Mixing
Is $frac{x^2}{x}$ continuous?
$begingroup$
This might seem as a weird and straight forward to answer question, but it really confuses me. As the title states, I'm wondering whether
$$f(x)=frac{x^2}{x}$$
is continuous. To be more precise, I'm wondering whether it's continuous at $x=0$. That question arose because I kind of have two contradicting ideas regarding that situation.
Contiuity definition
In university, one of the definitions we got was that a function is continuous in $x_0$ (here $0$) if (and only if) the following is true:
$$limlimits_{xsearrow x_0}f(x)=limlimits_{xnearrow x_0}f(x)=f(x_0)$$
The first two parts are obvious - of course they both approach $0$. But the last part is what's causing me trouble here - because it would yield to $frac{0^2}{0}=frac{0}{0}$ which is known not to be defined. So $f(x)$ wouldn't be continuous in $x_0=0$, right?
Arithmetic
The obvious thing to do though would be to rewrite $f(x)$ as
$$widetilde f(x)=x$$
Now it'd be perfectly clear that $f(0)=0$ just as well. So is the function continuous or is it not?
My approach
So I have an idea of what's probably happening here, but I'm really not sure about it, so I'd be happy for your help. In my understanding, $f(x)$ is defined as $f:mathbb{R}setminus{0}rightarrowmathbb R, xmapstofrac{x^2}{x}$, whereas $widetilde f(x)$ is defined as $widetilde f:mathbb Rrightarrowmathbb R, xmapsto x$. So both functions would be continuous, but on another domain. However, that's still weird, isn't it? I mean, didn't we only make equivalence transformations, meaning that all parts of the following would be true?
$$f(x)=widetilde f(x)qquadLongleftrightarrowqquad frac{x^2}{x}=x$$
$varepsilon-delta$ criteria
And then there's another criteria of continuity, called the $varepsilon-delta$ criteria. I'm not sure if I can explain it well in english, so I'll just write it down in predicate logic:
$$forallvarepsilon>0existsdelta>0forall xin D,|x-x_0|<delta:|f(x)-f(x_0)|<varepsilon$$
So this would once again apply for $widetilde f(x)$ and not for $f(x)$. But then again we learned a visualisation of this criteria, with a rectangle of dimensions $2deltatimes2varepsilon$ closing in around $x_0$ - this would work quite fine with both variations.
calculus limits continuity epsilon-delta
asked Jan 26 at 22:07
MetaColonMetaColon
1256
$endgroup$
add a comment |
$begingroup$
This might seem as a weird and straight forward to answer question, but it really confuses me. As the title states, I'm wondering whether
$$f(x)=frac{x^2}{x}$$
is continuous. To be more precise, I'm wondering whether it's continuous at $x=0$. That question arose because I kind of have two contradicting ideas regarding that situation.
Contiuity definition
In university, one of the definitions we got was that a function is continuous in $x_0$ (here $0$) if (and only if) the following is true:
$$limlimits_{xsearrow x_0}f(x)=limlimits_{xnearrow x_0}f(x)=f(x_0)$$
The first two parts are obvious - of course they both approach $0$. But the last part is what's causing me trouble here - because it would yield to $frac{0^2}{0}=frac{0}{0}$ which is known not to be defined. So $f(x)$ wouldn't be continuous in $x_0=0$, right?
Arithmetic
The obvious thing to do though would be to rewrite $f(x)$ as
$$widetilde f(x)=x$$
Now it'd be perfectly clear that $f(0)=0$ just as well. So is the function continuous or is it not?
My approach
So I have an idea of what's probably happening here, but I'm really not sure about it, so I'd be happy for your help. In my understanding, $f(x)$ is defined as $f:mathbb{R}setminus{0}rightarrowmathbb R, xmapstofrac{x^2}{x}$, whereas $widetilde f(x)$ is defined as $widetilde f:mathbb Rrightarrowmathbb R, xmapsto x$. So both functions would be continuous, but on another domain. However, that's still weird, isn't it? I mean, didn't we only make equivalence transformations, meaning that all parts of the following would be true?
$$f(x)=widetilde f(x)qquadLongleftrightarrowqquad frac{x^2}{x}=x$$
$varepsilon-delta$ criteria
And then there's another criteria of continuity, called the $varepsilon-delta$ criteria. I'm not sure if I can explain it well in english, so I'll just write it down in predicate logic:
$$forallvarepsilon>0existsdelta>0forall xin D,|x-x_0|<delta:|f(x)-f(x_0)|<varepsilon$$
So this would once again apply for $widetilde f(x)$ and not for $f(x)$. But then again we learned a visualisation of this criteria, with a rectangle of dimensions $2deltatimes2varepsilon$ closing in around $x_0$ - this would work quite fine with both variations.
calculus limits continuity epsilon-delta
asked Jan 26 at 22:07
MetaColonMetaColon
1256
$endgroup$
$begingroup$
The answer is straightforward: it's not continuous at $0$ because $0$ is not inside the domain of $f$. That's the end of it. I remember that this confused me too for a while.
$endgroup$
– rubik
Jan 26 at 22:13
$begingroup$
@rubik ok, but isn't dividing by $x$ an equivalent transformation? And if so, does that imply that equivalent transformations can alter the domain?
$endgroup$
– MetaColon
Jan 26 at 22:17
$begingroup$
It is not because you cannot divide by $x = 0$.
$endgroup$
– Klaus
Jan 26 at 22:18
$begingroup$
@Klaus ok than I now also know where my mistake lay.
$endgroup$
– MetaColon
Jan 26 at 22:19
add a comment |
$begingroup$
This might seem as a weird and straight forward to answer question, but it really confuses me. As the title states, I'm wondering whether
$$f(x)=frac{x^2}{x}$$
is continuous. To be more precise, I'm wondering whether it's continuous at $x=0$. That question arose because I kind of have two contradicting ideas regarding that situation.
Contiuity definition
In university, one of the definitions we got was that a function is continuous in $x_0$ (here $0$) if (and only if) the following is true:
$$limlimits_{xsearrow x_0}f(x)=limlimits_{xnearrow x_0}f(x)=f(x_0)$$
The first two parts are obvious - of course they both approach $0$. But the last part is what's causing me trouble here - because it would yield to $frac{0^2}{0}=frac{0}{0}$ which is known not to be defined. So $f(x)$ wouldn't be continuous in $x_0=0$, right?
