Is $limlimits_{xto x_0}f'(x)=f'(x_0)$?
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Let $f$ be a function defined in the open interval $(a,b)$ and let $x_0in(a,b)$. Suppose in addition that $f'(x)$ exists for all $x_0neq xin(a,b)$. Is the following statement true:
If $limlimits_{xto x_0}f'(x)$ exists, then $f'(x_0)$ exists and $limlimits_{xto x_0}f'(x)=f'(x_0)$.
Thanks!
calculus
$endgroup$
add a comment |
$begingroup$
Let $f$ be a function defined in the open interval $(a,b)$ and let $x_0in(a,b)$. Suppose in addition that $f'(x)$ exists for all $x_0neq xin(a,b)$. Is the following statement true:
If $limlimits_{xto x_0}f'(x)$ exists, then $f'(x_0)$ exists and $limlimits_{xto x_0}f'(x)=f'(x_0)$.
Thanks!
calculus
$endgroup$
$begingroup$
The statement in your question is saying that $f'(x)$ is continuous at some $x_0$. Now you must ask yourself if every derivative is continuous.
$endgroup$
– Masacroso
Jun 20 '16 at 11:36
1
$begingroup$
No, this is weaker than saying $f'$ is continuous.
$endgroup$
– GEdgar
Jun 20 '16 at 21:07
add a comment |
$begingroup$
Let $f$ be a function defined in the open interval $(a,b)$ and let $x_0in(a,b)$. Suppose in addition that $f'(x)$ exists for all $x_0neq xin(a,b)$. Is the following statement true:
If $limlimits_{xto x_0}f'(x)$ exists, then $f'(x_0)$ exists and $limlimits_{xto x_0}f'(x)=f'(x_0)$.
Thanks!
calculus
$endgroup$
Let $f$ be a function defined in the open interval $(a,b)$ and let $x_0in(a,b)$. Suppose in addition that $f'(x)$ exists for all $x_0neq xin(a,b)$. Is the following statement true:
If $limlimits_{xto x_0}f'(x)$ exists, then $f'(x_0)$ exists and $limlimits_{xto x_0}f'(x)=f'(x_0)$.
Thanks!
calculus
calculus
edited Jun 20 '16 at 20:16
snulty
2,86811338
2,86811338
asked Jun 20 '16 at 11:29
boazboaz
2,400614
2,400614
$begingroup$
The statement in your question is saying that $f'(x)$ is continuous at some $x_0$. Now you must ask yourself if every derivative is continuous.
$endgroup$
– Masacroso
Jun 20 '16 at 11:36
1
$begingroup$
No, this is weaker than saying $f'$ is continuous.
$endgroup$
– GEdgar
Jun 20 '16 at 21:07
add a comment |
$begingroup$
The statement in your question is saying that $f'(x)$ is continuous at some $x_0$. Now you must ask yourself if every derivative is continuous.
$endgroup$
– Masacroso
Jun 20 '16 at 11:36
1
$begingroup$
No, this is weaker than saying $f'$ is continuous.
$endgroup$
– GEdgar
Jun 20 '16 at 21:07
$begingroup$
The statement in your question is saying that $f'(x)$ is continuous at some $x_0$. Now you must ask yourself if every derivative is continuous.
$endgroup$
– Masacroso
Jun 20 '16 at 11:36
$begingroup$
The statement in your question is saying that $f'(x)$ is continuous at some $x_0$. Now you must ask yourself if every derivative is continuous.
$endgroup$
– Masacroso
Jun 20 '16 at 11:36
1
1
$begingroup$
No, this is weaker than saying $f'$ is continuous.
$endgroup$
– GEdgar
Jun 20 '16 at 21:07
$begingroup$
No, this is weaker than saying $f'$ is continuous.
$endgroup$
– GEdgar
Jun 20 '16 at 21:07
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
A qualified "yes": If $f$ is continuous at $x_{0}$, and if $limlimits_{x to x_{0}}f'(x) = L$ exists, then $f$ is differentiable at $x_{0}$, and $f'(x_{0}) = L$.
