Why is this not correct？
$begingroup$
We are assigned to deal with the following task.
Assume that $f(x)$ is derivable for any $x in mathbb{R}$. We want to
research $limlimits_{x to x_0}f'(x)$ where $x_0 in mathbb{R}$.
Notice that
begin{align*}
f'(x_0)=lim_{x to x_0}frac{f(x)-f(x_0)}{x-x_0}=lim_{x to x_0}frac{f'(xi)(x-x_0)}{x-x_0}=lim_{x to x_0}f'(xi),
end{align*}
where we applied Lagrange's Mean Value Theorem, and $ x_0 lessgtr xi lessgtr x.$ Since $xi$ is squeezed by $x_0$ and $x$, then $x to x_0$ implies $xi to x_0$. Thus
$$f'(x_0)=lim_{x to x_0}f'(xi)=lim_{xi to x_0}f'(xi).$$
What does this say? It shows that $f'(x)$ is always continuous at any point $x=x_0$, which is an absurd conclusion, because we know safely $f'(x)$ may probably has the discontinuity point (of the second kind). But where dose the mistake occur during the reasoning above?
calculus
asked Dec 15 '18 at 9:03
mengdie1982mengdie1982
4,932618
$endgroup$
|
show 1 more comment
$begingroup$
We are assigned to deal with the following task.
Assume that $f(x)$ is derivable for any $x in mathbb{R}$. We want to
research $limlimits_{x to x_0}f'(x)$ where $x_0 in mathbb{R}$.
Notice that
begin{align*}
f'(x_0)=lim_{x to x_0}frac{f(x)-f(x_0)}{x-x_0}=lim_{x to x_0}frac{f'(xi)(x-x_0)}{x-x_0}=lim_{x to x_0}f'(xi),
end{align*}
where we applied Lagrange's Mean Value Theorem, and $ x_0 lessgtr xi lessgtr x.$ Since $xi$ is squeezed by $x_0$ and $x$, then $x to x_0$ implies $xi to x_0$. Thus
$$f'(x_0)=lim_{x to x_0}f'(xi)=lim_{xi to x_0}f'(xi).$$
What does this say? It shows that $f'(x)$ is always continuous at any point $x=x_0$, which is an absurd conclusion, because we know safely $f'(x)$ may probably has the discontinuity point (of the second kind). But where dose the mistake occur during the reasoning above?
calculus
asked Dec 15 '18 at 9:03
mengdie1982mengdie1982
4,932618
$endgroup$
$begingroup$
Recall that $xi$ is actually a function of $x$. Thus, $lim_{xrightarrow x_0} f'(xi(x))$ does not imply $lim_{xi rightarrow x_0} f'(xi)$
$endgroup$
– Severin Schraven
Dec 15 '18 at 9:08
$begingroup$
it is just a $ksi$ not every $x$...
$endgroup$
– dmtri
Dec 15 '18 at 9:09
$begingroup$
@SeverinSchraven Do you know the squeeze theorem? $x_0 lessgtr xi(x) lessgtr x$. Let $x to x_0$，then $xi(x) to x_0$. It has no fault here.
$endgroup$
– mengdie1982
Dec 15 '18 at 9:25
2
$begingroup$
Reread my comment, I did NOT claim that $xi(x) rightarrow x_0$ is wrong. Your line of reasoning is the following: Let $$ f(x) = begin{cases} 0;& xin mathbb{Q} \ 1,& xin mathbb{R}setminus mathbb{Q} end{cases} $$ now take $x_0=0$ and $xi(x)$ be a rational number between $0$ and $x$. Clearly we have $lim_{xrightarrow 0} f(xi(x))=0$ and thus $lim_{xi rightarrow 0} f(xi)= lim_{x rightarrow 0} f(xi(x))$. But $lim_{xi rightarrow 0} f(xi)$ does not even exist. That is, because you do not hit every number with $xi(x)$.
$endgroup$
– Severin Schraven
Dec 15 '18 at 9:32
$begingroup$
@SeverinSchraven You have pointed out some essential thing！
$endgroup$
– mengdie1982
Dec 15 '18 at 9:36
|
show 1 more comment
$begingroup$
We are assigned to deal with the following task.
Assume that $f(x)$ is derivable for any $x in mathbb{R}$. We want to
research $limlimits_{x to x_0}f'(x)$ where $x_0 in mathbb{R}$.