Arithmetic
The obvious thing to do though would be to rewrite $f(x)$ as
$$widetilde f(x)=x$$
Now it'd be perfectly clear that $f(0)=0$ just as well. So is the function continuous or is it not?
My approach
So I have an idea of what's probably happening here, but I'm really not sure about it, so I'd be happy for your help. In my understanding, $f(x)$ is defined as $f:mathbb{R}setminus{0}rightarrowmathbb R, xmapstofrac{x^2}{x}$, whereas $widetilde f(x)$ is defined as $widetilde f:mathbb Rrightarrowmathbb R, xmapsto x$. So both functions would be continuous, but on another domain. However, that's still weird, isn't it? I mean, didn't we only make equivalence transformations, meaning that all parts of the following would be true?
$$f(x)=widetilde f(x)qquadLongleftrightarrowqquad frac{x^2}{x}=x$$
$varepsilon-delta$ criteria
And then there's another criteria of continuity, called the $varepsilon-delta$ criteria. I'm not sure if I can explain it well in english, so I'll just write it down in predicate logic:
$$forallvarepsilon>0existsdelta>0forall xin D,|x-x_0|<delta:|f(x)-f(x_0)|<varepsilon$$
So this would once again apply for $widetilde f(x)$ and not for $f(x)$. But then again we learned a visualisation of this criteria, with a rectangle of dimensions $2deltatimes2varepsilon$ closing in around $x_0$ - this would work quite fine with both variations.
calculus limits continuity epsilon-delta
asked Jan 26 at 22:07
MetaColonMetaColon
1256
$endgroup$
This might seem as a weird and straight forward to answer question, but it really confuses me. As the title states, I'm wondering whether
$$f(x)=frac{x^2}{x}$$
is continuous. To be more precise, I'm wondering whether it's continuous at $x=0$. That question arose because I kind of have two contradicting ideas regarding that situation.
Contiuity definition
In university, one of the definitions we got was that a function is continuous in $x_0$ (here $0$) if (and only if) the following is true:
$$limlimits_{xsearrow x_0}f(x)=limlimits_{xnearrow x_0}f(x)=f(x_0)$$
The first two parts are obvious - of course they both approach $0$. But the last part is what's causing me trouble here - because it would yield to $frac{0^2}{0}=frac{0}{0}$ which is known not to be defined. So $f(x)$ wouldn't be continuous in $x_0=0$, right?
Arithmetic
The obvious thing to do though would be to rewrite $f(x)$ as
$$widetilde f(x)=x$$
Now it'd be perfectly clear that $f(0)=0$ just as well. So is the function continuous or is it not?
My approach
So I have an idea of what's probably happening here, but I'm really not sure about it, so I'd be happy for your help. In my understanding, $f(x)$ is defined as $f:mathbb{R}setminus{0}rightarrowmathbb R, xmapstofrac{x^2}{x}$, whereas $widetilde f(x)$ is defined as $widetilde f:mathbb Rrightarrowmathbb R, xmapsto x$. So both functions would be continuous, but on another domain. However, that's still weird, isn't it? I mean, didn't we only make equivalence transformations, meaning that all parts of the following would be true?
$$f(x)=widetilde f(x)qquadLongleftrightarrowqquad frac{x^2}{x}=x$$
$varepsilon-delta$ criteria
And then there's another criteria of continuity, called the $varepsilon-delta$ criteria. I'm not sure if I can explain it well in english, so I'll just write it down in predicate logic:
$$forallvarepsilon>0existsdelta>0forall xin D,|x-x_0|<delta:|f(x)-f(x_0)|<varepsilon$$
So this would once again apply for $widetilde f(x)$ and not for $f(x)$. But then again we learned a visualisation of this criteria, with a rectangle of dimensions $2deltatimes2varepsilon$ closing in around $x_0$ - this would work quite fine with both variations.
calculus limits continuity epsilon-delta
calculus limits continuity epsilon-delta
asked Jan 26 at 22:07
MetaColonMetaColon
1256
asked Jan 26 at 22:07
MetaColonMetaColon
1256
asked Jan 26 at 22:07
MetaColonMetaColon
1256
asked Jan 26 at 22:07
MetaColonMetaColon
1256
asked Jan 26 at 22:07
MetaColonMetaColon
1256
1256
$begingroup$
The answer is straightforward: it's not continuous at $0$ because $0$ is not inside the domain of $f$. That's the end of it. I remember that this confused me too for a while.
$endgroup$
– rubik
Jan 26 at 22:13
$begingroup$
@rubik ok, but isn't dividing by $x$ an equivalent transformation? And if so, does that imply that equivalent transformations can alter the domain?
$endgroup$
– MetaColon
Jan 26 at 22:17
$begingroup$
It is not because you cannot divide by $x = 0$.
$endgroup$
– Klaus
Jan 26 at 22:18
$begingroup$
@Klaus ok than I now also know where my mistake lay.
$endgroup$
– MetaColon
Jan 26 at 22:19
add a comment |
$begingroup$
The answer is straightforward: it's not continuous at $0$ because $0$ is not inside the domain of $f$. That's the end of it. I remember that this confused me too for a while.
$endgroup$
– rubik
Jan 26 at 22:13
$begingroup$
@rubik ok, but isn't dividing by $x$ an equivalent transformation? And if so, does that imply that equivalent transformations can alter the domain?
$endgroup$
– MetaColon
Jan 26 at 22:17
$begingroup$
It is not because you cannot divide by $x = 0$.
$endgroup$
– Klaus
Jan 26 at 22:18
$begingroup$
@Klaus ok than I now also know where my mistake lay.
$endgroup$
– MetaColon
Jan 26 at 22:19
$begingroup$
The answer is straightforward: it's not continuous at $0$ because $0$ is not inside the domain of $f$. That's the end of it. I remember that this confused me too for a while.
$endgroup$
– rubik
Jan 26 at 22:13
$begingroup$
The answer is straightforward: it's not continuous at $0$ because $0$ is not inside the domain of $f$. That's the end of it. I remember that this confused me too for a while.
$endgroup$
– rubik
Jan 26 at 22:13
$begingroup$
@rubik ok, but isn't dividing by $x$ an equivalent transformation? And if so, does that imply that equivalent transformations can alter the domain?