Qualitatively, the derivative of a continuous function cannot have a removable discontinuity. (If you don't assume $f$ is continuous, then $f$ itself can have a removable or jump discontinuity.)
The claim follows from the Mean Value Theorem: If $delta > 0$ and $f'(x_{0} + h)$ is defined for $0 < |h| < delta$, then for each such $h$, the Mean Value Theorem (applied to $f$ on the closed interval with endpoints $x_{0}$ and $x_{0} + h$) says there is a $t$ between $x_{0}$ and $x_{0} + h$ such that
$$
frac{f(x_{0} + h) - f(x_{0})}{h} = f'(t).
$$
Since $|t - x_{0}| < |h|$, taking the limit as $h to 0$ forces $t - x_{0} to 0$, as well, so
$$
f'(x_{0}) = lim_{h to 0} frac{f(x_{0} + h) - f(x_{0})}{h}
= lim_{t to x_{0}} f'(t).
$$
(Continuity of $f$ was needed to invoke the Mean Value Theorem.)
$endgroup$
2
$begingroup$
Thanks Andrew, a great answer.
$endgroup$
– boaz
Jun 20 '16 at 12:04
$begingroup$
You're welcome. :) (And, fixed a couple of typos.)
$endgroup$
– Andrew D. Hwang
Jun 20 '16 at 20:03
1
$begingroup$
This is a useful fact to know when you do those popular tricky questions like: is $f(x)=x^2sin(x), x ne 0, f(0)=0$ differentiable everywhere?
$endgroup$
– GEdgar
Jun 20 '16 at 21:10
$begingroup$
Since you assume completeness, then you can also get to the same result by using de l'Hôpital's rule to solve the limit. However the statement works also in a non complete field such as Q, I think. Have a look at my my answer, if you like.
$endgroup$
– Matteo
Dec 18 '18 at 9:42
add a comment |
$begingroup$
No, it is not true. For a simple counterexample take the indicator function $chi_{[0,1)}$ in $xin(-1,1)$ and take $x_0$ to be 0.
$endgroup$
add a comment |
$begingroup$
Not neccessarily true. We need $f'$ to be continuous at $x_0$. Actually the conditions that $lim_{xto x_0} f'(x)$ and $f'(x_0)$ exist and they are equal to each other are the conditions for the function $f'$ to be continous.
Here's a counter-example: Consider the function:
$$f(x) = begin{cases} 0, & mbox{if } mbox{$x le 0$} \ x, & mbox{if } mbox{$x>0$} end{cases}$$
Obviously $f$ is defined on $mathbb{R}$, but the function isn't differentiable at $0$ and $f'$ is discontinuous at $0$. So by choosing $x_0 = 0$ all the conditions are satisfied, but the result isn't, hence the implication is wrong.
$endgroup$
$begingroup$
In the proposed example, $f'$ has one-sided limits, but does not have a two-sided limit...?
$endgroup$
– Andrew D. Hwang
Jun 20 '16 at 11:56
$begingroup$
@AndrewD.Hwang Yeah, both one-sided limits exist, but they are not equal to each other, hence the two-sided limit doesn't exists. At the end by checking the graph we can see that it's continuous, therefore both one-sided limits exist, but it's not "smooth" so we expect them not to be equal to each other.
$endgroup$
– Stefan4024
Jun 20 '16 at 11:59
1
$begingroup$
Please, change “$f'$ is continuous” into “$f$ is continuous at $x_0$”.
$endgroup$
– egreg
Jun 20 '16 at 21:25
1
$begingroup$
This is an incorrect answer to the question. $lim_{xto 0} f'(x)$ does not exist in your example.
$endgroup$
– zhw.
Jun 20 '16 at 21:39
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
A qualified "yes": If $f$ is continuous at $x_{0}$, and if $limlimits_{x to x_{0}}f'(x) = L$ exists, then $f$ is differentiable at $x_{0}$, and $f'(x_{0}) = L$.
