If $lim limits_{x to c} f'(x) = l in Bbb R$. Does it mean $f$ is differentiable at $c$ and $f'(c) = l$.

up vote
0
down vote

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1

Suppose that $lim limits_{x to c} f'(x) = l in Bbb
R$.

1) Does it mean that $f$ is differentiable at $c$ and $f'(c) = l$?

2) Is it possible that $f'(c)$ does not exist?

3) What is the difference whether $f$ is continuous or not?

4) If $f$ is continuous, how to prove the statement?

real-analysis

share|cite|improve this question

edited Oct 31 at 9:58

Robert Z

90k1056128

asked Oct 31 at 9:27

kenkenb

15718

add a comment | 

up vote
0
down vote

favorite

1

Suppose that $lim limits_{x to c} f'(x) = l in Bbb
R$.

1) Does it mean that $f$ is differentiable at $c$ and $f'(c) = l$?

2) Is it possible that $f'(c)$ does not exist?

3) What is the difference whether $f$ is continuous or not?

4) If $f$ is continuous, how to prove the statement?

real-analysis

share|cite|improve this question

edited Oct 31 at 9:58

Robert Z

90k1056128

asked Oct 31 at 9:27

kenkenb

15718

add a comment | 

up vote
0
down vote

favorite

1

up vote
0
down vote

favorite

1

1

Suppose that $lim limits_{x to c} f'(x) = l in Bbb
R$.

1) Does it mean that $f$ is differentiable at $c$ and $f'(c) = l$?

2) Is it possible that $f'(c)$ does not exist?

3) What is the difference whether $f$ is continuous or not?

4) If $f$ is continuous, how to prove the statement?

real-analysis

share|cite|improve this question

edited Oct 31 at 9:58

Robert Z

90k1056128

asked Oct 31 at 9:27

kenkenb

15718

Suppose that $lim limits_{x to c} f'(x) = l in Bbb
R$.

1) Does it mean that $f$ is differentiable at $c$ and $f'(c) = l$?

2) Is it possible that $f'(c)$ does not exist?

3) What is the difference whether $f$ is continuous or not?

4) If $f$ is continuous, how to prove the statement?

real-analysis

real-analysis

share|cite|improve this question

edited Oct 31 at 9:58

Robert Z

90k1056128

asked Oct 31 at 9:27

kenkenb

15718

share|cite|improve this question

edited Oct 31 at 9:58

Robert Z

90k1056128

asked Oct 31 at 9:27

kenkenb

15718

share|cite|improve this question

share|cite|improve this question

edited Oct 31 at 9:58

Robert Z

90k1056128

edited Oct 31 at 9:58

Robert Z

90k1056128

edited Oct 31 at 9:58

Robert Z

90k1056128

90k1056128

asked Oct 31 at 9:27

kenkenb

15718

asked Oct 31 at 9:27

kenkenb

15718

asked Oct 31 at 9:27

kenkenb

15718

15718

add a comment | 

add a comment | 

2 Answers
2

active

oldest

votes

up vote
2
down vote



accepted

Yes, if $lim limits_{x to c} f'(x) = l in Bbb
R$ then $f$ is differentiable at $c$ provided that $f$ is continuous at $c$. Then by the Mean Value Theorem, there is $t_x$ between $x$ and $c$ such that
$$frac{f(x)-f(c)}{x-c}=f'(t_x).$$
It follows that, as $xto c$, we have $t_xto c$ and $f'(t_x)to l$.
Hence $f'(c)$ exists and it is equal to $l$.

Of course, if $f$ is not continuous at $c$ or if $f$ is not even defined at $c$, then $f$ is not differentiable at $c$. Take for example
$$f(x) = begin{cases}1 & text{if $x=0$}\ x & text{otherwise} end{cases}.$$
Then $lim_{xto 0}f'(x)=1$, but the $f$ is not differentiable at $0$:
$$lim_{xto 0^+}frac{f(x)-f(0)}{x-0}=lim_{xto 0^+}frac{x-1}{x}=-infty.$$

share|cite|improve this answer

edited Oct 31 at 9:49

answered Oct 31 at 9:33

Robert Z

90k1056128

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Could you give more hints?
  – kenkenb
  Oct 31 at 9:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Robert Z How do you get that $f'(t_x)to l$ as $t_x to c$ implies $f'(c) = l$ ? Wouldn't you require $f$ to be $C^1$ for that to happen?
  – Deepakms
  Oct 31 at 9:53
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Deepakms NO. It follows from the assumption that $lim limits_{x to c} f'(x) = l$.
  – Robert Z
  Oct 31 at 9:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @kenkenb Maybe you could show some effort and fill the gaps...
  – Robert Z
  Oct 31 at 10:16
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @ Robert Z ok. thanks for your effort anyway
  – kenkenb
  Oct 31 at 10:55
  
