Flaw in proof that a differentiable function has continuous derivative
Clash Royale CLAN TAG#URR8PPP
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Let f be a function differentiable on $(a,b)$ and continuous on $cin(a,b)$. If $c+h in (a,b)$ then by the mean value theorem $fracf(c+h)-f(c)h=f'(c+theta h)$ for $theta in [0,1]$. Let $h xrightarrow0$, then $f'(c+theta h) xrightarrow f'(c)$ by the above.
My reasoning is the following: $theta$ is a function of $h$, so in fact it is not true that that $theta(h_n)h_n$ represents any arbitrary sequence that tends to 0, so using this definition of limits(the sequence definition), what we have does not follow. EDIT: I am aware of the explicit counterexample $x^2sin(1/x)$ but that doesn't capture the full meat of the question.
real-analysis limits analysis derivatives convergence
asked Mar 4 at 17:42
user3184807user3184807
357110
$endgroup$
add a comment |
$begingroup$
Let f be a function differentiable on $(a,b)$ and continuous on $cin(a,b)$. If $c+h in (a,b)$ then by the mean value theorem $fracf(c+h)-f(c)h=f'(c+theta h)$ for $theta in [0,1]$. Let $h xrightarrow0$, then $f'(c+theta h) xrightarrow f'(c)$ by the above.
My reasoning is the following: $theta$ is a function of $h$, so in fact it is not true that that $theta(h_n)h_n$ represents any arbitrary sequence that tends to 0, so using this definition of limits(the sequence definition), what we have does not follow. EDIT: I am aware of the explicit counterexample $x^2sin(1/x)$ but that doesn't capture the full meat of the question.
real-analysis limits analysis derivatives convergence
asked Mar 4 at 17:42
user3184807user3184807
357110
$endgroup$
$begingroup$
Well under the conditions of your question, it is guaranteed (via mean value theorem) that there is a sequence $c_n$ of values tending to $c$ such that $f'(c_n) to f'(c) $. What is **not** guaranteed is that the same result holds for any/every sequence $c_nto c$ and thus continuity of $f'$ at $c$ can't be inferred.
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– Paramanand Singh
Mar 6 at 5:02
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That is essentially what my reasoning said
$endgroup$
– user3184807
Mar 6 at 19:07
add a comment |
$begingroup$
Let f be a function differentiable on $(a,b)$ and continuous on $cin(a,b)$. If $c+h in (a,b)$ then by the mean value theorem $fracf(c+h)-f(c)h=f'(c+theta h)$ for $theta in [0,1]$. Let $h xrightarrow0$, then $f'(c+theta h) xrightarrow f'(c)$ by the above.
My reasoning is the following: $theta$ is a function of $h$, so in fact it is not true that that $theta(h_n)h_n$ represents any arbitrary sequence that tends to 0, so using this definition of limits(the sequence definition), what we have does not follow. EDIT: I am aware of the explicit counterexample $x^2sin(1/x)$ but that doesn't capture the full meat of the question.
real-analysis limits analysis derivatives convergence
asked Mar 4 at 17:42
user3184807user3184807
357110
$endgroup$
Let f be a function differentiable on $(a,b)$ and continuous on $cin(a,b)$. If $c+h in (a,b)$ then by the mean value theorem $fracf(c+h)-f(c)h=f'(c+theta h)$ for $theta in [0,1]$. Let $h xrightarrow0$, then $f'(c+theta h) xrightarrow f'(c)$ by the above.
My reasoning is the following: $theta$ is a function of $h$, so in fact it is not true that that $theta(h_n)h_n$ represents any arbitrary sequence that tends to 0, so using this definition of limits(the sequence definition), what we have does not follow. EDIT: I am aware of the explicit counterexample $x^2sin(1/x)$ but that doesn't capture the full meat of the question.
real-analysis limits analysis derivatives convergence
real-analysis limits analysis derivatives convergence
asked Mar 4 at 17:42
user3184807user3184807
357110
asked Mar 4 at 17:42
user3184807user3184807
357110
edited Mar 4 at 18:17
user3184807
asked Mar 4 at 17:42
user3184807user3184807
357110
asked Mar 4 at 17:42
user3184807user3184807
357110
asked Mar 4 at 17:42
user3184807user3184807
357110
357110
$begingroup$
Well under the conditions of your question, it is guaranteed (via mean value theorem) that there is a sequence $c_n$ of values tending to $c$ such that $f'(c_n) to f'(c) $. What is **not** guaranteed is that the same result holds for any/every sequence $c_nto c$ and thus continuity of $f'$ at $c$ can't be inferred.
