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If $lim limits_{x to c} f'(x) = l in Bbb R$. Does it mean $f$ is differentiable at $c$ and $f'(c) = l$.
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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
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
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
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
up vote
0
down vote
favorite
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
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
edited Oct 31 at 9:58
Robert Z
90k1056128
asked Oct 31 at 9:27
kenkenb
15718
edited Oct 31 at 9:58
Robert Z
90k1056128
asked Oct 31 at 9:27
kenkenb
15718
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 $0$:
$$lim_{xto 0^+}frac{f(x)-f(0)}{x-0}=lim_{xto 0^+}frac{x-1}{x}=-infty.$$
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.
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 |
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 $0$:
$$lim_{xto 0^+}frac{f(x)-f(0)}{x-0}=lim_{xto 0^+}frac{x-1}{x}=-infty.$$
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 $0$:
$$lim_{xto 0^+}frac{f(x)-f(0)}{x-0}=lim_{xto 0^+}frac{x-1}{x}=-infty.$$
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 $0$:
$$lim_{xto 0^+}frac{f(x)-f(0)}{x-0}=lim_{xto 0^+}frac{x-1}{x}=-infty.$$
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 $0$:
$$lim_{xto 0^+}frac{f(x)-f(0)}{x-0}=lim_{xto 0^+}frac{x-1}{x}=-infty.$$
answered Oct 31 at 9:33
Robert Z
90k1056128
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.
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.
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.
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.
answered Oct 31 at 9:32
Deepakms
368114
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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