Mixing
Continuity of 1/x/x [duplicate]
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Is $f(x) = x/x$ the same as $f(x) = 1$?
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We know that if $f(x)$ and $g(x)$ are continuous in a domain then $f(x)/g(x)$ is continuous in the domain except for those elements in the domain for which $g(x) = 0$. If I take two functions $f(x) = x$ and $g(x) = 1/x$, then $f(x)$ is continuous for all $R$ and $g(x)$ is continuous for $R-0$ but if I take $f(x)/g(x)$ which is $x/1/x$ or $x^2$ then it is continuous for all $R$ even when $0$ was not in the domain of $g(x)$. Is this correct?
continuity
asked Jul 20 at 7:40
Jahanpanah
141
marked as duplicate by Jam, Arthur, Martin R, Henrik, Parcly Taxel Jul 20 at 11:04
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |Â
up vote
0
down vote
favorite
This question already has an answer here:
Is $f(x) = x/x$ the same as $f(x) = 1$?
3 answers
We know that if $f(x)$ and $g(x)$ are continuous in a domain then $f(x)/g(x)$ is continuous in the domain except for those elements in the domain for which $g(x) = 0$. If I take two functions $f(x) = x$ and $g(x) = 1/x$, then $f(x)$ is continuous for all $R$ and $g(x)$ is continuous for $R-0$ but if I take $f(x)/g(x)$ which is $x/1/x$ or $x^2$ then it is continuous for all $R$ even when $0$ was not in the domain of $g(x)$. Is this correct?
continuity
asked Jul 20 at 7:40
Jahanpanah
141
marked as duplicate by Jam, Arthur, Martin R, Henrik, Parcly Taxel Jul 20 at 11:04
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
2
Not exactly because $fracxfrac1x = x^2$ only when $x neq 0$, so at $x=0$, $fracf(x)g(x)$ is undefined
– Osama Ghani
Jul 20 at 7:46
To elaborate, $f(x)/g(x)$ is undefined at $0$ but has a removable discontinuity. So if it were defined at $0$, the discontinuity could be removed.
– Jam
Jul 20 at 7:47
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
This question already has an answer here:
Is $f(x) = x/x$ the same as $f(x) = 1$?
3 answers
We know that if $f(x)$ and $g(x)$ are continuous in a domain then $f(x)/g(x)$ is continuous in the domain except for those elements in the domain for which $g(x) = 0$. If I take two functions $f(x) = x$ and $g(x) = 1/x$, then $f(x)$ is continuous for all $R$ and $g(x)$ is continuous for $R-0$ but if I take $f(x)/g(x)$ which is $x/1/x$ or $x^2$ then it is continuous for all $R$ even when $0$ was not in the domain of $g(x)$. Is this correct?
continuity
asked Jul 20 at 7:40
Jahanpanah
141
This question already has an answer here:
Is $f(x) = x/x$ the same as $f(x) = 1$?
3 answers
We know that if $f(x)$ and $g(x)$ are continuous in a domain then $f(x)/g(x)$ is continuous in the domain except for those elements in the domain for which $g(x) = 0$. If I take two functions $f(x) = x$ and $g(x) = 1/x$, then $f(x)$ is continuous for all $R$ and $g(x)$ is continuous for $R-0$ but if I take $f(x)/g(x)$ which is $x/1/x$ or $x^2$ then it is continuous for all $R$ even when $0$ was not in the domain of $g(x)$. Is this correct?
This question already has an answer here:
Is $f(x) = x/x$ the same as $f(x) = 1$?
3 answers
continuity
asked Jul 20 at 7:40
Jahanpanah
141
asked Jul 20 at 7:40
Jahanpanah
141
asked Jul 20 at 7:40
Jahanpanah
141
asked Jul 20 at 7:40
Jahanpanah
141
141
marked as duplicate by Jam, Arthur, Martin R, Henrik, Parcly Taxel Jul 20 at 11:04
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Jam, Arthur, Martin R, Henrik, Parcly Taxel Jul 20 at 11:04
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
2
Not exactly because $fracxfrac1x = x^2$ only when $x neq 0$, so at $x=0$, $fracf(x)g(x)$ is undefined
– Osama Ghani
Jul 20 at 7:46
To elaborate, $f(x)/g(x)$ is undefined at $0$ but has a removable discontinuity. So if it were defined at $0$, the discontinuity could be removed.
– Jam
Jul 20 at 7:47
add a comment |Â
2
Not exactly because $fracxfrac1x = x^2$ only when $x neq 0$, so at $x=0$, $fracf(x)g(x)$ is undefined
– Osama Ghani
Jul 20 at 7:46
To elaborate, $f(x)/g(x)$ is undefined at $0$ but has a removable discontinuity. So if it were defined at $0$, the discontinuity could be removed.
– Jam
Jul 20 at 7:47
2
2
Not exactly because $fracxfrac1x = x^2$ only when $x neq 0$, so at $x=0$, $fracf(x)g(x)$ is undefined
– Osama Ghani
Jul 20 at 7:46
Not exactly because $fracxfrac1x = x^2$ only when $x neq 0$, so at $x=0$, $fracf(x)g(x)$ is undefined
– Osama Ghani
Jul 20 at 7:46
To elaborate, $f(x)/g(x)$ is undefined at $0$ but has a removable discontinuity. So if it were defined at $0$, the discontinuity could be removed.
– Jam
Jul 20 at 7:47
To elaborate, $f(x)/g(x)$ is undefined at $0$ but has a removable discontinuity. So if it were defined at $0$, the discontinuity could be removed.