  
  

  
  
  
  

add a comment | 

up vote
2
down vote

It is possible that $f'(c)$ does not even exist. The existence of a limit of derivative does not even guarantee continuity of $f$ at $c$. For example consider a constant function with a jump discontinuity at $c$.
$$F(x) = begin{cases}0 & x neq 1 \ 1 & x=1end{cases}$$ is not differentiable at 1 even though the limit of derivative exist when $x$ tends to 1.

share|cite|improve this answer

edited Oct 31 at 9:39

answered Oct 31 at 9:32

Deepakms

368114

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  In fact , it is not even mean that $f$ is defined at $c$
  – Hagen von Eitzen
  Oct 31 at 10:12
  

  
  
  
  

add a comment | 

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote



accepted

Yes, if $lim limits_{x to c} f'(x) = l in Bbb
R$ then $f$ is differentiable at $c$ provided that $f$ is continuous at $c$. Then by the Mean Value Theorem, there is $t_x$ between $x$ and $c$ such that
$$frac{f(x)-f(c)}{x-c}=f'(t_x).$$
It follows that, as $xto c$, we have $t_xto c$ and $f'(t_x)to l$.
Hence $f'(c)$ exists and it is equal to $l$.

Of course, if $f$ is not continuous at $c$ or if $f$ is not even defined at $c$, then $f$ is not differentiable at $c$. Take for example
$$f(x) = begin{cases}1 & text{if $x=0$}\ x & text{otherwise} end{cases}.$$
Then $lim_{xto 0}f'(x)=1$, but the $f$ is not differentiable at $0$:
$$lim_{xto 0^+}frac{f(x)-f(0)}{x-0}=lim_{xto 0^+}frac{x-1}{x}=-infty.$$

share|cite|improve this answer

edited Oct 31 at 9:49

answered Oct 31 at 9:33

Robert Z

90k1056128

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Could you give more hints?
  – kenkenb
  Oct 31 at 9:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Robert Z How do you get that $f'(t_x)to l$ as $t_x to c$ implies $f'(c) = l$ ? Wouldn't you require $f$ to be $C^1$ for that to happen?
  – Deepakms
  Oct 31 at 9:53
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Deepakms NO. It follows from the assumption that $lim limits_{x to c} f'(x) = l$.
  – Robert Z
  Oct 31 at 9:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @kenkenb Maybe you could show some effort and fill the gaps...
  – Robert Z
  Oct 31 at 10:16
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @ Robert Z ok. thanks for your effort anyway
  – kenkenb
  Oct 31 at 10:55
  
  
  

  
  
  
  

add a comment | 

up vote
2
down vote



accepted

Yes, if $lim limits_{x to c} f'(x) = l in Bbb
R$ then $f$ is differentiable at $c$ provided that $f$ is continuous at $c$. Then by the Mean Value Theorem, there is $t_x$ between $x$ and $c$ such that
$$frac{f(x)-f(c)}{x-c}=f'(t_x).$$
It follows that, as $xto c$, we have $t_xto c$ and $f'(t_x)to l$.
Hence $f'(c)$ exists and it is equal to $l$.

Of course, if $f$ is not continuous at $c$ or if $f$ is not even defined at $c$, then $f$ is not differentiable at $c$. Take for example
$$f(x) = begin{cases}1 & text{if $x=0$}\ x & text{otherwise} end{cases}.$$
Then $lim_{xto 0}f'(x)=1$, but the $f$ is not differentiable at $0$:
$$lim_{xto 0^+}frac{f(x)-f(0)}{x-0}=lim_{xto 0^+}frac{x-1}{x}=-infty.$$

share|cite|improve this answer

edited Oct 31 at 9:49

answered Oct 31 at 9:33

Robert Z

90k1056128

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Could you give more hints?
  – kenkenb
  Oct 31 at 9:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Robert Z How do you get that $f'(t_x)to l$ as $t_x to c$ implies $f'(c) = l$ ? Wouldn't you require $f$ to be $C^1$ for that to happen?
  – Deepakms
  Oct 31 at 9:53
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Deepakms NO. It follows from the assumption that $lim limits_{x to c} f'(x) = l$.
  – Robert Z
  Oct 31 at 9:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @kenkenb Maybe you could show some effort and fill the gaps...
  – Robert Z
  Oct 31 at 10:16
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @ Robert Z ok. thanks for your effort anyway
  – kenkenb
  Oct 31 at 10:55
  