$endgroup$
– Paramanand Singh
Mar 6 at 5:02
$begingroup$
That is essentially what my reasoning said
$endgroup$
– user3184807
Mar 6 at 19:07
add a comment |
$begingroup$
Well under the conditions of your question, it is guaranteed (via mean value theorem) that there is a sequence $c_n$ of values tending to $c$ such that $f'(c_n) to f'(c) $. What is **not** guaranteed is that the same result holds for any/every sequence $c_nto c$ and thus continuity of $f'$ at $c$ can't be inferred.
$endgroup$
– Paramanand Singh
Mar 6 at 5:02
$begingroup$
That is essentially what my reasoning said
$endgroup$
– user3184807
Mar 6 at 19:07
$begingroup$
Well under the conditions of your question, it is guaranteed (via mean value theorem) that there is a sequence $c_n$ of values tending to $c$ such that $f'(c_n) to f'(c) $. What is **not** guaranteed is that the same result holds for any/every sequence $c_nto c$ and thus continuity of $f'$ at $c$ can't be inferred.
$endgroup$
– Paramanand Singh
Mar 6 at 5:02
$begingroup$
Well under the conditions of your question, it is guaranteed (via mean value theorem) that there is a sequence $c_n$ of values tending to $c$ such that $f'(c_n) to f'(c) $. What is **not** guaranteed is that the same result holds for any/every sequence $c_nto c$ and thus continuity of $f'$ at $c$ can't be inferred.
$endgroup$
– Paramanand Singh
Mar 6 at 5:02
$begingroup$
That is essentially what my reasoning said
$endgroup$
– user3184807
Mar 6 at 19:07
$begingroup$
That is essentially what my reasoning said
$endgroup$
– user3184807
Mar 6 at 19:07
add a comment |
3 Answers
3
active
oldest
votes
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It is not true that $theta$ need be a function of $h,$ because it may not be uniquely specified. There could be many $theta$'s that work for a given $h.$
We only know $f'(y)to f'(c)$ as $yto c$ within the set of $y=x+theta h$ that arise in the MVT process you described. But the set of such $y$ may not equal any deleted neighborhood of $c,$ as the example $x^2sin(1/x)$ shows.
answered Mar 4 at 18:23
zhw.zhw.
74.7k43275
$endgroup$
$begingroup$
So we can make $theta$ a function by choosing a value of $h$ and then say the union of all the sequences $theta(h_n)h_n$ where $h_n xrightarrow0$ over all functions $theta$ need not be all sequences tending to 0?
$endgroup$
– user3184807
Mar 4 at 18:27
1
$begingroup$
That's a lot of choosing, and the axiom of choice may even come in here. But yes, the sequences we get through the MVT may not be all necessary sequences. In the case of $x^2sin(1/x)$ the sequence $1/(2npi)to 0,$ but this sequence does not arise as $theta(h_n)h_n$ in the MVT process.
$endgroup$
– zhw.
Mar 4 at 18:35
add a comment |
$begingroup$
If you look at $f(x)=x^2 sin(1/x)$ on $(0,1]$ and $f(0)=0$, you have $f'(x)=2xsin(1/x)-cos(1/x)$ on $(0,1]$ and $f'(0)=0$. Thus MVT tells you that for every $h in (0,1)$ there exists $theta(h) in (0,1)$ with
$$hsin(1/h)=2htheta(h)sin(1/(htheta(h)))-cos(1/(htheta(h))).$$
Notice that once $h$ is small enough, $htheta(h)$ is forced to stay relatively close to the zeros of $cos(1/x)$, since
$$|cos(1/(htheta(h)))|=|hsin(1/h)-2htheta(h)sin(1/(htheta(h)))|<3h.$$
This means that the MVT is only providing you with information about $f'(x)$ approaching $f'(0)$ along sequences which stay sufficiently close to the zeros of $cos(1/x)$. If you instead look at something like $f' left ( frac1pi n right )$ you find that it doesn't go to zero. There is no contradiction because the MVT is never giving you information about those points.
answered Mar 4 at 18:18
IanIan
68.9k25392
$endgroup$
$begingroup$
So does this essentially come down to the idea that $theta(h_n)h_n$ cannot always be made to be any arbitrary sequence tending to 0?
$endgroup$
– user3184807
Mar 4 at 18:21
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@user3184807 Yes, that is essentially what happens.