$endgroup$
– MetaColon
Jan 26 at 22:17
$begingroup$
@rubik ok, but isn't dividing by $x$ an equivalent transformation? And if so, does that imply that equivalent transformations can alter the domain?
$endgroup$
– MetaColon
Jan 26 at 22:17
$begingroup$
It is not because you cannot divide by $x = 0$.
$endgroup$
– Klaus
Jan 26 at 22:18
$begingroup$
It is not because you cannot divide by $x = 0$.
$endgroup$
– Klaus
Jan 26 at 22:18
$begingroup$
@Klaus ok than I now also know where my mistake lay.
$endgroup$
– MetaColon
Jan 26 at 22:19
$begingroup$
@Klaus ok than I now also know where my mistake lay.
$endgroup$
– MetaColon
Jan 26 at 22:19
add a comment |
8 Answers
8
active
oldest
votes
$begingroup$
It is not continuous at $0$ because it is not defined at $0$ but it can be extended by continuity at $0$.
answered Jan 26 at 22:08
Tsemo AristideTsemo Aristide
59.7k11446
$endgroup$
$begingroup$
To be quite honest, this doesn't help me at all
$endgroup$
– MetaColon
Jan 26 at 22:09
2
$begingroup$
But this is the right answer. For a function to be continuous at point a, the value of f(a) must be the limit of f(x) when x goes to a. In your case there is no f(0), so it cannot be continuous there. If you define f(0)=0 then you are good
$endgroup$
– Shaq
Jan 26 at 22:11
$begingroup$
@Shaq but imagine you don't define that. Isn't it still exactly the same as $widetilde f$?
$endgroup$
– MetaColon
Jan 26 at 22:12
3
$begingroup$
$f$ and $tilde{f}$ are not the same functions because the domains are different!
$endgroup$
– Klaus
Jan 26 at 22:14
1
$begingroup$
@MetaColon No, it's not. The original function cannot be defined at zero as it is, whereas $;tilde f;$ is defined everywhere. This is the right answer.
$endgroup$
– DonAntonio
Jan 26 at 22:14
|
show 1 more comment
$begingroup$
Actually, if we are going to be rigorous, then the question makes no sense. That's so because you did not state what is the domain of $f$. If you choose to decide that the domain is the largest subset of $mathbb R$ for which the expression $dfrac{x^2}x$ makes sense (which is $mathbb{R}setminus{0}$), then the answer now becomes: it makes no sense to ask whether $f$ is continuous at $0$ since continuity is defined only for points of the domain.
However, if the domain of $f$ is $mathbb{R}setminus{0}$, then you can extend $f$ to a continuous function from $mathbb R$ into $mathbb R$.
answered Jan 26 at 22:13
José Carlos SantosJosé Carlos Santos
169k23132238
$endgroup$
$begingroup$
I know you meant this, but I want to add: if the domain is $Bbb R setminus 0$, then for this particular $f$, there is a continuous extension to $Bbb R$. In general, not all functions on the first set admit continuous extension to the second. Example: $g(x) = frac{1}{x}$.
$endgroup$
– John Hughes
Jan 26 at 22:23
add a comment |
$begingroup$
$$ x text{ continuous at $0$ } iff lim_{xto 0^-}f(x)=lim_{xto 0^+}f(x)=f(0)$$
But $f(0)$ doesn't exist, so this doesn't hold.
answered Jan 26 at 22:13
Rhys HughesRhys Hughes
7,0351630
$endgroup$
add a comment |
$begingroup$
The definition of continuity ends with $cdots = f(x_0).$
Since $f(0)$ doesn't exist, the function is not continuous there. The graph of the function is a straight line of slope $1$ through the origin, but has a hole at the origin.
To be continuous at a point, there are three things necessary:
- The limit exists.
- The function exists.
- The above two are equal.
answered Jan 26 at 22:13
B. GoddardB. Goddard
19.7k21442
$endgroup$
add a comment |
$begingroup$
You are asking if the function is continuous at $x=0$. The definition of being "continuous at $x=0$" means that $$lim_{tto0}f(t)=f(0)$$ With this function, $lim_{tto0}f(t)=0$ and $f(0)$ is undefined. So $$lim_{tto0}f(t)neq f(0)$$ and therefore the function is not continuous at $x=0$.
answered Jan 26 at 22:13
alex.jordanalex.jordan
39.5k560122
$endgroup$
add a comment |
$begingroup$
When you give a definition of a function, it is down to you to make clear what the intended domain of that function is. There is a common convention that if you write:
$$
f(x) = t(x)
$$
where $t(x)$ is some algebraic expression in the variable $x$, then the domain of $f$ is restricted to values of $x$ for which deriving the value of $t(x)$ is well-defined, e.g., doesn't involve division by zero.
This convention is very vague and in itself ill-defined. E.g., the domain of the function $r$ defined by
$$
r(x) = sqrt{x}
$$
will be at most $Bbb{R}_{ge0}$ if the range of $r$ is expected to be $Bbb{R}$, but could be $Bbb{C}$ if the range of $r$ is expected to be $Bbb{C}$.
So when you ask about whether $f(x) = frac{x^2}{x}$ is continuous, you are:
- firstly, making a category error (an equation is not a function) and,
- secondly, if we forget the category error and assume you mean the function $f$ defined by the equation, not giving enough information about the function in question.
A rigorous definition of your function would say precisely what its domain is intended to be.
answered Jan 26 at 22:40
Rob ArthanRob Arthan
29.5k42967
$endgroup$
1
$begingroup$
What exactly do you mean by category error?
$endgroup$
– MetaColon
Jan 26 at 22:43
1
$begingroup$
$f(x) = x^2/x$ is an equation in which $f$ appears as a function. Your question is asking if the equation is continuous, but it's actually the function that may or may not be continuous. A category error is applying a notion (like continuity) to the wrong kind of thing (like an equation rather than a function).
$endgroup$
– Rob Arthan
Jan 26 at 22:50
add a comment |
$begingroup$
The function is undefined at $x=0$ so can't be continuous there.
The definition of the function can be extended so that the function is given a value at $x=0$ in a way which does make the extended function continuous there.
If the function were extended by giving a value at $x=0$ which does not make it continuous, we would say that it had a "removable discontinuity" at $x=0$. The discontinuity would be removed by giving it "the correct value" at an isolated point.
