Why is this function only differentiable at zero?
$begingroup$
In my analysis class, we had the following function:
$$f: mathbbR to mathbbR: x mapsto begincases
x^2 & text if x in mathbbQ\ -x^2 & textelsewhere.endcases$$
At zero, I can see that this function is differentiable: indeed, we have that
$$lim_h to 0Big|fracf(h) -f (0)hBig| = lim_h to 0|h| = 0$$
by continuity of the absolute-value function and therefore, the limit of $h to 0$ of the quotients also equals zero.
My course notes then state that this function is nowhere differentiable except at zero. At this point, I only know about continuity and the definition of differentiability. Another example of my course notes used a sequence converging to zero, such that the quotient in the definition of derivative does not converge or has no limit. Therefore, I tried doing the same, but without result... Any hints on how to solve this?
Note that I do know that this function is not continuous (except at zero), but I have not seen the result (differentiable implies continuous) yet, so I'm sure the lecture notes do not intend to use this result...
calculus limits derivatives
edited Aug 28 '18 at 11:09
Nosrati
26.5k62354
asked Aug 28 '18 at 9:36
StudentStudent
2,1741727
$endgroup$
add a comment |
$begingroup$
In my analysis class, we had the following function:
$$f: mathbbR to mathbbR: x mapsto begincases
x^2 & text if x in mathbbQ\ -x^2 & textelsewhere.endcases$$
At zero, I can see that this function is differentiable: indeed, we have that
$$lim_h to 0Big|fracf(h) -f (0)hBig| = lim_h to 0|h| = 0$$
by continuity of the absolute-value function and therefore, the limit of $h to 0$ of the quotients also equals zero.
My course notes then state that this function is nowhere differentiable except at zero. At this point, I only know about continuity and the definition of differentiability. Another example of my course notes used a sequence converging to zero, such that the quotient in the definition of derivative does not converge or has no limit. Therefore, I tried doing the same, but without result... Any hints on how to solve this?
Note that I do know that this function is not continuous (except at zero), but I have not seen the result (differentiable implies continuous) yet, so I'm sure the lecture notes do not intend to use this result...
calculus limits derivatives
edited Aug 28 '18 at 11:09
Nosrati
26.5k62354
asked Aug 28 '18 at 9:36
StudentStudent
2,1741727
$endgroup$
$begingroup$
What happens if you take a rational sequence converging to $0 neq x in mathbbQ$? What if you take a non-rational sequence? How does this change when $x notin mathbbQ$?
$endgroup$
– Hetebrij
Aug 28 '18 at 9:40
$begingroup$
"I have not seen the result (differentiable implies continuous) yet, so I'm sure the lecture notes do not intend to use this result..." Sorry but this sentence is just absurd. You should know that differentiability is stronger than continuity and how to prove that it is.
$endgroup$
– Did
Aug 28 '18 at 10:01
2
$begingroup$
@Did: this result is only proven in the section after this example. I tried to indicate that I suspected the intended proof of this example does not use this result, nothing more, nothing less...
$endgroup$
– Student
Aug 28 '18 at 10:04
$begingroup$
By the way it is much easier to prove that the function is discontinuous at all points except $0$ and thus the method at the end of your question is the most natural one to handle this.
$endgroup$
– Paramanand Singh
Aug 29 '18 at 1:55
add a comment |
$begingroup$
In my analysis class, we had the following function:
$$f: mathbbR to mathbbR: x mapsto begincases
x^2 & text if x in mathbbQ\ -x^2 & textelsewhere.endcases$$
At zero, I can see that this function is differentiable: indeed, we have that
$$lim_h to 0Big|fracf(h) -f (0)hBig| = lim_h to 0|h| = 0$$
by continuity of the absolute-value function and therefore, the limit of $h to 0$ of the quotients also equals zero.