Notice that
begin{align*}
f'(x_0)=lim_{x to x_0}frac{f(x)-f(x_0)}{x-x_0}=lim_{x to x_0}frac{f'(xi)(x-x_0)}{x-x_0}=lim_{x to x_0}f'(xi),
end{align*}
where we applied Lagrange's Mean Value Theorem, and $ x_0 lessgtr xi lessgtr x.$ Since $xi$ is squeezed by $x_0$ and $x$, then $x to x_0$ implies $xi to x_0$. Thus
$$f'(x_0)=lim_{x to x_0}f'(xi)=lim_{xi to x_0}f'(xi).$$
What does this say? It shows that $f'(x)$ is always continuous at any point $x=x_0$, which is an absurd conclusion, because we know safely $f'(x)$ may probably has the discontinuity point (of the second kind). But where dose the mistake occur during the reasoning above?
calculus
asked Dec 15 '18 at 9:03
mengdie1982mengdie1982
4,932618
$endgroup$
We are assigned to deal with the following task.
Assume that $f(x)$ is derivable for any $x in mathbb{R}$. We want to
research $limlimits_{x to x_0}f'(x)$ where $x_0 in mathbb{R}$.
Notice that
begin{align*}
f'(x_0)=lim_{x to x_0}frac{f(x)-f(x_0)}{x-x_0}=lim_{x to x_0}frac{f'(xi)(x-x_0)}{x-x_0}=lim_{x to x_0}f'(xi),
end{align*}
where we applied Lagrange's Mean Value Theorem, and $ x_0 lessgtr xi lessgtr x.$ Since $xi$ is squeezed by $x_0$ and $x$, then $x to x_0$ implies $xi to x_0$. Thus
$$f'(x_0)=lim_{x to x_0}f'(xi)=lim_{xi to x_0}f'(xi).$$
What does this say? It shows that $f'(x)$ is always continuous at any point $x=x_0$, which is an absurd conclusion, because we know safely $f'(x)$ may probably has the discontinuity point (of the second kind). But where dose the mistake occur during the reasoning above?
calculus
calculus
asked Dec 15 '18 at 9:03
mengdie1982mengdie1982
4,932618
asked Dec 15 '18 at 9:03
mengdie1982mengdie1982
4,932618
asked Dec 15 '18 at 9:03
mengdie1982mengdie1982
4,932618
asked Dec 15 '18 at 9:03
mengdie1982mengdie1982
4,932618
asked Dec 15 '18 at 9:03
mengdie1982mengdie1982
4,932618
4,932618
$begingroup$
Recall that $xi$ is actually a function of $x$. Thus, $lim_{xrightarrow x_0} f'(xi(x))$ does not imply $lim_{xi rightarrow x_0} f'(xi)$
$endgroup$
– Severin Schraven
Dec 15 '18 at 9:08
$begingroup$
it is just a $ksi$ not every $x$...
$endgroup$
– dmtri
Dec 15 '18 at 9:09
$begingroup$
@SeverinSchraven Do you know the squeeze theorem? $x_0 lessgtr xi(x) lessgtr x$. Let $x to x_0$，then $xi(x) to x_0$. It has no fault here.
$endgroup$
– mengdie1982
Dec 15 '18 at 9:25
2
$begingroup$
Reread my comment, I did NOT claim that $xi(x) rightarrow x_0$ is wrong. Your line of reasoning is the following: Let $$ f(x) = begin{cases} 0;& xin mathbb{Q} \ 1,& xin mathbb{R}setminus mathbb{Q} end{cases} $$ now take $x_0=0$ and $xi(x)$ be a rational number between $0$ and $x$. Clearly we have $lim_{xrightarrow 0} f(xi(x))=0$ and thus $lim_{xi rightarrow 0} f(xi)= lim_{x rightarrow 0} f(xi(x))$. But $lim_{xi rightarrow 0} f(xi)$ does not even exist. That is, because you do not hit every number with $xi(x)$.
$endgroup$
– Severin Schraven
Dec 15 '18 at 9:32
$begingroup$
@SeverinSchraven You have pointed out some essential thing！
$endgroup$
– mengdie1982
Dec 15 '18 at 9:36
|
show 1 more comment
$begingroup$
Recall that $xi$ is actually a function of $x$. Thus, $lim_{xrightarrow x_0} f'(xi(x))$ does not imply $lim_{xi rightarrow x_0} f'(xi)$
$endgroup$
– Severin Schraven
Dec 15 '18 at 9:08
$begingroup$
it is just a $ksi$ not every $x$...