Qualitatively, the derivative of a continuous function cannot have a removable discontinuity. (If you don't assume $f$ is continuous, then $f$ itself can have a removable or jump discontinuity.)
The claim follows from the Mean Value Theorem: If $delta > 0$ and $f'(x_{0} + h)$ is defined for $0 < |h| < delta$, then for each such $h$, the Mean Value Theorem (applied to $f$ on the closed interval with endpoints $x_{0}$ and $x_{0} + h$) says there is a $t$ between $x_{0}$ and $x_{0} + h$ such that
$$
frac{f(x_{0} + h) - f(x_{0})}{h} = f'(t).
$$
Since $|t - x_{0}| < |h|$, taking the limit as $h to 0$ forces $t - x_{0} to 0$, as well, so
$$
f'(x_{0}) = lim_{h to 0} frac{f(x_{0} + h) - f(x_{0})}{h}
= lim_{t to x_{0}} f'(t).
$$
(Continuity of $f$ was needed to invoke the Mean Value Theorem.)
$endgroup$
2
$begingroup$
Thanks Andrew, a great answer.
$endgroup$
– boaz
Jun 20 '16 at 12:04
$begingroup$
You're welcome. :) (And, fixed a couple of typos.)
$endgroup$
– Andrew D. Hwang
Jun 20 '16 at 20:03
1
$begingroup$
This is a useful fact to know when you do those popular tricky questions like: is $f(x)=x^2sin(x), x ne 0, f(0)=0$ differentiable everywhere?
$endgroup$
– GEdgar
Jun 20 '16 at 21:10
$begingroup$
Since you assume completeness, then you can also get to the same result by using de l'Hôpital's rule to solve the limit. However the statement works also in a non complete field such as Q, I think. Have a look at my my answer, if you like.
$endgroup$
– Matteo
Dec 18 '18 at 9:42
add a comment |
$begingroup$
A qualified "yes": If $f$ is continuous at $x_{0}$, and if $limlimits_{x to x_{0}}f'(x) = L$ exists, then $f$ is differentiable at $x_{0}$, and $f'(x_{0}) = L$.
Qualitatively, the derivative of a continuous function cannot have a removable discontinuity. (If you don't assume $f$ is continuous, then $f$ itself can have a removable or jump discontinuity.)
The claim follows from the Mean Value Theorem: If $delta > 0$ and $f'(x_{0} + h)$ is defined for $0 < |h| < delta$, then for each such $h$, the Mean Value Theorem (applied to $f$ on the closed interval with endpoints $x_{0}$ and $x_{0} + h$) says there is a $t$ between $x_{0}$ and $x_{0} + h$ such that
$$
frac{f(x_{0} + h) - f(x_{0})}{h} = f'(t).
$$
Since $|t - x_{0}| < |h|$, taking the limit as $h to 0$ forces $t - x_{0} to 0$, as well, so
$$
f'(x_{0}) = lim_{h to 0} frac{f(x_{0} + h) - f(x_{0})}{h}
= lim_{t to x_{0}} f'(t).
$$
(Continuity of $f$ was needed to invoke the Mean Value Theorem.)
$endgroup$
2
$begingroup$
Thanks Andrew, a great answer.
$endgroup$
– boaz
Jun 20 '16 at 12:04
$begingroup$
You're welcome. :) (And, fixed a couple of typos.)
$endgroup$
– Andrew D. Hwang
Jun 20 '16 at 20:03
1
$begingroup$
This is a useful fact to know when you do those popular tricky questions like: is $f(x)=x^2sin(x), x ne 0, f(0)=0$ differentiable everywhere?
$endgroup$
– GEdgar
Jun 20 '16 at 21:10
$begingroup$
Since you assume completeness, then you can also get to the same result by using de l'Hôpital's rule to solve the limit. However the statement works also in a non complete field such as Q, I think. Have a look at my my answer, if you like.
