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When we check for the continuity of a function of one variable, we check the left hand side and right hand side limit of the function about the point in question, clearly this point lies inside the domain, what I mean is that if we are checking the continuity at $c$ then $a\leq c\leq b$, extending the same idea to functions with two variables, when we check the function's behavior/continuity at some point, we check it along two different paths and I have seen some examples where the author considers X-axis, Y-axis, $y=x$, $y=kx^2$ etc, my question is, these paths have to be included in the region under investigation, what I understand is that if we have a function of two variables, first we will have to draw the region defined by the function for various values of the independent variables $x$ and $y$ and then choose two different paths included in that region.enter image description here

As shown in the above diagram, if I get the region as shown by the Blue area enclosed by the line $y=x$ and $y=0$ from $x=2$ to $x \to \infty$, the curve $y=x^2$ is not included in the region so I can not use it to test the continuity, but in the example that I have seen author does not mention that the path he is choosing lies inside the region.

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It is extremely difficult to check continuity of a $2$-variable function by checking "along paths". In order to be able to deduce continuity at a point by checking continuity along paths, you must check the limit along every possible path that converges to the point and is contained in the domain. What is behind this is that you can check continuity by checking that $\lim f(x_n,y_n) = f(x,y)$ for any sequence $(x_n,y_n)$ that converges to $(x,y)$. In the one variable case, we can do it by just checking left and right, but in two variables there are way too many to check. – Arturo Magidin Apr 14 '12 at 4:29
    
To complement what @Arturo said: if you can find two paths along which the limits are different, then you can conclude that the function is not continuous at the point. That’s why you so often see examples with limits calculated along two paths. But agreement along two paths, or two hundred, doesn’t ensure agreement along all possible sequences converging to the point. – Brian M. Scott Apr 14 '12 at 4:32
    
Where are you trying to test continuity using $y=x^2$? No point of that curve lies inside your domain, so that curve/path does not approach any point in which you would be trying to test continuity... – Arturo Magidin Apr 14 '12 at 4:34
    
@ArturoMagidin, you said "you must check the limit along every possible path that converges to the point and is contained in the domain", Now, at $(0,0)$, many path converge, like $y=x, y=x^2$, so I have two path, but what is the check that both paths are included in the region under consideration? – Vikram Apr 14 '12 at 4:55
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@Vikram: No, that is not the reason your set-up is confusing. Your set-up is confusing because you explicitly state that you are considering a domain that does not include $(0,0)$, and yet you also claim to be trying to find the limit as $(x,y)$ approaches $(0,0)$. You are not just confused about what you are trying to do, you are confused about what it is you are confused about. You explicitly claim in your post that you are considering as domain only the blue region. The blue region does not contain $(0,0)$. So it's not a problem of 2d vs 3d. – Arturo Magidin Apr 14 '12 at 19:41

In 1-D the epsilon-delta definition of continuity at a point $x_0$ is $\forall \epsilon>0$, $\exists \delta>0$ such that $|f(x_0+\epsilon)-f(x_0)|<\delta$.

You said

extending the same idea to functions with two variables, when we check the function's behavior/continuity at some point, we check it along two different paths

but actually the way to extend the idea is to consider a ϵ-width ball around $\vec{x_0}$. Then the epsilon-delta definition will be $\forall \vec{\epsilon}$ with $|\vec{\epsilon}|>0$, $\exists \delta>0$ such that $\|f(\vec{x_0}+\vec{\epsilon})-f(\vec{x_0})\|<\delta$.

In English, the continuity holds in every direction around the point $\vec{x_0}$.

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