  
  

  
  
  
  

add a comment | 

up vote
2
down vote



accepted

up vote
2
down vote



accepted

Yes, if $lim limits_{x to c} f'(x) = l in Bbb
R$ then $f$ is differentiable at $c$ provided that $f$ is continuous at $c$. Then by the Mean Value Theorem, there is $t_x$ between $x$ and $c$ such that
$$frac{f(x)-f(c)}{x-c}=f'(t_x).$$
It follows that, as $xto c$, we have $t_xto c$ and $f'(t_x)to l$.
Hence $f'(c)$ exists and it is equal to $l$.

Of course, if $f$ is not continuous at $c$ or if $f$ is not even defined at $c$, then $f$ is not differentiable at $c$. Take for example
$$f(x) = begin{cases}1 & text{if $x=0$}\ x & text{otherwise} end{cases}.$$
Then $lim_{xto 0}f'(x)=1$, but the $f$ is not differentiable at $0$:
$$lim_{xto 0^+}frac{f(x)-f(0)}{x-0}=lim_{xto 0^+}frac{x-1}{x}=-infty.$$

share|cite|improve this answer

edited Oct 31 at 9:49

answered Oct 31 at 9:33

Robert Z

90k1056128

Yes, if $lim limits_{x to c} f'(x) = l in Bbb
R$ then $f$ is differentiable at $c$ provided that $f$ is continuous at $c$. Then by the Mean Value Theorem, there is $t_x$ between $x$ and $c$ such that
$$frac{f(x)-f(c)}{x-c}=f'(t_x).$$
It follows that, as $xto c$, we have $t_xto c$ and $f'(t_x)to l$.
Hence $f'(c)$ exists and it is equal to $l$.

Of course, if $f$ is not continuous at $c$ or if $f$ is not even defined at $c$, then $f$ is not differentiable at $c$. Take for example
$$f(x) = begin{cases}1 & text{if $x=0$}\ x & text{otherwise} end{cases}.$$
Then $lim_{xto 0}f'(x)=1$, but the $f$ is not differentiable at $0$:
$$lim_{xto 0^+}frac{f(x)-f(0)}{x-0}=lim_{xto 0^+}frac{x-1}{x}=-infty.$$

share|cite|improve this answer

edited Oct 31 at 9:49

answered Oct 31 at 9:33

Robert Z

90k1056128

share|cite|improve this answer

share|cite|improve this answer

edited Oct 31 at 9:49

edited Oct 31 at 9:49

edited Oct 31 at 9:49

answered Oct 31 at 9:33

Robert Z

90k1056128

answered Oct 31 at 9:33

Robert Z

90k1056128

answered Oct 31 at 9:33

Robert Z

90k1056128

90k1056128

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Could you give more hints?
  – kenkenb
  Oct 31 at 9:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Robert Z How do you get that $f'(t_x)to l$ as $t_x to c$ implies $f'(c) = l$ ? Wouldn't you require $f$ to be $C^1$ for that to happen?
  – Deepakms
  Oct 31 at 9:53
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Deepakms NO. It follows from the assumption that $lim limits_{x to c} f'(x) = l$.
  – Robert Z
  Oct 31 at 9:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @kenkenb Maybe you could show some effort and fill the gaps...
  – Robert Z
  Oct 31 at 10:16
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @ Robert Z ok. thanks for your effort anyway
  – kenkenb
  Oct 31 at 10:55
  
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Could you give more hints?
  – kenkenb
  Oct 31 at 9:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Robert Z How do you get that $f'(t_x)to l$ as $t_x to c$ implies $f'(c) = l$ ? Wouldn't you require $f$ to be $C^1$ for that to happen?
  – Deepakms
  Oct 31 at 9:53
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Deepakms NO. It follows from the assumption that $lim limits_{x to c} f'(x) = l$.
  – Robert Z
  Oct 31 at 9:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @kenkenb Maybe you could show some effort and fill the gaps...
  – Robert Z
  Oct 31 at 10:16
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @ Robert Z ok. thanks for your effort anyway
  – kenkenb
  Oct 31 at 10:55
  
  
  

  
  
  
  