My course notes then state that this function is nowhere differentiable except at zero. At this point, I only know about continuity and the definition of differentiability. Another example of my course notes used a sequence converging to zero, such that the quotient in the definition of derivative does not converge or has no limit. Therefore, I tried doing the same, but without result... Any hints on how to solve this?
Note that I do know that this function is not continuous (except at zero), but I have not seen the result (differentiable implies continuous) yet, so I'm sure the lecture notes do not intend to use this result...
calculus limits derivatives
edited Aug 28 '18 at 11:09
Nosrati
26.5k62354
asked Aug 28 '18 at 9:36
StudentStudent
2,1741727
$endgroup$
In my analysis class, we had the following function:
$$f: mathbbR to mathbbR: x mapsto begincases
x^2 & text if x in mathbbQ\ -x^2 & textelsewhere.endcases$$
At zero, I can see that this function is differentiable: indeed, we have that
$$lim_h to 0Big|fracf(h) -f (0)hBig| = lim_h to 0|h| = 0$$
by continuity of the absolute-value function and therefore, the limit of $h to 0$ of the quotients also equals zero.
My course notes then state that this function is nowhere differentiable except at zero. At this point, I only know about continuity and the definition of differentiability. Another example of my course notes used a sequence converging to zero, such that the quotient in the definition of derivative does not converge or has no limit. Therefore, I tried doing the same, but without result... Any hints on how to solve this?
Note that I do know that this function is not continuous (except at zero), but I have not seen the result (differentiable implies continuous) yet, so I'm sure the lecture notes do not intend to use this result...
calculus limits derivatives
calculus limits derivatives
edited Aug 28 '18 at 11:09
Nosrati
26.5k62354
asked Aug 28 '18 at 9:36
StudentStudent
2,1741727
edited Aug 28 '18 at 11:09
Nosrati
26.5k62354
asked Aug 28 '18 at 9:36
StudentStudent
2,1741727
edited Aug 28 '18 at 11:09
Nosrati
26.5k62354
edited Aug 28 '18 at 11:09
Nosrati
26.5k62354
edited Aug 28 '18 at 11:09
Nosrati
26.5k62354
26.5k62354
asked Aug 28 '18 at 9:36
StudentStudent
2,1741727
asked Aug 28 '18 at 9:36
StudentStudent
2,1741727
asked Aug 28 '18 at 9:36
StudentStudent
2,1741727
2,1741727
$begingroup$
What happens if you take a rational sequence converging to $0 neq x in mathbbQ$? What if you take a non-rational sequence? How does this change when $x notin mathbbQ$?
$endgroup$
– Hetebrij
Aug 28 '18 at 9:40
$begingroup$
"I have not seen the result (differentiable implies continuous) yet, so I'm sure the lecture notes do not intend to use this result..." Sorry but this sentence is just absurd. You should know that differentiability is stronger than continuity and how to prove that it is.
$endgroup$
– Did
Aug 28 '18 at 10:01
2
$begingroup$
@Did: this result is only proven in the section after this example. I tried to indicate that I suspected the intended proof of this example does not use this result, nothing more, nothing less...
$endgroup$
– Student
Aug 28 '18 at 10:04
$begingroup$
By the way it is much easier to prove that the function is discontinuous at all points except $0$ and thus the method at the end of your question is the most natural one to handle this.
$endgroup$
– Paramanand Singh
Aug 29 '18 at 1:55
add a comment |
$begingroup$
What happens if you take a rational sequence converging to $0 neq x in mathbbQ$? What if you take a non-rational sequence? How does this change when $x notin mathbbQ$?
$endgroup$
– Hetebrij
Aug 28 '18 at 9:40
$begingroup$
"I have not seen the result (differentiable implies continuous) yet, so I'm sure the lecture notes do not intend to use this result..." Sorry but this sentence is just absurd. You should know that differentiability is stronger than continuity and how to prove that it is.
$endgroup$
– Did
Aug 28 '18 at 10:01
2
$begingroup$
@Did: this result is only proven in the section after this example. I tried to indicate that I suspected the intended proof of this example does not use this result, nothing more, nothing less...
