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Why is this function only differentiable at zero?
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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 at 11:09
Nosrati
21.7k41746
asked Aug 28 at 9:36
Student
2,0571626
add a comment |Â
up vote
3
down vote
favorite
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 at 11:09
Nosrati
21.7k41746
asked Aug 28 at 9:36
Student
2,0571626
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 $x notin mathbbQ$?
– Hetebrij
Aug 28 at 9:40
"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.
– Did
Aug 28 at 10:01
2
@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...
– Student
Aug 28 at 10:04
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.
– Paramanand Singh
Aug 29 at 1:55
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
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 at 11:09
Nosrati
21.7k41746
asked Aug 28 at 9:36
Student
2,0571626
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 at 11:09
Nosrati
21.7k41746
asked Aug 28 at 9:36
Student
2,0571626
edited Aug 28 at 11:09
Nosrati
21.7k41746
edited Aug 28 at 11:09
Nosrati
21.7k41746
edited Aug 28 at 11:09
Nosrati
21.7k41746
21.7k41746
asked Aug 28 at 9:36
Student
2,0571626
asked Aug 28 at 9:36
Student
2,0571626
asked Aug 28 at 9:36
Student
2,0571626
2,0571626
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 $x notin mathbbQ$?
– Hetebrij
Aug 28 at 9:40
"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.
– Did
Aug 28 at 10:01
2
@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...
– Student
Aug 28 at 10:04
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.
– Paramanand Singh
Aug 29 at 1:55
add a comment |Â
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 $x notin mathbbQ$?
– Hetebrij
Aug 28 at 9:40
"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.
– Did
Aug 28 at 10:01
2
@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...
– Student
Aug 28 at 10:04
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.
– Paramanand Singh
Aug 29 at 1:55
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 $x notin mathbbQ$?
– Hetebrij
Aug 28 at 9:40
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 $x notin mathbbQ$?
– Hetebrij
Aug 28 at 9:40
"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.
– Did
Aug 28 at 10:01
"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.
– Did
Aug 28 at 10:01
2
2
@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...
– Student
Aug 28 at 10:04
@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...
– Student
Aug 28 at 10:04
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.
– Paramanand Singh
Aug 29 at 1:55
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.
– Paramanand Singh
Aug 29 at 1:55
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
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 at 9:53
Yves Daoust
113k665208
1
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 $mathbb R$).
– José Carlos Santos
Aug 28 at 10:11
@JoséCarlosSantos: you are right, fixing.
– Yves Daoust
Aug 28 at 10:15
2
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.
– Joonas Ilmavirta
Aug 28 at 13:29
Why do the limits not exist? At a glance it looks like reasonable logic to me...
– Chris
Aug 28 at 13:32
1
@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".
– Yves Daoust
Aug 28 at 13:59
add a comment |Â
up vote
4
down vote
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 at 9:47
José Carlos Santos
120k16101182
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
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 at 9:53
Yves Daoust
113k665208
1
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 $mathbb R$).
– José Carlos Santos
Aug 28 at 10:11
@JoséCarlosSantos: you are right, fixing.
– Yves Daoust
Aug 28 at 10:15
2
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.
– Joonas Ilmavirta
Aug 28 at 13:29
Why do the limits not exist? At a glance it looks like reasonable logic to me...
– Chris
Aug 28 at 13:32
1
@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".
– Yves Daoust
Aug 28 at 13:59
add a comment |Â
up vote
2
down vote
accepted
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 at 9:53
Yves Daoust
113k665208
1
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 $mathbb R$).
– José Carlos Santos
Aug 28 at 10:11
@JoséCarlosSantos: you are right, fixing.
– Yves Daoust
Aug 28 at 10:15
2
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.
– Joonas Ilmavirta
Aug 28 at 13:29
Why do the limits not exist? At a glance it looks like reasonable logic to me...
– Chris
Aug 28 at 13:32
1
@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".
– Yves Daoust
Aug 28 at 13:59
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
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 at 9:53
Yves Daoust
113k665208
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 at 9:53
Yves Daoust
113k665208
edited Aug 28 at 10:18
answered Aug 28 at 9:53
Yves Daoust
113k665208
answered Aug 28 at 9:53
Yves Daoust
113k665208
answered Aug 28 at 9:53
Yves Daoust
113k665208
113k665208
1
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 $mathbb R$).
– José Carlos Santos
Aug 28 at 10:11
@JoséCarlosSantos: you are right, fixing.
– Yves Daoust
Aug 28 at 10:15
2
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.
– Joonas Ilmavirta
Aug 28 at 13:29
Why do the limits not exist? At a glance it looks like reasonable logic to me...
– Chris
Aug 28 at 13:32
1
@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".
– Yves Daoust
Aug 28 at 13:59
add a comment |Â
1
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 $mathbb R$).
– José Carlos Santos
Aug 28 at 10:11
@JoséCarlosSantos: you are right, fixing.
– Yves Daoust
Aug 28 at 10:15
2
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.
– Joonas Ilmavirta
Aug 28 at 13:29
Why do the limits not exist? At a glance it looks like reasonable logic to me...
– Chris
Aug 28 at 13:32
1
@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".
– Yves Daoust
Aug 28 at 13:59
1
1
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 $mathbb R$).
– José Carlos Santos
Aug 28 at 10:11
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 $mathbb R$).
– José Carlos Santos
Aug 28 at 10:11
@JoséCarlosSantos: you are right, fixing.
– Yves Daoust
Aug 28 at 10:15
@JoséCarlosSantos: you are right, fixing.
– Yves Daoust
Aug 28 at 10:15
2
2
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.
– Joonas Ilmavirta
Aug 28 at 13:29
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.
– Joonas Ilmavirta
Aug 28 at 13:29
Why do the limits not exist? At a glance it looks like reasonable logic to me...
– Chris
Aug 28 at 13:32
Why do the limits not exist? At a glance it looks like reasonable logic to me...
– Chris
Aug 28 at 13:32
1
1
@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".
– Yves Daoust
Aug 28 at 13:59
@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".
– Yves Daoust
Aug 28 at 13:59
add a comment |Â
up vote
4
down vote
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 at 9:47
José Carlos Santos
120k16101182
add a comment |Â
up vote
4
down vote
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 at 9:47
José Carlos Santos
120k16101182
add a comment |Â
up vote
4
down vote
up vote
4
down vote
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 at 9:47
José Carlos Santos
120k16101182
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 at 9:47
José Carlos Santos
120k16101182
answered Aug 28 at 9:47
José Carlos Santos
120k16101182
answered Aug 28 at 9:47
José Carlos Santos
120k16101182
answered Aug 28 at 9:47
José Carlos Santos
120k16101182
120k16101182
add a comment |Â
add a comment |Â
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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 $x notin mathbbQ$?
– Hetebrij
Aug 28 at 9:40
"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.
– Did
Aug 28 at 10:01
2
@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...
– Student
Aug 28 at 10:04
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.
– Paramanand Singh
Aug 29 at 1:55