Sometimes this language is used more informally to say "the definition can be repaired or extended" in the case where the function is not properly defined at a some isolated point.
answered Jan 26 at 22:25
Mark BennetMark Bennet
81.7k984182
$endgroup$
add a comment |
$begingroup$
The question is not well-posed. It only makes sense to consider continuity at points in the domain of the function. If we view the function as $fcolon mathbb{R}setminus {0}to mathbb{R}$, then the function is continuous everywhere. This function has an extension $hat{f}colon mathbb{R}to mathbb{R}$ which is continuous at zero. Simply define $hat{f}(0)=0$.
answered Jan 26 at 22:50
Foobaz JohnFoobaz John
22.8k41452
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3088833%2fis-fracx2x-continuous%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
8 Answers
8
active
oldest
votes
8 Answers
8
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It is not continuous at $0$ because it is not defined at $0$ but it can be extended by continuity at $0$.
answered Jan 26 at 22:08
Tsemo AristideTsemo Aristide
59.7k11446
$endgroup$
$begingroup$
To be quite honest, this doesn't help me at all
$endgroup$
– MetaColon
Jan 26 at 22:09
2
$begingroup$
But this is the right answer. For a function to be continuous at point a, the value of f(a) must be the limit of f(x) when x goes to a. In your case there is no f(0), so it cannot be continuous there. If you define f(0)=0 then you are good
$endgroup$
– Shaq
Jan 26 at 22:11
$begingroup$
@Shaq but imagine you don't define that. Isn't it still exactly the same as $widetilde f$?
$endgroup$
– MetaColon
Jan 26 at 22:12
3
$begingroup$
$f$ and $tilde{f}$ are not the same functions because the domains are different!
$endgroup$
– Klaus
Jan 26 at 22:14
1
$begingroup$
@MetaColon No, it's not. The original function cannot be defined at zero as it is, whereas $;tilde f;$ is defined everywhere. This is the right answer.
$endgroup$
– DonAntonio
Jan 26 at 22:14
|
show 1 more comment
$begingroup$
It is not continuous at $0$ because it is not defined at $0$ but it can be extended by continuity at $0$.
answered Jan 26 at 22:08
Tsemo AristideTsemo Aristide
59.7k11446
$endgroup$
$begingroup$
To be quite honest, this doesn't help me at all
$endgroup$
– MetaColon
Jan 26 at 22:09
2
$begingroup$
But this is the right answer. For a function to be continuous at point a, the value of f(a) must be the limit of f(x) when x goes to a. In your case there is no f(0), so it cannot be continuous there. If you define f(0)=0 then you are good
$endgroup$
– Shaq
Jan 26 at 22:11
$begingroup$
@Shaq but imagine you don't define that. Isn't it still exactly the same as $widetilde f$?
$endgroup$
– MetaColon
Jan 26 at 22:12
3
$begingroup$
$f$ and $tilde{f}$ are not the same functions because the domains are different!
$endgroup$
– Klaus
Jan 26 at 22:14
1
$begingroup$
@MetaColon No, it's not. The original function cannot be defined at zero as it is, whereas $;tilde f;$ is defined everywhere. This is the right answer.
$endgroup$
– DonAntonio
Jan 26 at 22:14
|
show 1 more comment
$begingroup$
It is not continuous at $0$ because it is not defined at $0$ but it can be extended by continuity at $0$.
answered Jan 26 at 22:08
Tsemo AristideTsemo Aristide
59.7k11446
$endgroup$
It is not continuous at $0$ because it is not defined at $0$ but it can be extended by continuity at $0$.
answered Jan 26 at 22:08
Tsemo AristideTsemo Aristide
59.7k11446
answered Jan 26 at 22:08
Tsemo AristideTsemo Aristide
59.7k11446
answered Jan 26 at 22:08
Tsemo AristideTsemo Aristide
59.7k11446
answered Jan 26 at 22:08
Tsemo AristideTsemo Aristide
59.7k11446
59.7k11446
$begingroup$
To be quite honest, this doesn't help me at all
$endgroup$
– MetaColon
Jan 26 at 22:09
2
$begingroup$
But this is the right answer. For a function to be continuous at point a, the value of f(a) must be the limit of f(x) when x goes to a. In your case there is no f(0), so it cannot be continuous there. If you define f(0)=0 then you are good
$endgroup$
– Shaq
Jan 26 at 22:11
$begingroup$
@Shaq but imagine you don't define that. Isn't it still exactly the same as $widetilde f$?
$endgroup$
– MetaColon
Jan 26 at 22:12
3
$begingroup$
$f$ and $tilde{f}$ are not the same functions because the domains are different!
$endgroup$
– Klaus
Jan 26 at 22:14
1
$begingroup$
@MetaColon No, it's not. The original function cannot be defined at zero as it is, whereas $;tilde f;$ is defined everywhere. This is the right answer.
$endgroup$
– DonAntonio
Jan 26 at 22:14
|
show 1 more comment
$begingroup$
To be quite honest, this doesn't help me at all
$endgroup$
– MetaColon
Jan 26 at 22:09
2
$begingroup$
But this is the right answer. For a function to be continuous at point a, the value of f(a) must be the limit of f(x) when x goes to a. In your case there is no f(0), so it cannot be continuous there. If you define f(0)=0 then you are good
$endgroup$
– Shaq
Jan 26 at 22:11
$begingroup$
@Shaq but imagine you don't define that. Isn't it still exactly the same as $widetilde f$?
$endgroup$
– MetaColon
Jan 26 at 22:12
3
$begingroup$
$f$ and $tilde{f}$ are not the same functions because the domains are different!
$endgroup$
– Klaus
Jan 26 at 22:14
1
$begingroup$
@MetaColon No, it's not. The original function cannot be defined at zero as it is, whereas $;tilde f;$ is defined everywhere. This is the right answer.
$endgroup$
– DonAntonio
Jan 26 at 22:14
$begingroup$
To be quite honest, this doesn't help me at all
$endgroup$
– MetaColon
Jan 26 at 22:09
$begingroup$
To be quite honest, this doesn't help me at all
$endgroup$
– MetaColon
Jan 26 at 22:09
2
2
$begingroup$
But this is the right answer. For a function to be continuous at point a, the value of f(a) must be the limit of f(x) when x goes to a. In your case there is no f(0), so it cannot be continuous there. If you define f(0)=0 then you are good
$endgroup$
– Shaq
Jan 26 at 22:11
$begingroup$
But this is the right answer. For a function to be continuous at point a, the value of f(a) must be the limit of f(x) when x goes to a. In your case there is no f(0), so it cannot be continuous there. If you define f(0)=0 then you are good
$endgroup$
– Shaq
Jan 26 at 22:11
$begingroup$
@Shaq but imagine you don't define that. Isn't it still exactly the same as $widetilde f$?