$endgroup$
– Matteo
Dec 18 '18 at 9:42
add a comment |
$begingroup$
A qualified "yes": If $f$ is continuous at $x_{0}$, and if $limlimits_{x to x_{0}}f'(x) = L$ exists, then $f$ is differentiable at $x_{0}$, and $f'(x_{0}) = L$.
Qualitatively, the derivative of a continuous function cannot have a removable discontinuity. (If you don't assume $f$ is continuous, then $f$ itself can have a removable or jump discontinuity.)
The claim follows from the Mean Value Theorem: If $delta > 0$ and $f'(x_{0} + h)$ is defined for $0 < |h| < delta$, then for each such $h$, the Mean Value Theorem (applied to $f$ on the closed interval with endpoints $x_{0}$ and $x_{0} + h$) says there is a $t$ between $x_{0}$ and $x_{0} + h$ such that
$$
frac{f(x_{0} + h) - f(x_{0})}{h} = f'(t).
$$
Since $|t - x_{0}| < |h|$, taking the limit as $h to 0$ forces $t - x_{0} to 0$, as well, so
$$
f'(x_{0}) = lim_{h to 0} frac{f(x_{0} + h) - f(x_{0})}{h}
= lim_{t to x_{0}} f'(t).
$$
(Continuity of $f$ was needed to invoke the Mean Value Theorem.)
$endgroup$
A qualified "yes": If $f$ is continuous at $x_{0}$, and if $limlimits_{x to x_{0}}f'(x) = L$ exists, then $f$ is differentiable at $x_{0}$, and $f'(x_{0}) = L$.
Qualitatively, the derivative of a continuous function cannot have a removable discontinuity. (If you don't assume $f$ is continuous, then $f$ itself can have a removable or jump discontinuity.)
The claim follows from the Mean Value Theorem: If $delta > 0$ and $f'(x_{0} + h)$ is defined for $0 < |h| < delta$, then for each such $h$, the Mean Value Theorem (applied to $f$ on the closed interval with endpoints $x_{0}$ and $x_{0} + h$) says there is a $t$ between $x_{0}$ and $x_{0} + h$ such that
$$
frac{f(x_{0} + h) - f(x_{0})}{h} = f'(t).
$$
Since $|t - x_{0}| < |h|$, taking the limit as $h to 0$ forces $t - x_{0} to 0$, as well, so
$$
f'(x_{0}) = lim_{h to 0} frac{f(x_{0} + h) - f(x_{0})}{h}
= lim_{t to x_{0}} f'(t).
$$
(Continuity of $f$ was needed to invoke the Mean Value Theorem.)
edited Jun 20 '16 at 20:02
answered Jun 20 '16 at 11:53
Andrew D. HwangAndrew D. Hwang
53.1k448114
53.1k448114
2
$begingroup$
Thanks Andrew, a great answer.
$endgroup$
– boaz
Jun 20 '16 at 12:04
$begingroup$
You're welcome. :) (And, fixed a couple of typos.)
$endgroup$
– Andrew D. Hwang
Jun 20 '16 at 20:03
1
$begingroup$
This is a useful fact to know when you do those popular tricky questions like: is $f(x)=x^2sin(x), x ne 0, f(0)=0$ differentiable everywhere?
$endgroup$
– GEdgar
Jun 20 '16 at 21:10
$begingroup$
Since you assume completeness, then you can also get to the same result by using de l'Hôpital's rule to solve the limit. However the statement works also in a non complete field such as Q, I think. Have a look at my my answer, if you like.
$endgroup$
– Matteo
Dec 18 '18 at 9:42
add a comment |
2
$begingroup$
Thanks Andrew, a great answer.
$endgroup$
– boaz
Jun 20 '16 at 12:04
$begingroup$
You're welcome. :) (And, fixed a couple of typos.)