– Jam
Jul 20 at 7:47
add a comment |Â
1 Answer
1
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A function is not just defined by its mapping rule but also by the domain and codomain!
As mentioned in the comments, you have to check the domain carefully. If you have two functions $f:Dto mathbb R$ and $g:Etomathbb R$ then $h:=fracfg$ is defined on
$$
xin Dcap E~:~g(x)neq 0.
$$
In your case $D=mathbb R$ and $E=mathbb Rsetminus 0$. Because $g(x)neq 0$ holds for all $xin E$, you get $Dcap E=E$ as your domain of the quotient $h$. Further $h$ is continuous at each point of $E$ as composition of continuous functions.
Nevertheless, there are a lot of examples, where the quotient $h$ can be continuously extended to a larger domain $Fsupset E$. In your case, you can extend $h$ continuously to whole $mathbb R$. But keep in mind, that the extension is formally not the same as the quotient $h$.
answered Jul 20 at 8:06
Mundron Schmidt
7,1162727
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
A function is not just defined by its mapping rule but also by the domain and codomain!
As mentioned in the comments, you have to check the domain carefully. If you have two functions $f:Dto mathbb R$ and $g:Etomathbb R$ then $h:=fracfg$ is defined on
$$
xin Dcap E~:~g(x)neq 0.
$$
In your case $D=mathbb R$ and $E=mathbb Rsetminus 0$. Because $g(x)neq 0$ holds for all $xin E$, you get $Dcap E=E$ as your domain of the quotient $h$. Further $h$ is continuous at each point of $E$ as composition of continuous functions.
Nevertheless, there are a lot of examples, where the quotient $h$ can be continuously extended to a larger domain $Fsupset E$. In your case, you can extend $h$ continuously to whole $mathbb R$. But keep in mind, that the extension is formally not the same as the quotient $h$.
answered Jul 20 at 8:06
Mundron Schmidt
7,1162727
add a comment |Â
up vote
2
down vote
A function is not just defined by its mapping rule but also by the domain and codomain!
As mentioned in the comments, you have to check the domain carefully. If you have two functions $f:Dto mathbb R$ and $g:Etomathbb R$ then $h:=fracfg$ is defined on
$$
xin Dcap E~:~g(x)neq 0.
$$
In your case $D=mathbb R$ and $E=mathbb Rsetminus 0$. Because $g(x)neq 0$ holds for all $xin E$, you get $Dcap E=E$ as your domain of the quotient $h$. Further $h$ is continuous at each point of $E$ as composition of continuous functions.
Nevertheless, there are a lot of examples, where the quotient $h$ can be continuously extended to a larger domain $Fsupset E$. In your case, you can extend $h$ continuously to whole $mathbb R$. But keep in mind, that the extension is formally not the same as the quotient $h$.
answered Jul 20 at 8:06
Mundron Schmidt
7,1162727
add a comment |Â
up vote
2
down vote
up vote
2
down vote
A function is not just defined by its mapping rule but also by the domain and codomain!
As mentioned in the comments, you have to check the domain carefully. If you have two functions $f:Dto mathbb R$ and $g:Etomathbb R$ then $h:=fracfg$ is defined on
$$
xin Dcap E~:~g(x)neq 0.
$$
In your case $D=mathbb R$ and $E=mathbb Rsetminus 0$. Because $g(x)neq 0$ holds for all $xin E$, you get $Dcap E=E$ as your domain of the quotient $h$. Further $h$ is continuous at each point of $E$ as composition of continuous functions.
Nevertheless, there are a lot of examples, where the quotient $h$ can be continuously extended to a larger domain $Fsupset E$. In your case, you can extend $h$ continuously to whole $mathbb R$. But keep in mind, that the extension is formally not the same as the quotient $h$.
answered Jul 20 at 8:06
Mundron Schmidt
7,1162727
A function is not just defined by its mapping rule but also by the domain and codomain!
As mentioned in the comments, you have to check the domain carefully. If you have two functions $f:Dto mathbb R$ and $g:Etomathbb R$ then $h:=fracfg$ is defined on
$$
xin Dcap E~:~g(x)neq 0.
$$
In your case $D=mathbb R$ and $E=mathbb Rsetminus 0$. Because $g(x)neq 0$ holds for all $xin E$, you get $Dcap E=E$ as your domain of the quotient $h$. Further $h$ is continuous at each point of $E$ as composition of continuous functions.
Nevertheless, there are a lot of examples, where the quotient $h$ can be continuously extended to a larger domain $Fsupset E$. In your case, you can extend $h$ continuously to whole $mathbb R$. But keep in mind, that the extension is formally not the same as the quotient $h$.
answered Jul 20 at 8:06
Mundron Schmidt
7,1162727
answered Jul 20 at 8:06
Mundron Schmidt
7,1162727
answered Jul 20 at 8:06
Mundron Schmidt
7,1162727
answered Jul 20 at 8:06
Mundron Schmidt
7,1162727
7,1162727
add a comment |Â
add a comment |Â
2
Not exactly because $fracxfrac1x = x^2$ only when $x neq 0$, so at $x=0$, $fracf(x)g(x)$ is undefined
– Osama Ghani
Jul 20 at 7:46
To elaborate, $f(x)/g(x)$ is undefined at $0$ but has a removable discontinuity. So if it were defined at $0$, the discontinuity could be removed.
– Jam
Jul 20 at 7:47