$endgroup$
– MetaColon
Jan 26 at 22:12
$begingroup$
@Shaq but imagine you don't define that. Isn't it still exactly the same as $widetilde f$?
$endgroup$
– MetaColon
Jan 26 at 22:12
3
3
$begingroup$
$f$ and $tilde{f}$ are not the same functions because the domains are different!
$endgroup$
– Klaus
Jan 26 at 22:14
$begingroup$
$f$ and $tilde{f}$ are not the same functions because the domains are different!
$endgroup$
– Klaus
Jan 26 at 22:14
1
1
$begingroup$
@MetaColon No, it's not. The original function cannot be defined at zero as it is, whereas $;tilde f;$ is defined everywhere. This is the right answer.
$endgroup$
– DonAntonio
Jan 26 at 22:14
$begingroup$
@MetaColon No, it's not. The original function cannot be defined at zero as it is, whereas $;tilde f;$ is defined everywhere. This is the right answer.
$endgroup$
– DonAntonio
Jan 26 at 22:14
|
show 1 more comment
$begingroup$
Actually, if we are going to be rigorous, then the question makes no sense. That's so because you did not state what is the domain of $f$. If you choose to decide that the domain is the largest subset of $mathbb R$ for which the expression $dfrac{x^2}x$ makes sense (which is $mathbb{R}setminus{0}$), then the answer now becomes: it makes no sense to ask whether $f$ is continuous at $0$ since continuity is defined only for points of the domain.
However, if the domain of $f$ is $mathbb{R}setminus{0}$, then you can extend $f$ to a continuous function from $mathbb R$ into $mathbb R$.
answered Jan 26 at 22:13
José Carlos SantosJosé Carlos Santos
169k23132238
$endgroup$
$begingroup$
I know you meant this, but I want to add: if the domain is $Bbb R setminus 0$, then for this particular $f$, there is a continuous extension to $Bbb R$. In general, not all functions on the first set admit continuous extension to the second. Example: $g(x) = frac{1}{x}$.
$endgroup$
– John Hughes
Jan 26 at 22:23
add a comment |
$begingroup$
Actually, if we are going to be rigorous, then the question makes no sense. That's so because you did not state what is the domain of $f$. If you choose to decide that the domain is the largest subset of $mathbb R$ for which the expression $dfrac{x^2}x$ makes sense (which is $mathbb{R}setminus{0}$), then the answer now becomes: it makes no sense to ask whether $f$ is continuous at $0$ since continuity is defined only for points of the domain.
However, if the domain of $f$ is $mathbb{R}setminus{0}$, then you can extend $f$ to a continuous function from $mathbb R$ into $mathbb R$.
answered Jan 26 at 22:13
José Carlos SantosJosé Carlos Santos
169k23132238
$endgroup$
$begingroup$
I know you meant this, but I want to add: if the domain is $Bbb R setminus 0$, then for this particular $f$, there is a continuous extension to $Bbb R$. In general, not all functions on the first set admit continuous extension to the second. Example: $g(x) = frac{1}{x}$.
$endgroup$
– John Hughes
Jan 26 at 22:23
add a comment |
$begingroup$
Actually, if we are going to be rigorous, then the question makes no sense. That's so because you did not state what is the domain of $f$. If you choose to decide that the domain is the largest subset of $mathbb R$ for which the expression $dfrac{x^2}x$ makes sense (which is $mathbb{R}setminus{0}$), then the answer now becomes: it makes no sense to ask whether $f$ is continuous at $0$ since continuity is defined only for points of the domain.
However, if the domain of $f$ is $mathbb{R}setminus{0}$, then you can extend $f$ to a continuous function from $mathbb R$ into $mathbb R$.
answered Jan 26 at 22:13
José Carlos SantosJosé Carlos Santos
169k23132238
$endgroup$
Actually, if we are going to be rigorous, then the question makes no sense. That's so because you did not state what is the domain of $f$. If you choose to decide that the domain is the largest subset of $mathbb R$ for which the expression $dfrac{x^2}x$ makes sense (which is $mathbb{R}setminus{0}$), then the answer now becomes: it makes no sense to ask whether $f$ is continuous at $0$ since continuity is defined only for points of the domain.
However, if the domain of $f$ is $mathbb{R}setminus{0}$, then you can extend $f$ to a continuous function from $mathbb R$ into $mathbb R$.
answered Jan 26 at 22:13
José Carlos SantosJosé Carlos Santos
169k23132238
answered Jan 26 at 22:13
José Carlos SantosJosé Carlos Santos
169k23132238
answered Jan 26 at 22:13
José Carlos SantosJosé Carlos Santos
169k23132238
answered Jan 26 at 22:13
José Carlos SantosJosé Carlos Santos
169k23132238
169k23132238
$begingroup$
I know you meant this, but I want to add: if the domain is $Bbb R setminus 0$, then for this particular $f$, there is a continuous extension to $Bbb R$. In general, not all functions on the first set admit continuous extension to the second. Example: $g(x) = frac{1}{x}$.
$endgroup$
– John Hughes
Jan 26 at 22:23
add a comment |
$begingroup$
I know you meant this, but I want to add: if the domain is $Bbb R setminus 0$, then for this particular $f$, there is a continuous extension to $Bbb R$. In general, not all functions on the first set admit continuous extension to the second. Example: $g(x) = frac{1}{x}$.
$endgroup$
– John Hughes
Jan 26 at 22:23
$begingroup$
I know you meant this, but I want to add: if the domain is $Bbb R setminus 0$, then for this particular $f$, there is a continuous extension to $Bbb R$. In general, not all functions on the first set admit continuous extension to the second. Example: $g(x) = frac{1}{x}$.
$endgroup$
– John Hughes
Jan 26 at 22:23
$begingroup$
I know you meant this, but I want to add: if the domain is $Bbb R setminus 0$, then for this particular $f$, there is a continuous extension to $Bbb R$. In general, not all functions on the first set admit continuous extension to the second. Example: $g(x) = frac{1}{x}$.