$endgroup$
– Student
Aug 28 '18 at 10:04
$begingroup$
By the way it is much easier to prove that the function is discontinuous at all points except $0$ and thus the method at the end of your question is the most natural one to handle this.
$endgroup$
– Paramanand Singh
Aug 29 '18 at 1:55
$begingroup$
What happens if you take a rational sequence converging to $0 neq x in mathbbQ$? What if you take a non-rational sequence? How does this change when $x notin mathbbQ$?
$endgroup$
– Hetebrij
Aug 28 '18 at 9:40
$begingroup$
What happens if you take a rational sequence converging to $0 neq x in mathbbQ$? What if you take a non-rational sequence? How does this change when $x notin mathbbQ$?
$endgroup$
– Hetebrij
Aug 28 '18 at 9:40
$begingroup$
"I have not seen the result (differentiable implies continuous) yet, so I'm sure the lecture notes do not intend to use this result..." Sorry but this sentence is just absurd. You should know that differentiability is stronger than continuity and how to prove that it is.
$endgroup$
– Did
Aug 28 '18 at 10:01
$begingroup$
"I have not seen the result (differentiable implies continuous) yet, so I'm sure the lecture notes do not intend to use this result..." Sorry but this sentence is just absurd. You should know that differentiability is stronger than continuity and how to prove that it is.
$endgroup$
– Did
Aug 28 '18 at 10:01
2
2
$begingroup$
@Did: this result is only proven in the section after this example. I tried to indicate that I suspected the intended proof of this example does not use this result, nothing more, nothing less...
$endgroup$
– Student
Aug 28 '18 at 10:04
$begingroup$
@Did: this result is only proven in the section after this example. I tried to indicate that I suspected the intended proof of this example does not use this result, nothing more, nothing less...
$endgroup$
– Student
Aug 28 '18 at 10:04
$begingroup$
By the way it is much easier to prove that the function is discontinuous at all points except $0$ and thus the method at the end of your question is the most natural one to handle this.
$endgroup$
– Paramanand Singh
Aug 29 '18 at 1:55
$begingroup$
By the way it is much easier to prove that the function is discontinuous at all points except $0$ and thus the method at the end of your question is the most natural one to handle this.
$endgroup$
– Paramanand Singh
Aug 29 '18 at 1:55
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
If you compute the limit
$$lim_xto x_0fracf(x)-f(x_0)x-x_0$$ using a subsequence of rationals and a subsequence of irrationals, you will converge to $2x_0$ and $-2x_0$ respectively, which are different numbers. This shows that the limit does not exist when $x_0ne0.$
Erratum:
As commented by José, the above reasoning is not right. If $x_0$ is rational, then the limit via the irrationals does not exist, and similarly if $x_0$ is rational the limit via the irrationals does not exist.
So in both cases, the limit does not exist.
answered Aug 28 '18 at 9:53
Yves DaoustYves Daoust
131k676229
$endgroup$
1
$begingroup$
This is not correct. If $x_0$ is irrational and if $(x_n)_ninmathbbN$ is a sequence of rational numbers such that $lim_ntoinftyx_n=x_0$, then the limit $lim_ntoinftyfracf(x_n)-f(x_0)x_n-x_0$ is not $pm2x_0$. It simply doesn't exist (in $mathbb R$).
$endgroup$
– José Carlos Santos
Aug 28 '18 at 10:11
$begingroup$
@JoséCarlosSantos: you are right, fixing.
$endgroup$
– Yves Daoust
Aug 28 '18 at 10:15
2
$begingroup$
Would it be possible to rewrite the answer with the right reasoning? The erroneous argument can be confusing to some, so I would prefer to see it removed rather than commented.
$endgroup$
– Joonas Ilmavirta
Aug 28 '18 at 13:29
$begingroup$
Why do the limits not exist? At a glance it looks like reasonable logic to me...