$endgroup$
– dmtri
Dec 15 '18 at 9:09
$begingroup$
@SeverinSchraven Do you know the squeeze theorem? $x_0 lessgtr xi(x) lessgtr x$. Let $x to x_0$，then $xi(x) to x_0$. It has no fault here.
$endgroup$
– mengdie1982
Dec 15 '18 at 9:25
2
$begingroup$
Reread my comment, I did NOT claim that $xi(x) rightarrow x_0$ is wrong. Your line of reasoning is the following: Let $$ f(x) = begin{cases} 0;& xin mathbb{Q} \ 1,& xin mathbb{R}setminus mathbb{Q} end{cases} $$ now take $x_0=0$ and $xi(x)$ be a rational number between $0$ and $x$. Clearly we have $lim_{xrightarrow 0} f(xi(x))=0$ and thus $lim_{xi rightarrow 0} f(xi)= lim_{x rightarrow 0} f(xi(x))$. But $lim_{xi rightarrow 0} f(xi)$ does not even exist. That is, because you do not hit every number with $xi(x)$.
$endgroup$
– Severin Schraven
Dec 15 '18 at 9:32
$begingroup$
@SeverinSchraven You have pointed out some essential thing！
$endgroup$
– mengdie1982
Dec 15 '18 at 9:36
$begingroup$
Recall that $xi$ is actually a function of $x$. Thus, $lim_{xrightarrow x_0} f'(xi(x))$ does not imply $lim_{xi rightarrow x_0} f'(xi)$
$endgroup$
– Severin Schraven
Dec 15 '18 at 9:08
$begingroup$
Recall that $xi$ is actually a function of $x$. Thus, $lim_{xrightarrow x_0} f'(xi(x))$ does not imply $lim_{xi rightarrow x_0} f'(xi)$
$endgroup$
– Severin Schraven
Dec 15 '18 at 9:08
$begingroup$
it is just a $ksi$ not every $x$...
$endgroup$
– dmtri
Dec 15 '18 at 9:09
$begingroup$
it is just a $ksi$ not every $x$...
$endgroup$
– dmtri
Dec 15 '18 at 9:09
$begingroup$
@SeverinSchraven Do you know the squeeze theorem? $x_0 lessgtr xi(x) lessgtr x$. Let $x to x_0$，then $xi(x) to x_0$. It has no fault here.
$endgroup$
– mengdie1982
Dec 15 '18 at 9:25
$begingroup$
@SeverinSchraven Do you know the squeeze theorem? $x_0 lessgtr xi(x) lessgtr x$. Let $x to x_0$，then $xi(x) to x_0$. It has no fault here.
$endgroup$
– mengdie1982
Dec 15 '18 at 9:25
2
2
$begingroup$
Reread my comment, I did NOT claim that $xi(x) rightarrow x_0$ is wrong. Your line of reasoning is the following: Let $$ f(x) = begin{cases} 0;& xin mathbb{Q} \ 1,& xin mathbb{R}setminus mathbb{Q} end{cases} $$ now take $x_0=0$ and $xi(x)$ be a rational number between $0$ and $x$. Clearly we have $lim_{xrightarrow 0} f(xi(x))=0$ and thus $lim_{xi rightarrow 0} f(xi)= lim_{x rightarrow 0} f(xi(x))$. But $lim_{xi rightarrow 0} f(xi)$ does not even exist. That is, because you do not hit every number with $xi(x)$.
$endgroup$
– Severin Schraven
Dec 15 '18 at 9:32
$begingroup$
Reread my comment, I did NOT claim that $xi(x) rightarrow x_0$ is wrong. Your line of reasoning is the following: Let $$ f(x) = begin{cases} 0;& xin mathbb{Q} \ 1,& xin mathbb{R}setminus mathbb{Q} end{cases} $$ now take $x_0=0$ and $xi(x)$ be a rational number between $0$ and $x$. Clearly we have $lim_{xrightarrow 0} f(xi(x))=0$ and thus $lim_{xi rightarrow 0} f(xi)= lim_{x rightarrow 0} f(xi(x))$. But $lim_{xi rightarrow 0} f(xi)$ does not even exist. That is, because you do not hit every number with $xi(x)$.