$endgroup$
– John Hughes
Jan 26 at 22:23
add a comment |
$begingroup$
$$ x text{ continuous at $0$ } iff lim_{xto 0^-}f(x)=lim_{xto 0^+}f(x)=f(0)$$
But $f(0)$ doesn't exist, so this doesn't hold.
answered Jan 26 at 22:13
Rhys HughesRhys Hughes
7,0351630
$endgroup$
add a comment |
$begingroup$
$$ x text{ continuous at $0$ } iff lim_{xto 0^-}f(x)=lim_{xto 0^+}f(x)=f(0)$$
But $f(0)$ doesn't exist, so this doesn't hold.
answered Jan 26 at 22:13
Rhys HughesRhys Hughes
7,0351630
$endgroup$
add a comment |
$begingroup$
$$ x text{ continuous at $0$ } iff lim_{xto 0^-}f(x)=lim_{xto 0^+}f(x)=f(0)$$
But $f(0)$ doesn't exist, so this doesn't hold.
answered Jan 26 at 22:13
Rhys HughesRhys Hughes
7,0351630
$endgroup$
$$ x text{ continuous at $0$ } iff lim_{xto 0^-}f(x)=lim_{xto 0^+}f(x)=f(0)$$
But $f(0)$ doesn't exist, so this doesn't hold.
answered Jan 26 at 22:13
Rhys HughesRhys Hughes
7,0351630
answered Jan 26 at 22:13
Rhys HughesRhys Hughes
7,0351630
answered Jan 26 at 22:13
Rhys HughesRhys Hughes
7,0351630
answered Jan 26 at 22:13
Rhys HughesRhys Hughes
7,0351630
7,0351630
add a comment |
add a comment |
$begingroup$
The definition of continuity ends with $cdots = f(x_0).$
Since $f(0)$ doesn't exist, the function is not continuous there. The graph of the function is a straight line of slope $1$ through the origin, but has a hole at the origin.
To be continuous at a point, there are three things necessary:
- The limit exists.
- The function exists.
- The above two are equal.
answered Jan 26 at 22:13
B. GoddardB. Goddard
19.7k21442
$endgroup$
add a comment |
$begingroup$
The definition of continuity ends with $cdots = f(x_0).$
Since $f(0)$ doesn't exist, the function is not continuous there. The graph of the function is a straight line of slope $1$ through the origin, but has a hole at the origin.
To be continuous at a point, there are three things necessary:
- The limit exists.
- The function exists.
- The above two are equal.
answered Jan 26 at 22:13
B. GoddardB. Goddard
19.7k21442
$endgroup$
add a comment |
$begingroup$
The definition of continuity ends with $cdots = f(x_0).$
Since $f(0)$ doesn't exist, the function is not continuous there. The graph of the function is a straight line of slope $1$ through the origin, but has a hole at the origin.
To be continuous at a point, there are three things necessary:
- The limit exists.
- The function exists.
- The above two are equal.
answered Jan 26 at 22:13
B. GoddardB. Goddard
19.7k21442
$endgroup$
The definition of continuity ends with $cdots = f(x_0).$
Since $f(0)$ doesn't exist, the function is not continuous there. The graph of the function is a straight line of slope $1$ through the origin, but has a hole at the origin.
To be continuous at a point, there are three things necessary:
- The limit exists.
- The function exists.
- The above two are equal.
answered Jan 26 at 22:13
B. GoddardB. Goddard
19.7k21442
answered Jan 26 at 22:13
B. GoddardB. Goddard
19.7k21442
answered Jan 26 at 22:13
B. GoddardB. Goddard
19.7k21442
answered Jan 26 at 22:13
B. GoddardB. Goddard
19.7k21442
19.7k21442
add a comment |
add a comment |
$begingroup$
You are asking if the function is continuous at $x=0$. The definition of being "continuous at $x=0$" means that $$lim_{tto0}f(t)=f(0)$$ With this function, $lim_{tto0}f(t)=0$ and $f(0)$ is undefined. So $$lim_{tto0}f(t)neq f(0)$$ and therefore the function is not continuous at $x=0$.
answered Jan 26 at 22:13
alex.jordanalex.jordan
39.5k560122
$endgroup$
add a comment |
$begingroup$
You are asking if the function is continuous at $x=0$. The definition of being "continuous at $x=0$" means that $$lim_{tto0}f(t)=f(0)$$ With this function, $lim_{tto0}f(t)=0$ and $f(0)$ is undefined. So $$lim_{tto0}f(t)neq f(0)$$ and therefore the function is not continuous at $x=0$.
answered Jan 26 at 22:13
alex.jordanalex.jordan
39.5k560122
$endgroup$
add a comment |
$begingroup$
You are asking if the function is continuous at $x=0$. The definition of being "continuous at $x=0$" means that $$lim_{tto0}f(t)=f(0)$$ With this function, $lim_{tto0}f(t)=0$ and $f(0)$ is undefined. So $$lim_{tto0}f(t)neq f(0)$$ and therefore the function is not continuous at $x=0$.
answered Jan 26 at 22:13
alex.jordanalex.jordan
39.5k560122
$endgroup$
You are asking if the function is continuous at $x=0$. The definition of being "continuous at $x=0$" means that $$lim_{tto0}f(t)=f(0)$$ With this function, $lim_{tto0}f(t)=0$ and $f(0)$ is undefined. So $$lim_{tto0}f(t)neq f(0)$$ and therefore the function is not continuous at $x=0$.
answered Jan 26 at 22:13
alex.jordanalex.jordan
39.5k560122
answered Jan 26 at 22:13
alex.jordanalex.jordan
39.5k560122
answered Jan 26 at 22:13
alex.jordanalex.jordan
39.5k560122
answered Jan 26 at 22:13
alex.jordanalex.jordan
39.5k560122
39.5k560122
add a comment |
add a comment |
$begingroup$
When you give a definition of a function, it is down to you to make clear what the intended domain of that function is. There is a common convention that if you write:
$$
f(x) = t(x)
$$
where $t(x)$ is some algebraic expression in the variable $x$, then the domain of $f$ is restricted to values of $x$ for which deriving the value of $t(x)$ is well-defined, e.g., doesn't involve division by zero.