$endgroup$
– Chris
Aug 28 '18 at 13:32
1
$begingroup$
@chris: so did it to me at first glance. But if $x_0$ is "of the other kind", $f(x_0)$ is "of the other sign".
$endgroup$
– Yves Daoust
Aug 28 '18 at 13:59
add a comment |
$begingroup$
Consider the point $1$, for instance. In order to prove that $f$ is not differentiable there, you have to prove that the limit $lim_xto1fracf(x)-1x-1$ doesn't exist. Note that there are numbers arbitrarily close to $1$ whose square is irrational. For such a point $x$, $f(x)$ is close to $-1$, and therefore $leftlvertfracf(x)-1x-1rightrvert$ must be a very large number. It is not hard to formalize this in order to prove that the limit $lim_xto1fracf(x)-1x-1$ doesn't exist indeed.
The argument is similar at the other points of $mathbbRsetminus0$.
answered Aug 28 '18 at 9:47
José Carlos SantosJosé Carlos Santos
170k23132238
$endgroup$
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%2f2897070%2fwhy-is-this-function-only-differentiable-at-zero%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If you compute the limit
$$lim_xto x_0fracf(x)-f(x_0)x-x_0$$ using a subsequence of rationals and a subsequence of irrationals, you will converge to $2x_0$ and $-2x_0$ respectively, which are different numbers. This shows that the limit does not exist when $x_0ne0.$
Erratum:
As commented by José, the above reasoning is not right. If $x_0$ is rational, then the limit via the irrationals does not exist, and similarly if $x_0$ is rational the limit via the irrationals does not exist.
So in both cases, the limit does not exist.
answered Aug 28 '18 at 9:53
Yves DaoustYves Daoust
131k676229
$endgroup$
1
$begingroup$
This is not correct. If $x_0$ is irrational and if $(x_n)_ninmathbbN$ is a sequence of rational numbers such that $lim_ntoinftyx_n=x_0$, then the limit $lim_ntoinftyfracf(x_n)-f(x_0)x_n-x_0$ is not $pm2x_0$. It simply doesn't exist (in $mathbb R$).
$endgroup$
– José Carlos Santos
Aug 28 '18 at 10:11
$begingroup$
@JoséCarlosSantos: you are right, fixing.
$endgroup$
– Yves Daoust
Aug 28 '18 at 10:15
2
$begingroup$
Would it be possible to rewrite the answer with the right reasoning? The erroneous argument can be confusing to some, so I would prefer to see it removed rather than commented.
$endgroup$
– Joonas Ilmavirta
Aug 28 '18 at 13:29
$begingroup$
Why do the limits not exist? At a glance it looks like reasonable logic to me...
$endgroup$
– Chris
Aug 28 '18 at 13:32
1
$begingroup$
@chris: so did it to me at first glance. But if $x_0$ is "of the other kind", $f(x_0)$ is "of the other sign".
$endgroup$
– Yves Daoust
Aug 28 '18 at 13:59
add a comment |
$begingroup$
If you compute the limit
$$lim_xto x_0fracf(x)-f(x_0)x-x_0$$ using a subsequence of rationals and a subsequence of irrationals, you will converge to $2x_0$ and $-2x_0$ respectively, which are different numbers. This shows that the limit does not exist when $x_0ne0.$
Erratum:
As commented by José, the above reasoning is not right. If $x_0$ is rational, then the limit via the irrationals does not exist, and similarly if $x_0$ is rational the limit via the irrationals does not exist.
So in both cases, the limit does not exist.
answered Aug 28 '18 at 9:53
Yves DaoustYves Daoust
131k676229
$endgroup$
1
$begingroup$
This is not correct. If $x_0$ is irrational and if $(x_n)_ninmathbbN$ is a sequence of rational numbers such that $lim_ntoinftyx_n=x_0$, then the limit $lim_ntoinftyfracf(x_n)-f(x_0)x_n-x_0$ is not $pm2x_0$. It simply doesn't exist (in $mathbb R$).