$endgroup$
– Severin Schraven
Dec 15 '18 at 9:32
$begingroup$
@SeverinSchraven You have pointed out some essential thing！
$endgroup$
– mengdie1982
Dec 15 '18 at 9:36
$begingroup$
@SeverinSchraven You have pointed out some essential thing！
$endgroup$
– mengdie1982
Dec 15 '18 at 9:36
|
show 1 more comment
3 Answers
3
active
oldest
votes
$begingroup$
Your reasoning only shows that if $lim_{x to x_0} f'(x)$ exists, then it's equal to $f'(x_0)$. But, as you say, it may not exist.
This is a rather classical exercise, which has been treated many times on this site, for example in this question.
answered Dec 15 '18 at 9:48
Hans LundmarkHans Lundmark
35.8k564115
$endgroup$
add a comment |
$begingroup$
Actually，we can make a comment for the reasoning like this:
The fact that $limlimits_{xi to x_0}f'(xi)$ exists dose not imply $limlimits_{x to x_0}f'(x)$ also exsits, because, according to Heine's Theorem, the latter one necessitates that $f'(x_n)$ converges for any sequence $x_n to x_0$. As we can see, $xi_n$ is only a specific sequence. Even though $f'(xi_n)$ is convergent, this is not enough to guarantee $f'(x_n)$ converges as well.
answered Dec 15 '18 at 11:36
mengdie1982mengdie1982
4,932618
$endgroup$
$begingroup$
You mean $lim_{xto x_0}f'(xi)$ instead of $lim_{xito x_0}f'(xi)$.
$endgroup$
– Paramanand Singh
Dec 20 '18 at 9:42
add a comment |
$begingroup$
Actually, you are right, this also called 'theorem of derivative limitation' in some textbooks.
If you already assumed that $f(x)$ is derivable for any $x in mathbb{R}$.
With an additional condition the limitation of derivative $lim_{x to x_{0}}f'(x)$ exist.
You can write your conclusion:
$$lim_{x to x_{0}}f'(x)=f'(x_{0})$$
The condition $f(x)$ is derivable for any $x in mathbb{R}$ can also be weaken to $f(x)$ is continuous in $U(x_{0})$ which is the neighborhood of $x_{0}$, and derivable in $U°(x_{0})$ which is the punctured neighborhood of $x_{0}$.
The only problem is that you have not assumed the limitation of your derivative should always exist, which means the derivative can not divergence to infinity, like you say sometimes it would be the discontinuity point of the second kind.
And this theorem can be simply remembered as in other 'not so strict (please notice the comment below)' ways like:
If a continuous function has (finite) derivative, its derivative function is also continuous. (but maybe not derivable)
answered Dec 15 '18 at 15:20
NanayajitzukiNanayajitzuki
3185
$endgroup$
$begingroup$
It would be fair to say that you're wrong. The fact $f(x)$ is derivable does not imply $f'(x)$ is continuous. Here is a counterexample: $$f(x)=begin{cases}x^2sindfrac{1}{x},&x neq 0,\0,&x=0.end{cases}.$$ You may verify that $f(x)$ is derivable over $(-infty,+infty)$，but $f'(x)$ is not continuous at $x=0$.
$endgroup$
– mengdie1982
Dec 15 '18 at 16:16
$begingroup$
But of course, if we add a condition that $limlimits_{x to x_0}f'(x)$ exists, then $limlimits_{x to x_0}f'(x)=f'(x_0)$. In another word, $f'(x)$ is continuous at $x=x_0$ now！
$endgroup$
– mengdie1982
Dec 15 '18 at 16:24
$begingroup$
@mengdie1982 You still remind a good example for me. Yes, I should explain that the 'not so strict' place this the limitation of derivative must be exist not only the derivative exist. Your comment point out it.
$endgroup$
– Nanayajitzuki
Dec 15 '18 at 16:30
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Your reasoning only shows that if $lim_{x to x_0} f'(x)$ exists, then it's equal to $f'(x_0)$. But, as you say, it may not exist.
This is a rather classical exercise, which has been treated many times on this site, for example in this question.
answered Dec 15 '18 at 9:48
Hans LundmarkHans Lundmark
35.8k564115
$endgroup$
add a comment |
$begingroup$
Your reasoning only shows that if $lim_{x to x_0} f'(x)$ exists, then it's equal to $f'(x_0)$. But, as you say, it may not exist.
This is a rather classical exercise, which has been treated many times on this site, for example in this question.
answered Dec 15 '18 at 9:48
Hans LundmarkHans Lundmark
35.8k564115
$endgroup$
add a comment |
$begingroup$
Your reasoning only shows that if $lim_{x to x_0} f'(x)$ exists, then it's equal to $f'(x_0)$. But, as you say, it may not exist.