This convention is very vague and in itself ill-defined. E.g., the domain of the function $r$ defined by
$$
r(x) = sqrt{x}
$$
will be at most $Bbb{R}_{ge0}$ if the range of $r$ is expected to be $Bbb{R}$, but could be $Bbb{C}$ if the range of $r$ is expected to be $Bbb{C}$.
So when you ask about whether $f(x) = frac{x^2}{x}$ is continuous, you are:
- firstly, making a category error (an equation is not a function) and,
- secondly, if we forget the category error and assume you mean the function $f$ defined by the equation, not giving enough information about the function in question.
A rigorous definition of your function would say precisely what its domain is intended to be.
answered Jan 26 at 22:40
Rob ArthanRob Arthan
29.5k42967
$endgroup$
1
$begingroup$
What exactly do you mean by category error?
$endgroup$
– MetaColon
Jan 26 at 22:43
1
$begingroup$
$f(x) = x^2/x$ is an equation in which $f$ appears as a function. Your question is asking if the equation is continuous, but it's actually the function that may or may not be continuous. A category error is applying a notion (like continuity) to the wrong kind of thing (like an equation rather than a function).
$endgroup$
– Rob Arthan
Jan 26 at 22:50
add a comment |
$begingroup$
When you give a definition of a function, it is down to you to make clear what the intended domain of that function is. There is a common convention that if you write:
$$
f(x) = t(x)
$$
where $t(x)$ is some algebraic expression in the variable $x$, then the domain of $f$ is restricted to values of $x$ for which deriving the value of $t(x)$ is well-defined, e.g., doesn't involve division by zero.
This convention is very vague and in itself ill-defined. E.g., the domain of the function $r$ defined by
$$
r(x) = sqrt{x}
$$
will be at most $Bbb{R}_{ge0}$ if the range of $r$ is expected to be $Bbb{R}$, but could be $Bbb{C}$ if the range of $r$ is expected to be $Bbb{C}$.
So when you ask about whether $f(x) = frac{x^2}{x}$ is continuous, you are:
- firstly, making a category error (an equation is not a function) and,
- secondly, if we forget the category error and assume you mean the function $f$ defined by the equation, not giving enough information about the function in question.
A rigorous definition of your function would say precisely what its domain is intended to be.
answered Jan 26 at 22:40
Rob ArthanRob Arthan
29.5k42967
$endgroup$
1
$begingroup$
What exactly do you mean by category error?
$endgroup$
– MetaColon
Jan 26 at 22:43
1
$begingroup$
$f(x) = x^2/x$ is an equation in which $f$ appears as a function. Your question is asking if the equation is continuous, but it's actually the function that may or may not be continuous. A category error is applying a notion (like continuity) to the wrong kind of thing (like an equation rather than a function).
$endgroup$
– Rob Arthan
Jan 26 at 22:50
add a comment |
$begingroup$
When you give a definition of a function, it is down to you to make clear what the intended domain of that function is. There is a common convention that if you write:
$$
f(x) = t(x)
$$
where $t(x)$ is some algebraic expression in the variable $x$, then the domain of $f$ is restricted to values of $x$ for which deriving the value of $t(x)$ is well-defined, e.g., doesn't involve division by zero.
This convention is very vague and in itself ill-defined. E.g., the domain of the function $r$ defined by
$$
r(x) = sqrt{x}
$$
will be at most $Bbb{R}_{ge0}$ if the range of $r$ is expected to be $Bbb{R}$, but could be $Bbb{C}$ if the range of $r$ is expected to be $Bbb{C}$.
So when you ask about whether $f(x) = frac{x^2}{x}$ is continuous, you are:
- firstly, making a category error (an equation is not a function) and,
- secondly, if we forget the category error and assume you mean the function $f$ defined by the equation, not giving enough information about the function in question.
A rigorous definition of your function would say precisely what its domain is intended to be.
answered Jan 26 at 22:40
Rob ArthanRob Arthan
29.5k42967
$endgroup$
When you give a definition of a function, it is down to you to make clear what the intended domain of that function is. There is a common convention that if you write:
$$
f(x) = t(x)
$$
where $t(x)$ is some algebraic expression in the variable $x$, then the domain of $f$ is restricted to values of $x$ for which deriving the value of $t(x)$ is well-defined, e.g., doesn't involve division by zero.
This convention is very vague and in itself ill-defined. E.g., the domain of the function $r$ defined by
$$
r(x) = sqrt{x}
$$
will be at most $Bbb{R}_{ge0}$ if the range of $r$ is expected to be $Bbb{R}$, but could be $Bbb{C}$ if the range of $r$ is expected to be $Bbb{C}$.
So when you ask about whether $f(x) = frac{x^2}{x}$ is continuous, you are:
- firstly, making a category error (an equation is not a function) and,
- secondly, if we forget the category error and assume you mean the function $f$ defined by the equation, not giving enough information about the function in question.
A rigorous definition of your function would say precisely what its domain is intended to be.
answered Jan 26 at 22:40
Rob ArthanRob Arthan
29.5k42967
edited Jan 26 at 23:39
answered Jan 26 at 22:40
Rob ArthanRob Arthan
29.5k42967
answered Jan 26 at 22:40
Rob ArthanRob Arthan
29.5k42967
answered Jan 26 at 22:40
Rob ArthanRob Arthan
29.5k42967
29.5k42967
1
$begingroup$
What exactly do you mean by category error?
$endgroup$
– MetaColon
Jan 26 at 22:43
1
$begingroup$
$f(x) = x^2/x$ is an equation in which $f$ appears as a function. Your question is asking if the equation is continuous, but it's actually the function that may or may not be continuous. A category error is applying a notion (like continuity) to the wrong kind of thing (like an equation rather than a function).
$endgroup$
– Rob Arthan
Jan 26 at 22:50
add a comment |
1
$begingroup$
What exactly do you mean by category error?
$endgroup$
– MetaColon
Jan 26 at 22:43
1
$begingroup$
$f(x) = x^2/x$ is an equation in which $f$ appears as a function. Your question is asking if the equation is continuous, but it's actually the function that may or may not be continuous. A category error is applying a notion (like continuity) to the wrong kind of thing (like an equation rather than a function).
$endgroup$
– Rob Arthan
Jan 26 at 22:50
1
1
$begingroup$
What exactly do you mean by category error?
$endgroup$
– MetaColon
Jan 26 at 22:43
$begingroup$
What exactly do you mean by category error?