$endgroup$
– Andrew D. Hwang
Jun 20 '16 at 20:03
1
$begingroup$
This is a useful fact to know when you do those popular tricky questions like: is $f(x)=x^2sin(x), x ne 0, f(0)=0$ differentiable everywhere?
$endgroup$
– GEdgar
Jun 20 '16 at 21:10
$begingroup$
Since you assume completeness, then you can also get to the same result by using de l'Hôpital's rule to solve the limit. However the statement works also in a non complete field such as Q, I think. Have a look at my my answer, if you like.
$endgroup$
– Matteo
Dec 18 '18 at 9:42
2
2
$begingroup$
Thanks Andrew, a great answer.
$endgroup$
– boaz
Jun 20 '16 at 12:04
$begingroup$
Thanks Andrew, a great answer.
$endgroup$
– boaz
Jun 20 '16 at 12:04
$begingroup$
You're welcome. :) (And, fixed a couple of typos.)
$endgroup$
– Andrew D. Hwang
Jun 20 '16 at 20:03
$begingroup$
You're welcome. :) (And, fixed a couple of typos.)
$endgroup$
– Andrew D. Hwang
Jun 20 '16 at 20:03
1
1
$begingroup$
This is a useful fact to know when you do those popular tricky questions like: is $f(x)=x^2sin(x), x ne 0, f(0)=0$ differentiable everywhere?
$endgroup$
– GEdgar
Jun 20 '16 at 21:10
$begingroup$
This is a useful fact to know when you do those popular tricky questions like: is $f(x)=x^2sin(x), x ne 0, f(0)=0$ differentiable everywhere?
$endgroup$
– GEdgar
Jun 20 '16 at 21:10
$begingroup$
Since you assume completeness, then you can also get to the same result by using de l'Hôpital's rule to solve the limit. However the statement works also in a non complete field such as Q, I think. Have a look at my my answer, if you like.
$endgroup$
– Matteo
Dec 18 '18 at 9:42
$begingroup$
Since you assume completeness, then you can also get to the same result by using de l'Hôpital's rule to solve the limit. However the statement works also in a non complete field such as Q, I think. Have a look at my my answer, if you like.
$endgroup$
– Matteo
Dec 18 '18 at 9:42
add a comment |
$begingroup$
No, it is not true. For a simple counterexample take the indicator function $chi_{[0,1)}$ in $xin(-1,1)$ and take $x_0$ to be 0.
$endgroup$
add a comment |
$begingroup$
No, it is not true. For a simple counterexample take the indicator function $chi_{[0,1)}$ in $xin(-1,1)$ and take $x_0$ to be 0.
$endgroup$
add a comment |
$begingroup$
No, it is not true. For a simple counterexample take the indicator function $chi_{[0,1)}$ in $xin(-1,1)$ and take $x_0$ to be 0.
$endgroup$
No, it is not true. For a simple counterexample take the indicator function $chi_{[0,1)}$ in $xin(-1,1)$ and take $x_0$ to be 0.
answered Jun 20 '16 at 11:33
user335721
add a comment |
add a comment |
$begingroup$
Not neccessarily true. We need $f'$ to be continuous at $x_0$. Actually the conditions that $lim_{xto x_0} f'(x)$ and $f'(x_0)$ exist and they are equal to each other are the conditions for the function $f'$ to be continous.
Here's a counter-example: Consider the function:
$$f(x) = begin{cases} 0, & mbox{if } mbox{$x le 0$} \ x, & mbox{if } mbox{$x>0$} end{cases}$$
Obviously $f$ is defined on $mathbb{R}$, but the function isn't differentiable at $0$ and $f'$ is discontinuous at $0$. So by choosing $x_0 = 0$ all the conditions are satisfied, but the result isn't, hence the implication is wrong.
$endgroup$
$begingroup$
In the proposed example, $f'$ has one-sided limits, but does not have a two-sided limit...?
