Continuity of $f(x,y)$ which is not defined along a path
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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 $0,0$ ?
multivariable-calculus continuity
asked Sep 2 at 9:04
jeea
46312
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up vote
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down vote
favorite
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 $0,0$ ?
multivariable-calculus continuity
asked Sep 2 at 9:04
jeea
46312
How have you shown 'easily' that $f$ is continuous along the path, $t to (t,t)$, say?
– Daniel Littlewood
Sep 2 at 9:23
For any other path, we simply have to use first definition, which goes to 0
– jeea
Sep 2 at 9:35
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?
– Daniel Littlewood
Sep 2 at 9:53
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
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 $0,0$ ?
multivariable-calculus continuity
asked Sep 2 at 9:04
jeea
46312
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 $0,0$ ?
multivariable-calculus continuity
multivariable-calculus continuity
asked Sep 2 at 9:04
jeea
46312
asked Sep 2 at 9:04
jeea
46312
asked Sep 2 at 9:04
jeea
46312
asked Sep 2 at 9:04
jeea
46312
asked Sep 2 at 9:04
jeea
46312
46312
How have you shown 'easily' that $f$ is continuous along the path, $t to (t,t)$, say?
– Daniel Littlewood
Sep 2 at 9:23
For any other path, we simply have to use first definition, which goes to 0
– jeea
Sep 2 at 9:35
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?
– Daniel Littlewood
Sep 2 at 9:53
add a comment |Â
How have you shown 'easily' that $f$ is continuous along the path, $t to (t,t)$, say?
– Daniel Littlewood
Sep 2 at 9:23
For any other path, we simply have to use first definition, which goes to 0
– jeea
Sep 2 at 9:35
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?
– Daniel Littlewood
Sep 2 at 9:53
How have you shown 'easily' that $f$ is continuous along the path, $t to (t,t)$, say?
– Daniel Littlewood
Sep 2 at 9:23
How have you shown 'easily' that $f$ is continuous along the path, $t to (t,t)$, say?
– Daniel Littlewood
Sep 2 at 9:23
For any other path, we simply have to use first definition, which goes to 0
– jeea
Sep 2 at 9:35
For any other path, we simply have to use first definition, which goes to 0
– jeea
Sep 2 at 9:35
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?
– Daniel Littlewood
Sep 2 at 9:53
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?
– Daniel Littlewood
Sep 2 at 9:53
add a comment |Â
1 Answer
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up vote
1
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For the continuity at $(0,0)$ it doesn't mind if $f(x,y)$ is not defined for $x=yneq 0$, indeed we have that trivially
$$lim_substack(x,y)to(0,0)\quad :xneq yx^2+y^2=0=f(0,0)$$
therefore by definition $f(x,y)$ is continuous at $(0,0)$.
answered Sep 2 at 9:12
gimusi
72.2k73888
1
How can you take the limit to $(0,0)$ if the function isn't defined on a neighbourhood of $0$?
– Daniel Littlewood
Sep 2 at 9:20
@DanielLittlewood Just take the path $x=t$ and $y=-t$ for example.
– gimusi
Sep 2 at 9:21
I think you are using a non-standard definition of continuity in $mathbbR^2$. c.f. (en.wikipedia.org/wiki/…)
– Daniel Littlewood
Sep 2 at 9:24
@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 $x=yneq 0$ it doesn't mind since we are removing those points from the domain. The case seems similar to this OP math.stackexchange.com/questions/2889055/…
– gimusi
Sep 2 at 9:39
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!
– Daniel Littlewood
Sep 2 at 9:52
 |Â
show 5 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
For the continuity at $(0,0)$ it doesn't mind if $f(x,y)$ is not defined for $x=yneq 0$, indeed we have that trivially
$$lim_substack(x,y)to(0,0)\quad :xneq yx^2+y^2=0=f(0,0)$$
therefore by definition $f(x,y)$ is continuous at $(0,0)$.
answered Sep 2 at 9:12
gimusi
72.2k73888
1
How can you take the limit to $(0,0)$ if the function isn't defined on a neighbourhood of $0$?
– Daniel Littlewood
Sep 2 at 9:20
@DanielLittlewood Just take the path $x=t$ and $y=-t$ for example.
– gimusi
Sep 2 at 9:21
I think you are using a non-standard definition of continuity in $mathbbR^2$. c.f. (en.wikipedia.org/wiki/…)
– Daniel Littlewood
Sep 2 at 9:24
@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 $x=yneq 0$ it doesn't mind since we are removing those points from the domain. The case seems similar to this OP math.stackexchange.com/questions/2889055/…
– gimusi
Sep 2 at 9:39
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!
– Daniel Littlewood
Sep 2 at 9:52
 |Â
show 5 more comments
up vote
1
down vote
For the continuity at $(0,0)$ it doesn't mind if $f(x,y)$ is not defined for $x=yneq 0$, indeed we have that trivially
$$lim_substack(x,y)to(0,0)\quad :xneq yx^2+y^2=0=f(0,0)$$
therefore by definition $f(x,y)$ is continuous at $(0,0)$.
answered Sep 2 at 9:12
gimusi
72.2k73888
1
How can you take the limit to $(0,0)$ if the function isn't defined on a neighbourhood of $0$?