$endgroup$
– José Carlos Santos
Aug 28 '18 at 10:11
$begingroup$
@JoséCarlosSantos: you are right, fixing.
$endgroup$
– Yves Daoust
Aug 28 '18 at 10:15
2
$begingroup$
Would it be possible to rewrite the answer with the right reasoning? The erroneous argument can be confusing to some, so I would prefer to see it removed rather than commented.
$endgroup$
– Joonas Ilmavirta
Aug 28 '18 at 13:29
$begingroup$
Why do the limits not exist? At a glance it looks like reasonable logic to me...
$endgroup$
– Chris
Aug 28 '18 at 13:32
1
$begingroup$
@chris: so did it to me at first glance. But if $x_0$ is "of the other kind", $f(x_0)$ is "of the other sign".
$endgroup$
– Yves Daoust
Aug 28 '18 at 13:59
add a comment |
$begingroup$
If you compute the limit
$$lim_xto x_0fracf(x)-f(x_0)x-x_0$$ using a subsequence of rationals and a subsequence of irrationals, you will converge to $2x_0$ and $-2x_0$ respectively, which are different numbers. This shows that the limit does not exist when $x_0ne0.$
Erratum:
As commented by José, the above reasoning is not right. If $x_0$ is rational, then the limit via the irrationals does not exist, and similarly if $x_0$ is rational the limit via the irrationals does not exist.
So in both cases, the limit does not exist.
answered Aug 28 '18 at 9:53
Yves DaoustYves Daoust
131k676229
$endgroup$
If you compute the limit
$$lim_xto x_0fracf(x)-f(x_0)x-x_0$$ using a subsequence of rationals and a subsequence of irrationals, you will converge to $2x_0$ and $-2x_0$ respectively, which are different numbers. This shows that the limit does not exist when $x_0ne0.$
Erratum:
As commented by José, the above reasoning is not right. If $x_0$ is rational, then the limit via the irrationals does not exist, and similarly if $x_0$ is rational the limit via the irrationals does not exist.
So in both cases, the limit does not exist.
answered Aug 28 '18 at 9:53
Yves DaoustYves Daoust
131k676229
edited Aug 28 '18 at 10:18
answered Aug 28 '18 at 9:53
Yves DaoustYves Daoust
131k676229
answered Aug 28 '18 at 9:53
Yves DaoustYves Daoust
131k676229
answered Aug 28 '18 at 9:53
Yves DaoustYves Daoust
131k676229
131k676229
1
$begingroup$
This is not correct. If $x_0$ is irrational and if $(x_n)_ninmathbbN$ is a sequence of rational numbers such that $lim_ntoinftyx_n=x_0$, then the limit $lim_ntoinftyfracf(x_n)-f(x_0)x_n-x_0$ is not $pm2x_0$. It simply doesn't exist (in $mathbb R$).
$endgroup$
– José Carlos Santos
Aug 28 '18 at 10:11
$begingroup$
@JoséCarlosSantos: you are right, fixing.
$endgroup$
– Yves Daoust
Aug 28 '18 at 10:15
2
$begingroup$
Would it be possible to rewrite the answer with the right reasoning? The erroneous argument can be confusing to some, so I would prefer to see it removed rather than commented.
$endgroup$
– Joonas Ilmavirta
Aug 28 '18 at 13:29
$begingroup$
Why do the limits not exist? At a glance it looks like reasonable logic to me...
$endgroup$
– Chris
Aug 28 '18 at 13:32
1
$begingroup$
@chris: so did it to me at first glance. But if $x_0$ is "of the other kind", $f(x_0)$ is "of the other sign".