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– Ian
Mar 4 at 18:22
add a comment |
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You are assuming what you want to prove: that $f’$ is continuous at $c$ when you say that $f'(c+theta h) to f'(c)$ as $hto0$.
answered Mar 4 at 17:53
Julián AguirreJulián Aguirre
69.5k24297
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1
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No, because OP defined $theta(h)$ through the MVT in the first place. The point is that you know that $lim_h to 0 f'(c+htheta(h))=f'(c)$, but why doesn't this imply $lim_h to 0 f'(c+h)=f'(c)$? Presumably this is because of some very bad property of $theta(h)$.
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– Ian
Mar 4 at 17:54
1
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That has nothing to do with muy answer.
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– Julián Aguirre
Mar 4 at 17:56
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Let me spell it out, then. By definition $lim_h to 0 fracf(c+h)-f(c)h=f'(c)$, and by MVT $fracf(c+h)-f(c)h=f'(c+htheta(h))$ for some $theta(h) in (0,1)$. That's all OP assumed. Then they send $h to 0$ in the MVT relation and find $lim_h to 0 f'(c+h theta(h))=f'(c)$. Now they're wondering why they can't conclude $lim_h to 0 f'(c+h)=f'(c)$.
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– Ian
Mar 4 at 17:57
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@Ian: The problem is that the limit may not exist. If it exists, then it must equal $f'(c)$ (as has been covered many times on this site already, for example here: math.stackexchange.com/questions/257907/…).
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– Hans Lundmark
Mar 4 at 18:04
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@HansLundmark OP knows that already as well, the question is about intuition. The answer, for the case of something like $f(x)=x^2sin(1/x)$, is that $htheta(h)$ has numerous discontinuities in a vicinity of $h=0$, so you don't get information about entire neighborhoods of $h=0$ from the MVT.
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– Ian
Mar 4 at 18:08
|
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3 Answers
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active
oldest
votes
3 Answers
3
active
oldest
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active
oldest
votes
$begingroup$
It is not true that $theta$ need be a function of $h,$ because it may not be uniquely specified. There could be many $theta$'s that work for a given $h.$
We only know $f'(y)to f'(c)$ as $yto c$ within the set of $y=x+theta h$ that arise in the MVT process you described. But the set of such $y$ may not equal any deleted neighborhood of $c,$ as the example $x^2sin(1/x)$ shows.
answered Mar 4 at 18:23
zhw.zhw.
74.7k43275
$endgroup$
$begingroup$
So we can make $theta$ a function by choosing a value of $h$ and then say the union of all the sequences $theta(h_n)h_n$ where $h_n xrightarrow0$ over all functions $theta$ need not be all sequences tending to 0?
$endgroup$
– user3184807
Mar 4 at 18:27
1
$begingroup$
That's a lot of choosing, and the axiom of choice may even come in here. But yes, the sequences we get through the MVT may not be all necessary sequences. In the case of $x^2sin(1/x)$ the sequence $1/(2npi)to 0,$ but this sequence does not arise as $theta(h_n)h_n$ in the MVT process.
$endgroup$
– zhw.
Mar 4 at 18:35
add a comment |
$begingroup$
It is not true that $theta$ need be a function of $h,$ because it may not be uniquely specified. There could be many $theta$'s that work for a given $h.$
We only know $f'(y)to f'(c)$ as $yto c$ within the set of $y=x+theta h$ that arise in the MVT process you described. But the set of such $y$ may not equal any deleted neighborhood of $c,$ as the example $x^2sin(1/x)$ shows.
answered Mar 4 at 18:23
zhw.zhw.
74.7k43275
$endgroup$
$begingroup$
So we can make $theta$ a function by choosing a value of $h$ and then say the union of all the sequences $theta(h_n)h_n$ where $h_n xrightarrow0$ over all functions $theta$ need not be all sequences tending to 0?
$endgroup$
– user3184807
Mar 4 at 18:27
1
$begingroup$
That's a lot of choosing, and the axiom of choice may even come in here. But yes, the sequences we get through the MVT may not be all necessary sequences. In the case of $x^2sin(1/x)$ the sequence $1/(2npi)to 0,$ but this sequence does not arise as $theta(h_n)h_n$ in the MVT process.
$endgroup$
– zhw.
Mar 4 at 18:35
add a comment |
$begingroup$
It is not true that $theta$ need be a function of $h,$ because it may not be uniquely specified. There could be many $theta$'s that work for a given $h.$
We only know $f'(y)to f'(c)$ as $yto c$ within the set of $y=x+theta h$ that arise in the MVT process you described. But the set of such $y$ may not equal any deleted neighborhood of $c,$ as the example $x^2sin(1/x)$ shows.
answered Mar 4 at 18:23
zhw.zhw.