$endgroup$
– Andrew D. Hwang
Jun 20 '16 at 11:56
$begingroup$
@AndrewD.Hwang Yeah, both one-sided limits exist, but they are not equal to each other, hence the two-sided limit doesn't exists. At the end by checking the graph we can see that it's continuous, therefore both one-sided limits exist, but it's not "smooth" so we expect them not to be equal to each other.
$endgroup$
– Stefan4024
Jun 20 '16 at 11:59
1
$begingroup$
Please, change “$f'$ is continuous” into “$f$ is continuous at $x_0$”.
$endgroup$
– egreg
Jun 20 '16 at 21:25
1
$begingroup$
This is an incorrect answer to the question. $lim_{xto 0} f'(x)$ does not exist in your example.
$endgroup$
– zhw.
Jun 20 '16 at 21:39
add a comment |
$begingroup$
Not neccessarily true. We need $f'$ to be continuous at $x_0$. Actually the conditions that $lim_{xto x_0} f'(x)$ and $f'(x_0)$ exist and they are equal to each other are the conditions for the function $f'$ to be continous.
Here's a counter-example: Consider the function:
$$f(x) = begin{cases} 0, & mbox{if } mbox{$x le 0$} \ x, & mbox{if } mbox{$x>0$} end{cases}$$
Obviously $f$ is defined on $mathbb{R}$, but the function isn't differentiable at $0$ and $f'$ is discontinuous at $0$. So by choosing $x_0 = 0$ all the conditions are satisfied, but the result isn't, hence the implication is wrong.
$endgroup$
$begingroup$
In the proposed example, $f'$ has one-sided limits, but does not have a two-sided limit...?
$endgroup$
– Andrew D. Hwang
Jun 20 '16 at 11:56
$begingroup$
@AndrewD.Hwang Yeah, both one-sided limits exist, but they are not equal to each other, hence the two-sided limit doesn't exists. At the end by checking the graph we can see that it's continuous, therefore both one-sided limits exist, but it's not "smooth" so we expect them not to be equal to each other.
$endgroup$
– Stefan4024
Jun 20 '16 at 11:59
1
$begingroup$
Please, change “$f'$ is continuous” into “$f$ is continuous at $x_0$”.
$endgroup$
– egreg
Jun 20 '16 at 21:25
1
$begingroup$
This is an incorrect answer to the question. $lim_{xto 0} f'(x)$ does not exist in your example.
$endgroup$
– zhw.
Jun 20 '16 at 21:39
add a comment |
$begingroup$
Not neccessarily true. We need $f'$ to be continuous at $x_0$. Actually the conditions that $lim_{xto x_0} f'(x)$ and $f'(x_0)$ exist and they are equal to each other are the conditions for the function $f'$ to be continous.
Here's a counter-example: Consider the function:
$$f(x) = begin{cases} 0, & mbox{if } mbox{$x le 0$} \ x, & mbox{if } mbox{$x>0$} end{cases}$$
Obviously $f$ is defined on $mathbb{R}$, but the function isn't differentiable at $0$ and $f'$ is discontinuous at $0$. So by choosing $x_0 = 0$ all the conditions are satisfied, but the result isn't, hence the implication is wrong.
$endgroup$
Not neccessarily true. We need $f'$ to be continuous at $x_0$. Actually the conditions that $lim_{xto x_0} f'(x)$ and $f'(x_0)$ exist and they are equal to each other are the conditions for the function $f'$ to be continous.
Here's a counter-example: Consider the function:
$$f(x) = begin{cases} 0, & mbox{if } mbox{$x le 0$} \ x, & mbox{if } mbox{$x>0$} end{cases}$$
Obviously $f$ is defined on $mathbb{R}$, but the function isn't differentiable at $0$ and $f'$ is discontinuous at $0$. So by choosing $x_0 = 0$ all the conditions are satisfied, but the result isn't, hence the implication is wrong.
edited Jun 20 '16 at 21:26
answered Jun 20 '16 at 11:35
Stefan4024Stefan4024
30.6k63479
30.6k63479
$begingroup$
In the proposed example, $f'$ has one-sided limits, but does not have a two-sided limit...?