$endgroup$
– Yves Daoust
Aug 28 '18 at 13:59
add a comment |
1
$begingroup$
This is not correct. If $x_0$ is irrational and if $(x_n)_ninmathbbN$ is a sequence of rational numbers such that $lim_ntoinftyx_n=x_0$, then the limit $lim_ntoinftyfracf(x_n)-f(x_0)x_n-x_0$ is not $pm2x_0$. It simply doesn't exist (in $mathbb R$).
$endgroup$
– José Carlos Santos
Aug 28 '18 at 10:11
$begingroup$
@JoséCarlosSantos: you are right, fixing.
$endgroup$
– Yves Daoust
Aug 28 '18 at 10:15
2
$begingroup$
Would it be possible to rewrite the answer with the right reasoning? The erroneous argument can be confusing to some, so I would prefer to see it removed rather than commented.
$endgroup$
– Joonas Ilmavirta
Aug 28 '18 at 13:29
$begingroup$
Why do the limits not exist? At a glance it looks like reasonable logic to me...
$endgroup$
– Chris
Aug 28 '18 at 13:32
1
$begingroup$
@chris: so did it to me at first glance. But if $x_0$ is "of the other kind", $f(x_0)$ is "of the other sign".
$endgroup$
– Yves Daoust
Aug 28 '18 at 13:59
1
1
$begingroup$
This is not correct. If $x_0$ is irrational and if $(x_n)_ninmathbbN$ is a sequence of rational numbers such that $lim_ntoinftyx_n=x_0$, then the limit $lim_ntoinftyfracf(x_n)-f(x_0)x_n-x_0$ is not $pm2x_0$. It simply doesn't exist (in $mathbb R$).
$endgroup$
– José Carlos Santos
Aug 28 '18 at 10:11
$begingroup$
This is not correct. If $x_0$ is irrational and if $(x_n)_ninmathbbN$ is a sequence of rational numbers such that $lim_ntoinftyx_n=x_0$, then the limit $lim_ntoinftyfracf(x_n)-f(x_0)x_n-x_0$ is not $pm2x_0$. It simply doesn't exist (in $mathbb R$).
$endgroup$
– José Carlos Santos
Aug 28 '18 at 10:11
$begingroup$
@JoséCarlosSantos: you are right, fixing.
$endgroup$
– Yves Daoust
Aug 28 '18 at 10:15
$begingroup$
@JoséCarlosSantos: you are right, fixing.
$endgroup$
– Yves Daoust
Aug 28 '18 at 10:15
2
2
$begingroup$
Would it be possible to rewrite the answer with the right reasoning? The erroneous argument can be confusing to some, so I would prefer to see it removed rather than commented.
$endgroup$
– Joonas Ilmavirta
Aug 28 '18 at 13:29
$begingroup$
Would it be possible to rewrite the answer with the right reasoning? The erroneous argument can be confusing to some, so I would prefer to see it removed rather than commented.
$endgroup$
– Joonas Ilmavirta
Aug 28 '18 at 13:29
$begingroup$
Why do the limits not exist? At a glance it looks like reasonable logic to me...
$endgroup$
– Chris
Aug 28 '18 at 13:32
$begingroup$
Why do the limits not exist? At a glance it looks like reasonable logic to me...
$endgroup$
– Chris
Aug 28 '18 at 13:32
1
1
$begingroup$
@chris: so did it to me at first glance. But if $x_0$ is "of the other kind", $f(x_0)$ is "of the other sign".
$endgroup$
– Yves Daoust
Aug 28 '18 at 13:59
$begingroup$
@chris: so did it to me at first glance. But if $x_0$ is "of the other kind", $f(x_0)$ is "of the other sign".
$endgroup$
– Yves Daoust
Aug 28 '18 at 13:59
add a comment |
$begingroup$
Consider the point $1$, for instance. In order to prove that $f$ is not differentiable there, you have to prove that the limit $lim_xto1fracf(x)-1x-1$ doesn't exist. Note that there are numbers arbitrarily close to $1$ whose square is irrational. For such a point $x$, $f(x)$ is close to $-1$, and therefore $leftlvertfracf(x)-1x-1rightrvert$ must be a very large number. It is not hard to formalize this in order to prove that the limit $lim_xto1fracf(x)-1x-1$ doesn't exist indeed.