74.7k43275
$endgroup$
It is not true that $theta$ need be a function of $h,$ because it may not be uniquely specified. There could be many $theta$'s that work for a given $h.$
We only know $f'(y)to f'(c)$ as $yto c$ within the set of $y=x+theta h$ that arise in the MVT process you described. But the set of such $y$ may not equal any deleted neighborhood of $c,$ as the example $x^2sin(1/x)$ shows.
answered Mar 4 at 18:23
zhw.zhw.
74.7k43275
edited Mar 4 at 22:09
answered Mar 4 at 18:23
zhw.zhw.
74.7k43275
answered Mar 4 at 18:23
zhw.zhw.
74.7k43275
answered Mar 4 at 18:23
zhw.zhw.
74.7k43275
74.7k43275
$begingroup$
So we can make $theta$ a function by choosing a value of $h$ and then say the union of all the sequences $theta(h_n)h_n$ where $h_n xrightarrow0$ over all functions $theta$ need not be all sequences tending to 0?
$endgroup$
– user3184807
Mar 4 at 18:27
1
$begingroup$
That's a lot of choosing, and the axiom of choice may even come in here. But yes, the sequences we get through the MVT may not be all necessary sequences. In the case of $x^2sin(1/x)$ the sequence $1/(2npi)to 0,$ but this sequence does not arise as $theta(h_n)h_n$ in the MVT process.
$endgroup$
– zhw.
Mar 4 at 18:35
add a comment |
$begingroup$
So we can make $theta$ a function by choosing a value of $h$ and then say the union of all the sequences $theta(h_n)h_n$ where $h_n xrightarrow0$ over all functions $theta$ need not be all sequences tending to 0?
$endgroup$
– user3184807
Mar 4 at 18:27
1
$begingroup$
That's a lot of choosing, and the axiom of choice may even come in here. But yes, the sequences we get through the MVT may not be all necessary sequences. In the case of $x^2sin(1/x)$ the sequence $1/(2npi)to 0,$ but this sequence does not arise as $theta(h_n)h_n$ in the MVT process.
$endgroup$
– zhw.
Mar 4 at 18:35
$begingroup$
So we can make $theta$ a function by choosing a value of $h$ and then say the union of all the sequences $theta(h_n)h_n$ where $h_n xrightarrow0$ over all functions $theta$ need not be all sequences tending to 0?
$endgroup$
– user3184807
Mar 4 at 18:27
$begingroup$
So we can make $theta$ a function by choosing a value of $h$ and then say the union of all the sequences $theta(h_n)h_n$ where $h_n xrightarrow0$ over all functions $theta$ need not be all sequences tending to 0?
$endgroup$
– user3184807
Mar 4 at 18:27
1
1
$begingroup$
That's a lot of choosing, and the axiom of choice may even come in here. But yes, the sequences we get through the MVT may not be all necessary sequences. In the case of $x^2sin(1/x)$ the sequence $1/(2npi)to 0,$ but this sequence does not arise as $theta(h_n)h_n$ in the MVT process.
$endgroup$
– zhw.
Mar 4 at 18:35
$begingroup$
That's a lot of choosing, and the axiom of choice may even come in here. But yes, the sequences we get through the MVT may not be all necessary sequences. In the case of $x^2sin(1/x)$ the sequence $1/(2npi)to 0,$ but this sequence does not arise as $theta(h_n)h_n$ in the MVT process.
$endgroup$
– zhw.
Mar 4 at 18:35
add a comment |
$begingroup$
If you look at $f(x)=x^2 sin(1/x)$ on $(0,1]$ and $f(0)=0$, you have $f'(x)=2xsin(1/x)-cos(1/x)$ on $(0,1]$ and $f'(0)=0$. Thus MVT tells you that for every $h in (0,1)$ there exists $theta(h) in (0,1)$ with
$$hsin(1/h)=2htheta(h)sin(1/(htheta(h)))-cos(1/(htheta(h))).$$
Notice that once $h$ is small enough, $htheta(h)$ is forced to stay relatively close to the zeros of $cos(1/x)$, since
$$|cos(1/(htheta(h)))|=|hsin(1/h)-2htheta(h)sin(1/(htheta(h)))|<3h.$$
This means that the MVT is only providing you with information about $f'(x)$ approaching $f'(0)$ along sequences which stay sufficiently close to the zeros of $cos(1/x)$. If you instead look at something like $f' left ( frac1pi n right )$ you find that it doesn't go to zero. There is no contradiction because the MVT is never giving you information about those points.
answered Mar 4 at 18:18
IanIan
68.9k25392
$endgroup$
$begingroup$
So does this essentially come down to the idea that $theta(h_n)h_n$ cannot always be made to be any arbitrary sequence tending to 0?