Could you give more hints?
– kenkenb
Oct 31 at 9:35

Could you give more hints?
– kenkenb
Oct 31 at 9:35

@Robert Z How do you get that $f'(t_x)to l$ as $t_x to c$ implies $f'(c) = l$ ? Wouldn't you require $f$ to be $C^1$ for that to happen?
– Deepakms
Oct 31 at 9:53

@Robert Z How do you get that $f'(t_x)to l$ as $t_x to c$ implies $f'(c) = l$ ? Wouldn't you require $f$ to be $C^1$ for that to happen?
– Deepakms
Oct 31 at 9:53

@Deepakms NO. It follows from the assumption that $lim limits_{x to c} f'(x) = l$.
– Robert Z
Oct 31 at 9:55

@Deepakms NO. It follows from the assumption that $lim limits_{x to c} f'(x) = l$.
– Robert Z
Oct 31 at 9:55

@kenkenb Maybe you could show some effort and fill the gaps...
– Robert Z
Oct 31 at 10:16

@kenkenb Maybe you could show some effort and fill the gaps...
– Robert Z
Oct 31 at 10:16

@ Robert Z ok. thanks for your effort anyway
– kenkenb
Oct 31 at 10:55

@ Robert Z ok. thanks for your effort anyway
– kenkenb
Oct 31 at 10:55

add a comment | 

up vote
2
down vote

It is possible that $f'(c)$ does not even exist. The existence of a limit of derivative does not even guarantee continuity of $f$ at $c$. For example consider a constant function with a jump discontinuity at $c$.
$$F(x) = begin{cases}0 & x neq 1 \ 1 & x=1end{cases}$$ is not differentiable at 1 even though the limit of derivative exist when $x$ tends to 1.

share|cite|improve this answer

edited Oct 31 at 9:39

answered Oct 31 at 9:32

Deepakms

368114

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  In fact , it is not even mean that $f$ is defined at $c$
  – Hagen von Eitzen
  Oct 31 at 10:12
  

  
  
  
  

add a comment | 

up vote
2
down vote

It is possible that $f'(c)$ does not even exist. The existence of a limit of derivative does not even guarantee continuity of $f$ at $c$. For example consider a constant function with a jump discontinuity at $c$.
$$F(x) = begin{cases}0 & x neq 1 \ 1 & x=1end{cases}$$ is not differentiable at 1 even though the limit of derivative exist when $x$ tends to 1.

share|cite|improve this answer

edited Oct 31 at 9:39

answered Oct 31 at 9:32

Deepakms

368114

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  In fact , it is not even mean that $f$ is defined at $c$
  – Hagen von Eitzen
  Oct 31 at 10:12
  

  
  
  
  

add a comment | 

up vote
2
down vote

up vote
2
down vote

It is possible that $f'(c)$ does not even exist. The existence of a limit of derivative does not even guarantee continuity of $f$ at $c$. For example consider a constant function with a jump discontinuity at $c$.
$$F(x) = begin{cases}0 & x neq 1 \ 1 & x=1end{cases}$$ is not differentiable at 1 even though the limit of derivative exist when $x$ tends to 1.

share|cite|improve this answer

edited Oct 31 at 9:39

answered Oct 31 at 9:32

Deepakms

368114

It is possible that $f'(c)$ does not even exist. The existence of a limit of derivative does not even guarantee continuity of $f$ at $c$. For example consider a constant function with a jump discontinuity at $c$.
$$F(x) = begin{cases}0 & x neq 1 \ 1 & x=1end{cases}$$ is not differentiable at 1 even though the limit of derivative exist when $x$ tends to 1.

share|cite|improve this answer

edited Oct 31 at 9:39

answered Oct 31 at 9:32

Deepakms

368114

share|cite|improve this answer

share|cite|improve this answer

edited Oct 31 at 9:39

edited Oct 31 at 9:39

edited Oct 31 at 9:39

answered Oct 31 at 9:32

Deepakms

368114

answered Oct 31 at 9:32

Deepakms

368114

answered Oct 31 at 9:32

Deepakms

368114

368114

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  In fact , it is not even mean that $f$ is defined at $c$
  – Hagen von Eitzen
  Oct 31 at 10:12
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  In fact , it is not even mean that $f$ is defined at $c$
  – Hagen von Eitzen
  Oct 31 at 10:12
  

  
  
  
  

2

2

In fact , it is not even mean that $f$ is defined at $c$
– Hagen von Eitzen
Oct 31 at 10:12

In fact , it is not even mean that $f$ is defined at $c$
– Hagen von Eitzen
Oct 31 at 10:12

add a comment | 

 

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