Consider $f(x,y) = x^2+y^2$ when $x-y \neq 0$ and $0$ if $(x,y) = (0,0)$.
This function is easily shown to be continuous along all paths, but along $x=y$ it is not defined!
So will it said to be continuous at $0,0$ ?
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How have you shown 'easily' that $f$ is continuous along the path, $t \to (t,t)$, say? – preferred_anon Sep 02 '18 at 09:23
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For any other path, we simply have to use first definition, which goes to 0 – jeea Sep 02 '18 at 09:35
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Yes of course, but you have said $f$ is continuous along all paths. The function $t \mapsto (t,t)$ is a path, but $f$ is not continuous along it. There are many other paths which intersect the ray $x=y$. Do you see the problem? – preferred_anon Sep 02 '18 at 09:53
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@jeea Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details HERE – user Oct 23 '18 at 20:49
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@gimusi sorry I totally forgot about that! – jeea Oct 24 '18 at 13:48
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For the continuity at $(0,0)$ it doesn't mind if $f(x,y)$ is not defined for $x=y\neq 0$, indeed we have that trivially
$$\lim_{\substack{(x,y)\to(0,0)}\\\quad \:x\neq y}x^2+y^2=0=f(0,0)$$
therefore by definition $f(x,y)$ is continuous at $(0,0)$.
user
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1How can you take the limit to $(0,0)$ if the function isn't defined on a neighbourhood of $0$? – preferred_anon Sep 02 '18 at 09:20
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I think you are using a non-standard definition of continuity in $\mathbb{R}^{2}$. c.f. (https://en.wikipedia.org/wiki/Limit_of_a_function#Functions_of_more_than_one_variable) – preferred_anon Sep 02 '18 at 09:24
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@DanielLittlewood Of course we are referring to different definition but I'm not sure that mine is the "non-standard" one. If f(x,y) is not defined for $x=y\neq 0$ it doesn't mind since we are removing those points from the domain. The case seems similar to this OP https://math.stackexchange.com/questions/2889055/what-is-lim-x-to-0-frac-sin-frac-1x-sin-frac-1-x-does-it-exist?noredirect=1&lq=1 – user Sep 02 '18 at 09:39
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The various definitions given in the answer you link are satisfactory, and establish that the limit equals $0$ (which is true), although I suspect that the OP may have been given a definition which assumes that the domain of $f$ should be a nbhd of $(0,0)$, which would explain the confusion. My serious objection is with your comment: continuity along a path does not imply continuity of the function! – preferred_anon Sep 02 '18 at 09:52
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@DanielLittlewood The limit is derived trivially since, under the condition $x\neq y$, as $(x,y)\to (0,0)$ we have that $x^2+y^2 \to 0$ for any path. – user Sep 02 '18 at 09:57
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I am still not clear about the part that along $x=y$ function is not defined, so can we even talk about continuity along that path? According to epsilon delta definition limit should exist and f should continuous – jeea Sep 02 '18 at 18:52
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@jeea I'm assuming that by your definition f(x,y) is defined on $$D={(x,y)\in \mathbb{R^2}:x\neq y}\cup{(0,0)}$$ In that context is meaningles to consider the continuity for the points $x=y\neq0$ since they are not included in the domain. – user Sep 02 '18 at 19:01
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@gimusi i was asking Daniel Littlewood, not you! I agree with you, I was asking for Daniels point and answer :) – jeea Sep 03 '18 at 06:09
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