$endgroup$
– MetaColon
Jan 26 at 22:43
1
1
$begingroup$
$f(x) = x^2/x$ is an equation in which $f$ appears as a function. Your question is asking if the equation is continuous, but it's actually the function that may or may not be continuous. A category error is applying a notion (like continuity) to the wrong kind of thing (like an equation rather than a function).
$endgroup$
– Rob Arthan
Jan 26 at 22:50
$begingroup$
$f(x) = x^2/x$ is an equation in which $f$ appears as a function. Your question is asking if the equation is continuous, but it's actually the function that may or may not be continuous. A category error is applying a notion (like continuity) to the wrong kind of thing (like an equation rather than a function).
$endgroup$
– Rob Arthan
Jan 26 at 22:50
add a comment |
$begingroup$
The function is undefined at $x=0$ so can't be continuous there.
The definition of the function can be extended so that the function is given a value at $x=0$ in a way which does make the extended function continuous there.
If the function were extended by giving a value at $x=0$ which does not make it continuous, we would say that it had a "removable discontinuity" at $x=0$. The discontinuity would be removed by giving it "the correct value" at an isolated point.
Sometimes this language is used more informally to say "the definition can be repaired or extended" in the case where the function is not properly defined at a some isolated point.
answered Jan 26 at 22:25
Mark BennetMark Bennet
81.7k984182
$endgroup$
add a comment |
$begingroup$
The function is undefined at $x=0$ so can't be continuous there.
The definition of the function can be extended so that the function is given a value at $x=0$ in a way which does make the extended function continuous there.
If the function were extended by giving a value at $x=0$ which does not make it continuous, we would say that it had a "removable discontinuity" at $x=0$. The discontinuity would be removed by giving it "the correct value" at an isolated point.
Sometimes this language is used more informally to say "the definition can be repaired or extended" in the case where the function is not properly defined at a some isolated point.
answered Jan 26 at 22:25
Mark BennetMark Bennet
81.7k984182
$endgroup$
add a comment |
$begingroup$
The function is undefined at $x=0$ so can't be continuous there.
The definition of the function can be extended so that the function is given a value at $x=0$ in a way which does make the extended function continuous there.
If the function were extended by giving a value at $x=0$ which does not make it continuous, we would say that it had a "removable discontinuity" at $x=0$. The discontinuity would be removed by giving it "the correct value" at an isolated point.
Sometimes this language is used more informally to say "the definition can be repaired or extended" in the case where the function is not properly defined at a some isolated point.
answered Jan 26 at 22:25
Mark BennetMark Bennet
81.7k984182
$endgroup$
The function is undefined at $x=0$ so can't be continuous there.
The definition of the function can be extended so that the function is given a value at $x=0$ in a way which does make the extended function continuous there.
If the function were extended by giving a value at $x=0$ which does not make it continuous, we would say that it had a "removable discontinuity" at $x=0$. The discontinuity would be removed by giving it "the correct value" at an isolated point.
Sometimes this language is used more informally to say "the definition can be repaired or extended" in the case where the function is not properly defined at a some isolated point.
answered Jan 26 at 22:25
Mark BennetMark Bennet
81.7k984182
answered Jan 26 at 22:25
Mark BennetMark Bennet
81.7k984182
answered Jan 26 at 22:25
Mark BennetMark Bennet
81.7k984182
answered Jan 26 at 22:25
Mark BennetMark Bennet
81.7k984182
81.7k984182
add a comment |
add a comment |
$begingroup$
The question is not well-posed. It only makes sense to consider continuity at points in the domain of the function. If we view the function as $fcolon mathbb{R}setminus {0}to mathbb{R}$, then the function is continuous everywhere. This function has an extension $hat{f}colon mathbb{R}to mathbb{R}$ which is continuous at zero. Simply define $hat{f}(0)=0$.
answered Jan 26 at 22:50
Foobaz JohnFoobaz John
22.8k41452
$endgroup$
add a comment |
$begingroup$
The question is not well-posed. It only makes sense to consider continuity at points in the domain of the function. If we view the function as $fcolon mathbb{R}setminus {0}to mathbb{R}$, then the function is continuous everywhere. This function has an extension $hat{f}colon mathbb{R}to mathbb{R}$ which is continuous at zero. Simply define $hat{f}(0)=0$.
answered Jan 26 at 22:50
Foobaz JohnFoobaz John
22.8k41452
$endgroup$
add a comment |
$begingroup$
The question is not well-posed. It only makes sense to consider continuity at points in the domain of the function. If we view the function as $fcolon mathbb{R}setminus {0}to mathbb{R}$, then the function is continuous everywhere. This function has an extension $hat{f}colon mathbb{R}to mathbb{R}$ which is continuous at zero. Simply define $hat{f}(0)=0$.
answered Jan 26 at 22:50
Foobaz JohnFoobaz John
22.8k41452
$endgroup$
The question is not well-posed. It only makes sense to consider continuity at points in the domain of the function. If we view the function as $fcolon mathbb{R}setminus {0}to mathbb{R}$, then the function is continuous everywhere. This function has an extension $hat{f}colon mathbb{R}to mathbb{R}$ which is continuous at zero. Simply define $hat{f}(0)=0$.
answered Jan 26 at 22:50
Foobaz JohnFoobaz John
22.8k41452
answered Jan 26 at 22:50
Foobaz JohnFoobaz John
22.8k41452
answered Jan 26 at 22:50
Foobaz JohnFoobaz John
22.8k41452
answered Jan 26 at 22:50
Foobaz JohnFoobaz John
22.8k41452
22.8k41452
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3088833%2fis-fracx2x-continuous%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
The answer is straightforward: it's not continuous at $0$ because $0$ is not inside the domain of $f$. That's the end of it. I remember that this confused me too for a while.
$endgroup$
– rubik
Jan 26 at 22:13
$begingroup$
@rubik ok, but isn't dividing by $x$ an equivalent transformation? And if so, does that imply that equivalent transformations can alter the domain?
$endgroup$
– MetaColon
Jan 26 at 22:17
$begingroup$
It is not because you cannot divide by $x = 0$.
$endgroup$
– Klaus
Jan 26 at 22:18
$begingroup$
@Klaus ok than I now also know where my mistake lay.
$endgroup$
– MetaColon
Jan 26 at 22:19