac {x^2-y^2}{x+y+1}$$ and I had to find whether it was continuous at $(x_0,y_0)=(0,0)$

Answer 1

That cannot be done, since $f$ is continuous at $(0,0)$. If it wasn't, that one could perhaps prove that by choosing two distinct straight lines passing through $(0,0)$ and showing that when $(x,y)$ appraches $(0,0)$ along one of them one gets a cetain limit and that when $(x,y)$ approaches $(0,0)$ along the other line then one gets another (distinct) limit.

Answer 2

Substituting $y=mx$ in, you get $$f(x,y(x))=\frac{(1-m^2)x^2}{x(1+m)+1}$$Taking the limit as $x\to 0$, you get $0$. If instead you got something that depends on $m$ when you take this limit, you could conclude that $f$ is discontinuous. Since this does not happen, you cannot infer anything from this result. This isn't enough to show it is continuous there either, since there could be some other function $y(x)$ along which we get a limit different to $0$.

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To prove it is continuous, take $y(x)$ to be an arbitrary curve, with $y(0)=0$. Then $$f(x,y(x))=\frac{x^2-[y(x)]^2}{x+y(x)+1}$$ Taking the limit as $x\to0$ gives $\frac01=0$. So no matter what path is chosen, the limit is always $0$. Hence it is continuous.