The argument is similar at the other points of $mathbbRsetminus0$.
answered Aug 28 '18 at 9:47
José Carlos SantosJosé Carlos Santos
170k23132238
$endgroup$
add a comment |
$begingroup$
Consider the point $1$, for instance. In order to prove that $f$ is not differentiable there, you have to prove that the limit $lim_xto1fracf(x)-1x-1$ doesn't exist. Note that there are numbers arbitrarily close to $1$ whose square is irrational. For such a point $x$, $f(x)$ is close to $-1$, and therefore $leftlvertfracf(x)-1x-1rightrvert$ must be a very large number. It is not hard to formalize this in order to prove that the limit $lim_xto1fracf(x)-1x-1$ doesn't exist indeed.
The argument is similar at the other points of $mathbbRsetminus0$.
answered Aug 28 '18 at 9:47
José Carlos SantosJosé Carlos Santos
170k23132238
$endgroup$
add a comment |
$begingroup$
Consider the point $1$, for instance. In order to prove that $f$ is not differentiable there, you have to prove that the limit $lim_xto1fracf(x)-1x-1$ doesn't exist. Note that there are numbers arbitrarily close to $1$ whose square is irrational. For such a point $x$, $f(x)$ is close to $-1$, and therefore $leftlvertfracf(x)-1x-1rightrvert$ must be a very large number. It is not hard to formalize this in order to prove that the limit $lim_xto1fracf(x)-1x-1$ doesn't exist indeed.
The argument is similar at the other points of $mathbbRsetminus0$.
answered Aug 28 '18 at 9:47
José Carlos SantosJosé Carlos Santos
170k23132238
$endgroup$
Consider the point $1$, for instance. In order to prove that $f$ is not differentiable there, you have to prove that the limit $lim_xto1fracf(x)-1x-1$ doesn't exist. Note that there are numbers arbitrarily close to $1$ whose square is irrational. For such a point $x$, $f(x)$ is close to $-1$, and therefore $leftlvertfracf(x)-1x-1rightrvert$ must be a very large number. It is not hard to formalize this in order to prove that the limit $lim_xto1fracf(x)-1x-1$ doesn't exist indeed.
The argument is similar at the other points of $mathbbRsetminus0$.
answered Aug 28 '18 at 9:47
José Carlos SantosJosé Carlos Santos
170k23132238
answered Aug 28 '18 at 9:47
José Carlos SantosJosé Carlos Santos
170k23132238
answered Aug 28 '18 at 9:47
José Carlos SantosJosé Carlos Santos
170k23132238
answered Aug 28 '18 at 9:47
José Carlos SantosJosé Carlos Santos
170k23132238
170k23132238
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%2f2897070%2fwhy-is-this-function-only-differentiable-at-zero%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
$begingroup$
What happens if you take a rational sequence converging to $0 neq x in mathbbQ$? What if you take a non-rational sequence? How does this change when $x notin mathbbQ$?
$endgroup$
– Hetebrij
Aug 28 '18 at 9:40
$begingroup$
"I have not seen the result (differentiable implies continuous) yet, so I'm sure the lecture notes do not intend to use this result..." Sorry but this sentence is just absurd. You should know that differentiability is stronger than continuity and how to prove that it is.
$endgroup$
– Did
Aug 28 '18 at 10:01
2
$begingroup$
@Did: this result is only proven in the section after this example. I tried to indicate that I suspected the intended proof of this example does not use this result, nothing more, nothing less...
$endgroup$
– Student
Aug 28 '18 at 10:04
$begingroup$
By the way it is much easier to prove that the function is discontinuous at all points except $0$ and thus the method at the end of your question is the most natural one to handle this.
$endgroup$
– Paramanand Singh
Aug 29 '18 at 1:55