This is a rather classical exercise, which has been treated many times on this site, for example in this question.
answered Dec 15 '18 at 9:48
Hans LundmarkHans Lundmark
35.8k564115
$endgroup$
Your reasoning only shows that if $lim_{x to x_0} f'(x)$ exists, then it's equal to $f'(x_0)$. But, as you say, it may not exist.
This is a rather classical exercise, which has been treated many times on this site, for example in this question.
answered Dec 15 '18 at 9:48
Hans LundmarkHans Lundmark
35.8k564115
answered Dec 15 '18 at 9:48
Hans LundmarkHans Lundmark
35.8k564115
answered Dec 15 '18 at 9:48
Hans LundmarkHans Lundmark
35.8k564115
answered Dec 15 '18 at 9:48
Hans LundmarkHans Lundmark
35.8k564115
35.8k564115
add a comment |
add a comment |
$begingroup$
Actually，we can make a comment for the reasoning like this:
The fact that $limlimits_{xi to x_0}f'(xi)$ exists dose not imply $limlimits_{x to x_0}f'(x)$ also exsits, because, according to Heine's Theorem, the latter one necessitates that $f'(x_n)$ converges for any sequence $x_n to x_0$. As we can see, $xi_n$ is only a specific sequence. Even though $f'(xi_n)$ is convergent, this is not enough to guarantee $f'(x_n)$ converges as well.
answered Dec 15 '18 at 11:36
mengdie1982mengdie1982
4,932618
$endgroup$
$begingroup$
You mean $lim_{xto x_0}f'(xi)$ instead of $lim_{xito x_0}f'(xi)$.
$endgroup$
– Paramanand Singh
Dec 20 '18 at 9:42
add a comment |
$begingroup$
Actually，we can make a comment for the reasoning like this:
The fact that $limlimits_{xi to x_0}f'(xi)$ exists dose not imply $limlimits_{x to x_0}f'(x)$ also exsits, because, according to Heine's Theorem, the latter one necessitates that $f'(x_n)$ converges for any sequence $x_n to x_0$. As we can see, $xi_n$ is only a specific sequence. Even though $f'(xi_n)$ is convergent, this is not enough to guarantee $f'(x_n)$ converges as well.
answered Dec 15 '18 at 11:36
mengdie1982mengdie1982
4,932618
$endgroup$
$begingroup$
You mean $lim_{xto x_0}f'(xi)$ instead of $lim_{xito x_0}f'(xi)$.
$endgroup$
– Paramanand Singh
Dec 20 '18 at 9:42
add a comment |
$begingroup$
Actually，we can make a comment for the reasoning like this:
The fact that $limlimits_{xi to x_0}f'(xi)$ exists dose not imply $limlimits_{x to x_0}f'(x)$ also exsits, because, according to Heine's Theorem, the latter one necessitates that $f'(x_n)$ converges for any sequence $x_n to x_0$. As we can see, $xi_n$ is only a specific sequence. Even though $f'(xi_n)$ is convergent, this is not enough to guarantee $f'(x_n)$ converges as well.
answered Dec 15 '18 at 11:36
mengdie1982mengdie1982
4,932618
$endgroup$
Actually，we can make a comment for the reasoning like this:
The fact that $limlimits_{xi to x_0}f'(xi)$ exists dose not imply $limlimits_{x to x_0}f'(x)$ also exsits, because, according to Heine's Theorem, the latter one necessitates that $f'(x_n)$ converges for any sequence $x_n to x_0$. As we can see, $xi_n$ is only a specific sequence. Even though $f'(xi_n)$ is convergent, this is not enough to guarantee $f'(x_n)$ converges as well.
answered Dec 15 '18 at 11:36
mengdie1982mengdie1982
4,932618
answered Dec 15 '18 at 11:36
mengdie1982mengdie1982
4,932618
answered Dec 15 '18 at 11:36
mengdie1982mengdie1982
4,932618
answered Dec 15 '18 at 11:36
mengdie1982mengdie1982
4,932618
4,932618
$begingroup$
You mean $lim_{xto x_0}f'(xi)$ instead of $lim_{xito x_0}f'(xi)$.
$endgroup$
– Paramanand Singh
Dec 20 '18 at 9:42
add a comment |
$begingroup$
You mean $lim_{xto x_0}f'(xi)$ instead of $lim_{xito x_0}f'(xi)$.