$endgroup$
– user3184807
Mar 4 at 18:21
$begingroup$
@user3184807 Yes, that is essentially what happens.
$endgroup$
– Ian
Mar 4 at 18:22
add a comment |
$begingroup$
If you look at $f(x)=x^2 sin(1/x)$ on $(0,1]$ and $f(0)=0$, you have $f'(x)=2xsin(1/x)-cos(1/x)$ on $(0,1]$ and $f'(0)=0$. Thus MVT tells you that for every $h in (0,1)$ there exists $theta(h) in (0,1)$ with
$$hsin(1/h)=2htheta(h)sin(1/(htheta(h)))-cos(1/(htheta(h))).$$
Notice that once $h$ is small enough, $htheta(h)$ is forced to stay relatively close to the zeros of $cos(1/x)$, since
$$|cos(1/(htheta(h)))|=|hsin(1/h)-2htheta(h)sin(1/(htheta(h)))|<3h.$$
This means that the MVT is only providing you with information about $f'(x)$ approaching $f'(0)$ along sequences which stay sufficiently close to the zeros of $cos(1/x)$. If you instead look at something like $f' left ( frac1pi n right )$ you find that it doesn't go to zero. There is no contradiction because the MVT is never giving you information about those points.
answered Mar 4 at 18:18
IanIan
68.9k25392
$endgroup$
$begingroup$
So does this essentially come down to the idea that $theta(h_n)h_n$ cannot always be made to be any arbitrary sequence tending to 0?
$endgroup$
– user3184807
Mar 4 at 18:21
$begingroup$
@user3184807 Yes, that is essentially what happens.
$endgroup$
– Ian
Mar 4 at 18:22
add a comment |
$begingroup$
If you look at $f(x)=x^2 sin(1/x)$ on $(0,1]$ and $f(0)=0$, you have $f'(x)=2xsin(1/x)-cos(1/x)$ on $(0,1]$ and $f'(0)=0$. Thus MVT tells you that for every $h in (0,1)$ there exists $theta(h) in (0,1)$ with
$$hsin(1/h)=2htheta(h)sin(1/(htheta(h)))-cos(1/(htheta(h))).$$
Notice that once $h$ is small enough, $htheta(h)$ is forced to stay relatively close to the zeros of $cos(1/x)$, since
$$|cos(1/(htheta(h)))|=|hsin(1/h)-2htheta(h)sin(1/(htheta(h)))|<3h.$$
This means that the MVT is only providing you with information about $f'(x)$ approaching $f'(0)$ along sequences which stay sufficiently close to the zeros of $cos(1/x)$. If you instead look at something like $f' left ( frac1pi n right )$ you find that it doesn't go to zero. There is no contradiction because the MVT is never giving you information about those points.
answered Mar 4 at 18:18
IanIan
68.9k25392
$endgroup$
If you look at $f(x)=x^2 sin(1/x)$ on $(0,1]$ and $f(0)=0$, you have $f'(x)=2xsin(1/x)-cos(1/x)$ on $(0,1]$ and $f'(0)=0$. Thus MVT tells you that for every $h in (0,1)$ there exists $theta(h) in (0,1)$ with
$$hsin(1/h)=2htheta(h)sin(1/(htheta(h)))-cos(1/(htheta(h))).$$
Notice that once $h$ is small enough, $htheta(h)$ is forced to stay relatively close to the zeros of $cos(1/x)$, since
$$|cos(1/(htheta(h)))|=|hsin(1/h)-2htheta(h)sin(1/(htheta(h)))|<3h.$$
This means that the MVT is only providing you with information about $f'(x)$ approaching $f'(0)$ along sequences which stay sufficiently close to the zeros of $cos(1/x)$. If you instead look at something like $f' left ( frac1pi n right )$ you find that it doesn't go to zero. There is no contradiction because the MVT is never giving you information about those points.
answered Mar 4 at 18:18
IanIan
68.9k25392
edited Mar 4 at 18:39
answered Mar 4 at 18:18
IanIan
68.9k25392
answered Mar 4 at 18:18
IanIan
68.9k25392
answered Mar 4 at 18:18
IanIan
68.9k25392
68.9k25392
$begingroup$
So does this essentially come down to the idea that $theta(h_n)h_n$ cannot always be made to be any arbitrary sequence tending to 0?