– Daniel Littlewood
Sep 2 at 9:20
@DanielLittlewood Just take the path $x=t$ and $y=-t$ for example.
– gimusi
Sep 2 at 9:21
I think you are using a non-standard definition of continuity in $mathbbR^2$. c.f. (en.wikipedia.org/wiki/…)
– Daniel Littlewood
Sep 2 at 9:24
@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 $x=yneq 0$ it doesn't mind since we are removing those points from the domain. The case seems similar to this OP math.stackexchange.com/questions/2889055/…
– gimusi
Sep 2 at 9:39
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!
– Daniel Littlewood
Sep 2 at 9:52
 |Â
show 5 more comments
up vote
1
down vote
up vote
1
down vote
For the continuity at $(0,0)$ it doesn't mind if $f(x,y)$ is not defined for $x=yneq 0$, indeed we have that trivially
$$lim_substack(x,y)to(0,0)\quad :xneq yx^2+y^2=0=f(0,0)$$
therefore by definition $f(x,y)$ is continuous at $(0,0)$.
answered Sep 2 at 9:12
gimusi
72.2k73888
For the continuity at $(0,0)$ it doesn't mind if $f(x,y)$ is not defined for $x=yneq 0$, indeed we have that trivially
$$lim_substack(x,y)to(0,0)\quad :xneq yx^2+y^2=0=f(0,0)$$
therefore by definition $f(x,y)$ is continuous at $(0,0)$.
answered Sep 2 at 9:12
gimusi
72.2k73888
edited Sep 2 at 10:00
answered Sep 2 at 9:12
gimusi
72.2k73888
answered Sep 2 at 9:12
gimusi
72.2k73888
answered Sep 2 at 9:12
gimusi
72.2k73888
72.2k73888
1
How can you take the limit to $(0,0)$ if the function isn't defined on a neighbourhood of $0$?
– Daniel Littlewood
Sep 2 at 9:20
@DanielLittlewood Just take the path $x=t$ and $y=-t$ for example.
– gimusi
Sep 2 at 9:21
I think you are using a non-standard definition of continuity in $mathbbR^2$. c.f. (en.wikipedia.org/wiki/…)
– Daniel Littlewood
Sep 2 at 9:24
@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 $x=yneq 0$ it doesn't mind since we are removing those points from the domain. The case seems similar to this OP math.stackexchange.com/questions/2889055/…
– gimusi
Sep 2 at 9:39
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!
– Daniel Littlewood
Sep 2 at 9:52
 |Â
show 5 more comments
1
How can you take the limit to $(0,0)$ if the function isn't defined on a neighbourhood of $0$?
– Daniel Littlewood
Sep 2 at 9:20
@DanielLittlewood Just take the path $x=t$ and $y=-t$ for example.
– gimusi
Sep 2 at 9:21
I think you are using a non-standard definition of continuity in $mathbbR^2$. c.f. (en.wikipedia.org/wiki/…)
– Daniel Littlewood
Sep 2 at 9:24
@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 $x=yneq 0$ it doesn't mind since we are removing those points from the domain. The case seems similar to this OP math.stackexchange.com/questions/2889055/…
– gimusi
Sep 2 at 9:39
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!
– Daniel Littlewood
Sep 2 at 9:52
1
1
How can you take the limit to $(0,0)$ if the function isn't defined on a neighbourhood of $0$?
– Daniel Littlewood
Sep 2 at 9:20
How can you take the limit to $(0,0)$ if the function isn't defined on a neighbourhood of $0$?
– Daniel Littlewood
Sep 2 at 9:20
@DanielLittlewood Just take the path $x=t$ and $y=-t$ for example.
– gimusi
Sep 2 at 9:21
@DanielLittlewood Just take the path $x=t$ and $y=-t$ for example.
– gimusi
Sep 2 at 9:21
I think you are using a non-standard definition of continuity in $mathbbR^2$. c.f. (en.wikipedia.org/wiki/…)
– Daniel Littlewood
Sep 2 at 9:24
I think you are using a non-standard definition of continuity in $mathbbR^2$. c.f. (en.wikipedia.org/wiki/…)
– Daniel Littlewood
Sep 2 at 9:24
@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 $x=yneq 0$ it doesn't mind since we are removing those points from the domain. The case seems similar to this OP math.stackexchange.com/questions/2889055/…
– gimusi
Sep 2 at 9:39
@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 $x=yneq 0$ it doesn't mind since we are removing those points from the domain. The case seems similar to this OP math.stackexchange.com/questions/2889055/…
– gimusi
Sep 2 at 9:39
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!
– Daniel Littlewood
Sep 2 at 9:52
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!
– Daniel Littlewood
Sep 2 at 9:52
 |Â
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How have you shown 'easily' that $f$ is continuous along the path, $t to (t,t)$, say?
– Daniel Littlewood
Sep 2 at 9:23
For any other path, we simply have to use first definition, which goes to 0
– jeea
Sep 2 at 9:35
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?
– Daniel Littlewood
Sep 2 at 9:53