$endgroup$
– Paramanand Singh
Dec 20 '18 at 9:42
$begingroup$
You mean $lim_{xto x_0}f'(xi)$ instead of $lim_{xito x_0}f'(xi)$.
$endgroup$
– Paramanand Singh
Dec 20 '18 at 9:42
$begingroup$
You mean $lim_{xto x_0}f'(xi)$ instead of $lim_{xito x_0}f'(xi)$.
$endgroup$
– Paramanand Singh
Dec 20 '18 at 9:42
add a comment |
$begingroup$
Actually, you are right, this also called 'theorem of derivative limitation' in some textbooks.
If you already assumed that $f(x)$ is derivable for any $x in mathbb{R}$.
With an additional condition the limitation of derivative $lim_{x to x_{0}}f'(x)$ exist.
You can write your conclusion:
$$lim_{x to x_{0}}f'(x)=f'(x_{0})$$
The condition $f(x)$ is derivable for any $x in mathbb{R}$ can also be weaken to $f(x)$ is continuous in $U(x_{0})$ which is the neighborhood of $x_{0}$, and derivable in $U°(x_{0})$ which is the punctured neighborhood of $x_{0}$.
The only problem is that you have not assumed the limitation of your derivative should always exist, which means the derivative can not divergence to infinity, like you say sometimes it would be the discontinuity point of the second kind.
And this theorem can be simply remembered as in other 'not so strict (please notice the comment below)' ways like:
If a continuous function has (finite) derivative, its derivative function is also continuous. (but maybe not derivable)
answered Dec 15 '18 at 15:20
NanayajitzukiNanayajitzuki
3185
$endgroup$
$begingroup$
It would be fair to say that you're wrong. The fact $f(x)$ is derivable does not imply $f'(x)$ is continuous. Here is a counterexample: $$f(x)=begin{cases}x^2sindfrac{1}{x},&x neq 0,\0,&x=0.end{cases}.$$ You may verify that $f(x)$ is derivable over $(-infty,+infty)$，but $f'(x)$ is not continuous at $x=0$.
$endgroup$
– mengdie1982
Dec 15 '18 at 16:16
$begingroup$
But of course, if we add a condition that $limlimits_{x to x_0}f'(x)$ exists, then $limlimits_{x to x_0}f'(x)=f'(x_0)$. In another word, $f'(x)$ is continuous at $x=x_0$ now！
$endgroup$
– mengdie1982
Dec 15 '18 at 16:24
$begingroup$
@mengdie1982 You still remind a good example for me. Yes, I should explain that the 'not so strict' place this the limitation of derivative must be exist not only the derivative exist. Your comment point out it.
$endgroup$
– Nanayajitzuki
Dec 15 '18 at 16:30
add a comment |
$begingroup$
Actually, you are right, this also called 'theorem of derivative limitation' in some textbooks.
If you already assumed that $f(x)$ is derivable for any $x in mathbb{R}$.
With an additional condition the limitation of derivative $lim_{x to x_{0}}f'(x)$ exist.
You can write your conclusion:
$$lim_{x to x_{0}}f'(x)=f'(x_{0})$$
The condition $f(x)$ is derivable for any $x in mathbb{R}$ can also be weaken to $f(x)$ is continuous in $U(x_{0})$ which is the neighborhood of $x_{0}$, and derivable in $U°(x_{0})$ which is the punctured neighborhood of $x_{0}$.
The only problem is that you have not assumed the limitation of your derivative should always exist, which means the derivative can not divergence to infinity, like you say sometimes it would be the discontinuity point of the second kind.
And this theorem can be simply remembered as in other 'not so strict (please notice the comment below)' ways like:
If a continuous function has (finite) derivative, its derivative function is also continuous. (but maybe not derivable)
answered Dec 15 '18 at 15:20
NanayajitzukiNanayajitzuki
3185
$endgroup$
$begingroup$
It would be fair to say that you're wrong. The fact $f(x)$ is derivable does not imply $f'(x)$ is continuous. Here is a counterexample: $$f(x)=begin{cases}x^2sindfrac{1}{x},&x neq 0,\0,&x=0.end{cases}.$$ You may verify that $f(x)$ is derivable over $(-infty,+infty)$，but $f'(x)$ is not continuous at $x=0$.