$endgroup$
– user3184807
Mar 4 at 18:21
$begingroup$
@user3184807 Yes, that is essentially what happens.
$endgroup$
– Ian
Mar 4 at 18:22
add a comment |
$begingroup$
So does this essentially come down to the idea that $theta(h_n)h_n$ cannot always be made to be any arbitrary sequence tending to 0?
$endgroup$
– user3184807
Mar 4 at 18:21
$begingroup$
@user3184807 Yes, that is essentially what happens.
$endgroup$
– Ian
Mar 4 at 18:22
$begingroup$
So does this essentially come down to the idea that $theta(h_n)h_n$ cannot always be made to be any arbitrary sequence tending to 0?
$endgroup$
– user3184807
Mar 4 at 18:21
$begingroup$
So does this essentially come down to the idea that $theta(h_n)h_n$ cannot always be made to be any arbitrary sequence tending to 0?
$endgroup$
– user3184807
Mar 4 at 18:21
$begingroup$
@user3184807 Yes, that is essentially what happens.
$endgroup$
– Ian
Mar 4 at 18:22
$begingroup$
@user3184807 Yes, that is essentially what happens.
$endgroup$
– Ian
Mar 4 at 18:22
add a comment |
$begingroup$
You are assuming what you want to prove: that $f’$ is continuous at $c$ when you say that $f'(c+theta h) to f'(c)$ as $hto0$.
answered Mar 4 at 17:53
Julián AguirreJulián Aguirre
69.5k24297
$endgroup$
1
$begingroup$
No, because OP defined $theta(h)$ through the MVT in the first place. The point is that you know that $lim_h to 0 f'(c+htheta(h))=f'(c)$, but why doesn't this imply $lim_h to 0 f'(c+h)=f'(c)$? Presumably this is because of some very bad property of $theta(h)$.
$endgroup$
– Ian
Mar 4 at 17:54
1
$begingroup$
That has nothing to do with muy answer.
$endgroup$
– Julián Aguirre
Mar 4 at 17:56
$begingroup$
Let me spell it out, then. By definition $lim_h to 0 fracf(c+h)-f(c)h=f'(c)$, and by MVT $fracf(c+h)-f(c)h=f'(c+htheta(h))$ for some $theta(h) in (0,1)$. That's all OP assumed. Then they send $h to 0$ in the MVT relation and find $lim_h to 0 f'(c+h theta(h))=f'(c)$. Now they're wondering why they can't conclude $lim_h to 0 f'(c+h)=f'(c)$.
$endgroup$
– Ian
Mar 4 at 17:57
$begingroup$
@Ian: The problem is that the limit may not exist. If it exists, then it must equal $f'(c)$ (as has been covered many times on this site already, for example here: math.stackexchange.com/questions/257907/…).
$endgroup$
– Hans Lundmark
Mar 4 at 18:04
$begingroup$
@HansLundmark OP knows that already as well, the question is about intuition. The answer, for the case of something like $f(x)=x^2sin(1/x)$, is that $htheta(h)$ has numerous discontinuities in a vicinity of $h=0$, so you don't get information about entire neighborhoods of $h=0$ from the MVT.
$endgroup$
– Ian
Mar 4 at 18:08
|
show 4 more comments
$begingroup$
You are assuming what you want to prove: that $f’$ is continuous at $c$ when you say that $f'(c+theta h) to f'(c)$ as $hto0$.
answered Mar 4 at 17:53
Julián AguirreJulián Aguirre
69.5k24297
$endgroup$
1
$begingroup$
No, because OP defined $theta(h)$ through the MVT in the first place. The point is that you know that $lim_h to 0 f'(c+htheta(h))=f'(c)$, but why doesn't this imply $lim_h to 0 f'(c+h)=f'(c)$? Presumably this is because of some very bad property of $theta(h)$.
$endgroup$
– Ian
Mar 4 at 17:54
1
$begingroup$
That has nothing to do with muy answer.
$endgroup$
– Julián Aguirre
Mar 4 at 17:56
$begingroup$
Let me spell it out, then. By definition $lim_h to 0 fracf(c+h)-f(c)h=f'(c)$, and by MVT $fracf(c+h)-f(c)h=f'(c+htheta(h))$ for some $theta(h) in (0,1)$. That's all OP assumed. Then they send $h to 0$ in the MVT relation and find $lim_h to 0 f'(c+h theta(h))=f'(c)$. Now they're wondering why they can't conclude $lim_h to 0 f'(c+h)=f'(c)$.