$endgroup$
– Andrew D. Hwang
Jun 20 '16 at 11:56
$begingroup$
@AndrewD.Hwang Yeah, both one-sided limits exist, but they are not equal to each other, hence the two-sided limit doesn't exists. At the end by checking the graph we can see that it's continuous, therefore both one-sided limits exist, but it's not "smooth" so we expect them not to be equal to each other.
$endgroup$
– Stefan4024
Jun 20 '16 at 11:59
1
$begingroup$
Please, change “$f'$ is continuous” into “$f$ is continuous at $x_0$”.
$endgroup$
– egreg
Jun 20 '16 at 21:25
1
$begingroup$
This is an incorrect answer to the question. $lim_{xto 0} f'(x)$ does not exist in your example.
$endgroup$
– zhw.
Jun 20 '16 at 21:39
add a comment |
$begingroup$
In the proposed example, $f'$ has one-sided limits, but does not have a two-sided limit...?
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– Andrew D. Hwang
Jun 20 '16 at 11:56
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@AndrewD.Hwang Yeah, both one-sided limits exist, but they are not equal to each other, hence the two-sided limit doesn't exists. At the end by checking the graph we can see that it's continuous, therefore both one-sided limits exist, but it's not "smooth" so we expect them not to be equal to each other.
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– Stefan4024
Jun 20 '16 at 11:59
1
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Please, change “$f'$ is continuous” into “$f$ is continuous at $x_0$”.
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– egreg
Jun 20 '16 at 21:25
1
$begingroup$
This is an incorrect answer to the question. $lim_{xto 0} f'(x)$ does not exist in your example.
$endgroup$
– zhw.
Jun 20 '16 at 21:39
$begingroup$
In the proposed example, $f'$ has one-sided limits, but does not have a two-sided limit...?
$endgroup$
– Andrew D. Hwang
Jun 20 '16 at 11:56
$begingroup$
In the proposed example, $f'$ has one-sided limits, but does not have a two-sided limit...?
$endgroup$
– Andrew D. Hwang
Jun 20 '16 at 11:56
$begingroup$
@AndrewD.Hwang Yeah, both one-sided limits exist, but they are not equal to each other, hence the two-sided limit doesn't exists. At the end by checking the graph we can see that it's continuous, therefore both one-sided limits exist, but it's not "smooth" so we expect them not to be equal to each other.
$endgroup$
– Stefan4024
Jun 20 '16 at 11:59
$begingroup$
@AndrewD.Hwang Yeah, both one-sided limits exist, but they are not equal to each other, hence the two-sided limit doesn't exists. At the end by checking the graph we can see that it's continuous, therefore both one-sided limits exist, but it's not "smooth" so we expect them not to be equal to each other.
$endgroup$
– Stefan4024
Jun 20 '16 at 11:59
1
1
$begingroup$
Please, change “$f'$ is continuous” into “$f$ is continuous at $x_0$”.
$endgroup$
– egreg
Jun 20 '16 at 21:25
$begingroup$
Please, change “$f'$ is continuous” into “$f$ is continuous at $x_0$”.
$endgroup$
– egreg
Jun 20 '16 at 21:25
1
1
$begingroup$
This is an incorrect answer to the question. $lim_{xto 0} f'(x)$ does not exist in your example.
$endgroup$
– zhw.
Jun 20 '16 at 21:39
$begingroup$
This is an incorrect answer to the question. $lim_{xto 0} f'(x)$ does not exist in your example.
$endgroup$
– zhw.
Jun 20 '16 at 21:39
add a comment |
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$begingroup$
The statement in your question is saying that $f'(x)$ is continuous at some $x_0$. Now you must ask yourself if every derivative is continuous.
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– Masacroso
Jun 20 '16 at 11:36
1
$begingroup$
No, this is weaker than saying $f'$ is continuous.
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– GEdgar
Jun 20 '16 at 21:07