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– mengdie1982
Dec 15 '18 at 16:16
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But of course, if we add a condition that $limlimits_{x to x_0}f'(x)$ exists, then $limlimits_{x to x_0}f'(x)=f'(x_0)$. In another word, $f'(x)$ is continuous at $x=x_0$ now！
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– mengdie1982
Dec 15 '18 at 16:24
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@mengdie1982 You still remind a good example for me. Yes, I should explain that the 'not so strict' place this the limitation of derivative must be exist not only the derivative exist. Your comment point out it.
$endgroup$
– Nanayajitzuki
Dec 15 '18 at 16:30
add a comment |
$begingroup$
Actually, you are right, this also called 'theorem of derivative limitation' in some textbooks.
If you already assumed that $f(x)$ is derivable for any $x in mathbb{R}$.
With an additional condition the limitation of derivative $lim_{x to x_{0}}f'(x)$ exist.
You can write your conclusion:
$$lim_{x to x_{0}}f'(x)=f'(x_{0})$$
The condition $f(x)$ is derivable for any $x in mathbb{R}$ can also be weaken to $f(x)$ is continuous in $U(x_{0})$ which is the neighborhood of $x_{0}$, and derivable in $U°(x_{0})$ which is the punctured neighborhood of $x_{0}$.
The only problem is that you have not assumed the limitation of your derivative should always exist, which means the derivative can not divergence to infinity, like you say sometimes it would be the discontinuity point of the second kind.
And this theorem can be simply remembered as in other 'not so strict (please notice the comment below)' ways like:
If a continuous function has (finite) derivative, its derivative function is also continuous. (but maybe not derivable)
answered Dec 15 '18 at 15:20
NanayajitzukiNanayajitzuki
3185
$endgroup$
Actually, you are right, this also called 'theorem of derivative limitation' in some textbooks.
If you already assumed that $f(x)$ is derivable for any $x in mathbb{R}$.
With an additional condition the limitation of derivative $lim_{x to x_{0}}f'(x)$ exist.
You can write your conclusion:
$$lim_{x to x_{0}}f'(x)=f'(x_{0})$$
The condition $f(x)$ is derivable for any $x in mathbb{R}$ can also be weaken to $f(x)$ is continuous in $U(x_{0})$ which is the neighborhood of $x_{0}$, and derivable in $U°(x_{0})$ which is the punctured neighborhood of $x_{0}$.
The only problem is that you have not assumed the limitation of your derivative should always exist, which means the derivative can not divergence to infinity, like you say sometimes it would be the discontinuity point of the second kind.
And this theorem can be simply remembered as in other 'not so strict (please notice the comment below)' ways like:
If a continuous function has (finite) derivative, its derivative function is also continuous. (but maybe not derivable)
answered Dec 15 '18 at 15:20
NanayajitzukiNanayajitzuki
3185
edited Dec 15 '18 at 16:54
answered Dec 15 '18 at 15:20
NanayajitzukiNanayajitzuki
3185
answered Dec 15 '18 at 15:20
NanayajitzukiNanayajitzuki
3185
answered Dec 15 '18 at 15:20
NanayajitzukiNanayajitzuki
3185
3185
$begingroup$
It would be fair to say that you're wrong. The fact $f(x)$ is derivable does not imply $f'(x)$ is continuous. Here is a counterexample: $$f(x)=begin{cases}x^2sindfrac{1}{x},&x neq 0,\0,&x=0.end{cases}.$$ You may verify that $f(x)$ is derivable over $(-infty,+infty)$，but $f'(x)$ is not continuous at $x=0$.
$endgroup$
– mengdie1982
Dec 15 '18 at 16:16
$begingroup$
But of course, if we add a condition that $limlimits_{x to x_0}f'(x)$ exists, then $limlimits_{x to x_0}f'(x)=f'(x_0)$. In another word, $f'(x)$ is continuous at $x=x_0$ now！
$endgroup$
– mengdie1982
Dec 15 '18 at 16:24
$begingroup$
@mengdie1982 You still remind a good example for me. Yes, I should explain that the 'not so strict' place this the limitation of derivative must be exist not only the derivative exist. Your comment point out it.
$endgroup$
– Nanayajitzuki
Dec 15 '18 at 16:30
add a comment |
$begingroup$
It would be fair to say that you're wrong. The fact $f(x)$ is derivable does not imply $f'(x)$ is continuous. Here is a counterexample: $$f(x)=begin{cases}x^2sindfrac{1}{x},&x neq 0,\0,&x=0.end{cases}.$$ You may verify that $f(x)$ is derivable over $(-infty,+infty)$，but $f'(x)$ is not continuous at $x=0$.