$endgroup$
– Ian
Mar 4 at 17:57
$begingroup$
@Ian: The problem is that the limit may not exist. If it exists, then it must equal $f'(c)$ (as has been covered many times on this site already, for example here: math.stackexchange.com/questions/257907/…).
$endgroup$
– Hans Lundmark
Mar 4 at 18:04
$begingroup$
@HansLundmark OP knows that already as well, the question is about intuition. The answer, for the case of something like $f(x)=x^2sin(1/x)$, is that $htheta(h)$ has numerous discontinuities in a vicinity of $h=0$, so you don't get information about entire neighborhoods of $h=0$ from the MVT.
$endgroup$
– Ian
Mar 4 at 18:08
|
show 4 more comments
$begingroup$
You are assuming what you want to prove: that $f’$ is continuous at $c$ when you say that $f'(c+theta h) to f'(c)$ as $hto0$.
answered Mar 4 at 17:53
Julián AguirreJulián Aguirre
69.5k24297
$endgroup$
You are assuming what you want to prove: that $f’$ is continuous at $c$ when you say that $f'(c+theta h) to f'(c)$ as $hto0$.
answered Mar 4 at 17:53
Julián AguirreJulián Aguirre
69.5k24297
answered Mar 4 at 17:53
Julián AguirreJulián Aguirre
69.5k24297
answered Mar 4 at 17:53
Julián AguirreJulián Aguirre
69.5k24297
answered Mar 4 at 17:53
Julián AguirreJulián Aguirre
69.5k24297
69.5k24297
1
$begingroup$
No, because OP defined $theta(h)$ through the MVT in the first place. The point is that you know that $lim_h to 0 f'(c+htheta(h))=f'(c)$, but why doesn't this imply $lim_h to 0 f'(c+h)=f'(c)$? Presumably this is because of some very bad property of $theta(h)$.
$endgroup$
– Ian
Mar 4 at 17:54
1
$begingroup$
That has nothing to do with muy answer.
$endgroup$
– Julián Aguirre
Mar 4 at 17:56
$begingroup$
Let me spell it out, then. By definition $lim_h to 0 fracf(c+h)-f(c)h=f'(c)$, and by MVT $fracf(c+h)-f(c)h=f'(c+htheta(h))$ for some $theta(h) in (0,1)$. That's all OP assumed. Then they send $h to 0$ in the MVT relation and find $lim_h to 0 f'(c+h theta(h))=f'(c)$. Now they're wondering why they can't conclude $lim_h to 0 f'(c+h)=f'(c)$.
$endgroup$
– Ian
Mar 4 at 17:57
$begingroup$
@Ian: The problem is that the limit may not exist. If it exists, then it must equal $f'(c)$ (as has been covered many times on this site already, for example here: math.stackexchange.com/questions/257907/…).
$endgroup$
– Hans Lundmark
Mar 4 at 18:04
$begingroup$
@HansLundmark OP knows that already as well, the question is about intuition. The answer, for the case of something like $f(x)=x^2sin(1/x)$, is that $htheta(h)$ has numerous discontinuities in a vicinity of $h=0$, so you don't get information about entire neighborhoods of $h=0$ from the MVT.
$endgroup$
– Ian
Mar 4 at 18:08
|
show 4 more comments
1
$begingroup$
No, because OP defined $theta(h)$ through the MVT in the first place. The point is that you know that $lim_h to 0 f'(c+htheta(h))=f'(c)$, but why doesn't this imply $lim_h to 0 f'(c+h)=f'(c)$? Presumably this is because of some very bad property of $theta(h)$.
$endgroup$
– Ian
Mar 4 at 17:54
1
$begingroup$
That has nothing to do with muy answer.
$endgroup$
– Julián Aguirre
Mar 4 at 17:56
$begingroup$
Let me spell it out, then. By definition $lim_h to 0 fracf(c+h)-f(c)h=f'(c)$, and by MVT $fracf(c+h)-f(c)h=f'(c+htheta(h))$ for some $theta(h) in (0,1)$. That's all OP assumed. Then they send $h to 0$ in the MVT relation and find $lim_h to 0 f'(c+h theta(h))=f'(c)$. Now they're wondering why they can't conclude $lim_h to 0 f'(c+h)=f'(c)$.