$endgroup$
– mengdie1982
Dec 15 '18 at 16:16
$begingroup$
But of course, if we add a condition that $limlimits_{x to x_0}f'(x)$ exists, then $limlimits_{x to x_0}f'(x)=f'(x_0)$. In another word, $f'(x)$ is continuous at $x=x_0$ now！
$endgroup$
– mengdie1982
Dec 15 '18 at 16:24
$begingroup$
@mengdie1982 You still remind a good example for me. Yes, I should explain that the 'not so strict' place this the limitation of derivative must be exist not only the derivative exist. Your comment point out it.
$endgroup$
– Nanayajitzuki
Dec 15 '18 at 16:30
$begingroup$
It would be fair to say that you're wrong. The fact $f(x)$ is derivable does not imply $f'(x)$ is continuous. Here is a counterexample: $$f(x)=begin{cases}x^2sindfrac{1}{x},&x neq 0,\0,&x=0.end{cases}.$$ You may verify that $f(x)$ is derivable over $(-infty,+infty)$，but $f'(x)$ is not continuous at $x=0$.
$endgroup$
– mengdie1982
Dec 15 '18 at 16:16
$begingroup$
It would be fair to say that you're wrong. The fact $f(x)$ is derivable does not imply $f'(x)$ is continuous. Here is a counterexample: $$f(x)=begin{cases}x^2sindfrac{1}{x},&x neq 0,\0,&x=0.end{cases}.$$ You may verify that $f(x)$ is derivable over $(-infty,+infty)$，but $f'(x)$ is not continuous at $x=0$.
$endgroup$
– mengdie1982
Dec 15 '18 at 16:16
$begingroup$
But of course, if we add a condition that $limlimits_{x to x_0}f'(x)$ exists, then $limlimits_{x to x_0}f'(x)=f'(x_0)$. In another word, $f'(x)$ is continuous at $x=x_0$ now！
$endgroup$
– mengdie1982
Dec 15 '18 at 16:24
$begingroup$
But of course, if we add a condition that $limlimits_{x to x_0}f'(x)$ exists, then $limlimits_{x to x_0}f'(x)=f'(x_0)$. In another word, $f'(x)$ is continuous at $x=x_0$ now！
$endgroup$
– mengdie1982
Dec 15 '18 at 16:24
$begingroup$
@mengdie1982 You still remind a good example for me. Yes, I should explain that the 'not so strict' place this the limitation of derivative must be exist not only the derivative exist. Your comment point out it.
$endgroup$
– Nanayajitzuki
Dec 15 '18 at 16:30
$begingroup$
@mengdie1982 You still remind a good example for me. Yes, I should explain that the 'not so strict' place this the limitation of derivative must be exist not only the derivative exist. Your comment point out it.
$endgroup$
– Nanayajitzuki
Dec 15 '18 at 16:30
add a comment |
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$begingroup$
Recall that $xi$ is actually a function of $x$. Thus, $lim_{xrightarrow x_0} f'(xi(x))$ does not imply $lim_{xi rightarrow x_0} f'(xi)$
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– Severin Schraven
Dec 15 '18 at 9:08
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it is just a $ksi$ not every $x$...
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– dmtri
Dec 15 '18 at 9:09
$begingroup$
@SeverinSchraven Do you know the squeeze theorem? $x_0 lessgtr xi(x) lessgtr x$. Let $x to x_0$，then $xi(x) to x_0$. It has no fault here.
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– mengdie1982
Dec 15 '18 at 9:25
2
$begingroup$
Reread my comment, I did NOT claim that $xi(x) rightarrow x_0$ is wrong. Your line of reasoning is the following: Let $$ f(x) = begin{cases} 0;& xin mathbb{Q} \ 1,& xin mathbb{R}setminus mathbb{Q} end{cases} $$ now take $x_0=0$ and $xi(x)$ be a rational number between $0$ and $x$. Clearly we have $lim_{xrightarrow 0} f(xi(x))=0$ and thus $lim_{xi rightarrow 0} f(xi)= lim_{x rightarrow 0} f(xi(x))$. But $lim_{xi rightarrow 0} f(xi)$ does not even exist. That is, because you do not hit every number with $xi(x)$.
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– Severin Schraven
Dec 15 '18 at 9:32
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@SeverinSchraven You have pointed out some essential thing！
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– mengdie1982
Dec 15 '18 at 9:36