$endgroup$
– Ian
Mar 4 at 17:57
$begingroup$
@Ian: The problem is that the limit may not exist. If it exists, then it must equal $f'(c)$ (as has been covered many times on this site already, for example here: math.stackexchange.com/questions/257907/…).
$endgroup$
– Hans Lundmark
Mar 4 at 18:04
$begingroup$
@HansLundmark OP knows that already as well, the question is about intuition. The answer, for the case of something like $f(x)=x^2sin(1/x)$, is that $htheta(h)$ has numerous discontinuities in a vicinity of $h=0$, so you don't get information about entire neighborhoods of $h=0$ from the MVT.
$endgroup$
– Ian
Mar 4 at 18:08
1
1
$begingroup$
No, because OP defined $theta(h)$ through the MVT in the first place. The point is that you know that $lim_h to 0 f'(c+htheta(h))=f'(c)$, but why doesn't this imply $lim_h to 0 f'(c+h)=f'(c)$? Presumably this is because of some very bad property of $theta(h)$.
$endgroup$
– Ian
Mar 4 at 17:54
$begingroup$
No, because OP defined $theta(h)$ through the MVT in the first place. The point is that you know that $lim_h to 0 f'(c+htheta(h))=f'(c)$, but why doesn't this imply $lim_h to 0 f'(c+h)=f'(c)$? Presumably this is because of some very bad property of $theta(h)$.
$endgroup$
– Ian
Mar 4 at 17:54
1
1
$begingroup$
That has nothing to do with muy answer.
$endgroup$
– Julián Aguirre
Mar 4 at 17:56
$begingroup$
That has nothing to do with muy answer.
$endgroup$
– Julián Aguirre
Mar 4 at 17:56
$begingroup$
Let me spell it out, then. By definition $lim_h to 0 fracf(c+h)-f(c)h=f'(c)$, and by MVT $fracf(c+h)-f(c)h=f'(c+htheta(h))$ for some $theta(h) in (0,1)$. That's all OP assumed. Then they send $h to 0$ in the MVT relation and find $lim_h to 0 f'(c+h theta(h))=f'(c)$. Now they're wondering why they can't conclude $lim_h to 0 f'(c+h)=f'(c)$.
$endgroup$
– Ian
Mar 4 at 17:57
$begingroup$
Let me spell it out, then. By definition $lim_h to 0 fracf(c+h)-f(c)h=f'(c)$, and by MVT $fracf(c+h)-f(c)h=f'(c+htheta(h))$ for some $theta(h) in (0,1)$. That's all OP assumed. Then they send $h to 0$ in the MVT relation and find $lim_h to 0 f'(c+h theta(h))=f'(c)$. Now they're wondering why they can't conclude $lim_h to 0 f'(c+h)=f'(c)$.
$endgroup$
– Ian
Mar 4 at 17:57
$begingroup$
@Ian: The problem is that the limit may not exist. If it exists, then it must equal $f'(c)$ (as has been covered many times on this site already, for example here: math.stackexchange.com/questions/257907/…).
$endgroup$
– Hans Lundmark
Mar 4 at 18:04
$begingroup$
@Ian: The problem is that the limit may not exist. If it exists, then it must equal $f'(c)$ (as has been covered many times on this site already, for example here: math.stackexchange.com/questions/257907/…).
$endgroup$
– Hans Lundmark
Mar 4 at 18:04
$begingroup$
@HansLundmark OP knows that already as well, the question is about intuition. The answer, for the case of something like $f(x)=x^2sin(1/x)$, is that $htheta(h)$ has numerous discontinuities in a vicinity of $h=0$, so you don't get information about entire neighborhoods of $h=0$ from the MVT.
$endgroup$
– Ian
Mar 4 at 18:08
$begingroup$
@HansLundmark OP knows that already as well, the question is about intuition. The answer, for the case of something like $f(x)=x^2sin(1/x)$, is that $htheta(h)$ has numerous discontinuities in a vicinity of $h=0$, so you don't get information about entire neighborhoods of $h=0$ from the MVT.
$endgroup$
– Ian
Mar 4 at 18:08
|
show 4 more comments
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$begingroup$
Well under the conditions of your question, it is guaranteed (via mean value theorem) that there is a sequence $c_n$ of values tending to $c$ such that $f'(c_n) to f'(c) $. What is **not** guaranteed is that the same result holds for any/every sequence $c_nto c$ and thus continuity of $f'$ at $c$ can't be inferred.
$endgroup$
– Paramanand Singh
Mar 6 at 5:02
$begingroup$
That is essentially what my reasoning said
$endgroup$
– user3184807
Mar 6 at 19:07