Error in the derivative proof
$begingroup$
Assume $h$ is a function on an open interval $K$, and differentiable on $K$. Therefore $h'$ is cont on $K$.
The faulty proof goes as follows:
Let $ain K$
By the definition of the derivative:
$$h'(a)=lim_{xto a}frac{h(x)-h(a)}{x-a}$$
Since $h$ is cont, limit of numerator is $0$, same goes for denominator. Since $h$ is diff, we can apply LH rule.
$$h'(a)=lim_{xto a}frac{h(x)-h(a)}{x-a}to(LH)to lim_{xto a}frac{h'(x)-0}{1-0}=lim_{xto a}h'(x)$$
We have proven that $h'(a)=lim_{xto a}h'(x)$. By defn, $h'$ is cont.
There is an error in the proof, and I also need to fix the proof.
I don't see something right away.
My guess is that $x-aneq 0$, but I am not sure how to prove this.
limits derivatives proof-verification fake-proofs
asked Jan 8 at 20:21
K Split XK Split X
4,19611131
$endgroup$
add a comment |
$begingroup$
Assume $h$ is a function on an open interval $K$, and differentiable on $K$. Therefore $h'$ is cont on $K$.
The faulty proof goes as follows:
Let $ain K$
By the definition of the derivative:
$$h'(a)=lim_{xto a}frac{h(x)-h(a)}{x-a}$$
Since $h$ is cont, limit of numerator is $0$, same goes for denominator. Since $h$ is diff, we can apply LH rule.
$$h'(a)=lim_{xto a}frac{h(x)-h(a)}{x-a}to(LH)to lim_{xto a}frac{h'(x)-0}{1-0}=lim_{xto a}h'(x)$$
We have proven that $h'(a)=lim_{xto a}h'(x)$. By defn, $h'$ is cont.
There is an error in the proof, and I also need to fix the proof.
I don't see something right away.
My guess is that $x-aneq 0$, but I am not sure how to prove this.
limits derivatives proof-verification fake-proofs
asked Jan 8 at 20:21
K Split XK Split X
4,19611131
$endgroup$
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Aloizio Macedo♦
Jan 10 at 5:58
add a comment |
$begingroup$
Assume $h$ is a function on an open interval $K$, and differentiable on $K$. Therefore $h'$ is cont on $K$.
The faulty proof goes as follows:
Let $ain K$
By the definition of the derivative:
$$h'(a)=lim_{xto a}frac{h(x)-h(a)}{x-a}$$
Since $h$ is cont, limit of numerator is $0$, same goes for denominator. Since $h$ is diff, we can apply LH rule.
$$h'(a)=lim_{xto a}frac{h(x)-h(a)}{x-a}to(LH)to lim_{xto a}frac{h'(x)-0}{1-0}=lim_{xto a}h'(x)$$
We have proven that $h'(a)=lim_{xto a}h'(x)$. By defn, $h'$ is cont.
There is an error in the proof, and I also need to fix the proof.
I don't see something right away.
My guess is that $x-aneq 0$, but I am not sure how to prove this.
limits derivatives proof-verification fake-proofs
asked Jan 8 at 20:21
K Split XK Split X
4,19611131
$endgroup$
Assume $h$ is a function on an open interval $K$, and differentiable on $K$. Therefore $h'$ is cont on $K$.
The faulty proof goes as follows:
Let $ain K$
By the definition of the derivative:
$$h'(a)=lim_{xto a}frac{h(x)-h(a)}{x-a}$$
Since $h$ is cont, limit of numerator is $0$, same goes for denominator. Since $h$ is diff, we can apply LH rule.
$$h'(a)=lim_{xto a}frac{h(x)-h(a)}{x-a}to(LH)to lim_{xto a}frac{h'(x)-0}{1-0}=lim_{xto a}h'(x)$$
We have proven that $h'(a)=lim_{xto a}h'(x)$. By defn, $h'$ is cont.
There is an error in the proof, and I also need to fix the proof.
I don't see something right away.
My guess is that $x-aneq 0$, but I am not sure how to prove this.
limits derivatives proof-verification fake-proofs
limits derivatives proof-verification fake-proofs
asked Jan 8 at 20:21
K Split XK Split X
4,19611131
asked Jan 8 at 20:21
K Split XK Split X
4,19611131
edited Jan 8 at 20:34
K Split X
asked Jan 8 at 20:21
K Split XK Split X
4,19611131
asked Jan 8 at 20:21
K Split XK Split X
4,19611131
asked Jan 8 at 20:21
K Split XK Split X
4,19611131
4,19611131
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Aloizio Macedo♦
Jan 10 at 5:58
add a comment |
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Aloizio Macedo♦
Jan 10 at 5:58
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Aloizio Macedo♦
Jan 10 at 5:58
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Aloizio Macedo♦
Jan 10 at 5:58
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
L'Hospital's rule says that if $f'/g'$ converges to a limit $A$ (and some other assumptions are fulfilled), then also $f/g to A$.
In your “proof”, $displaystylelim_{x to a} h'(x)$ need not exist.
(But if it does, then the claim is true: Prove that $f'(a)=lim_{xrightarrow a}f'(x)$.)
answered Jan 8 at 20:42
Hans LundmarkHans Lundmark
35.3k564114
$endgroup$
$begingroup$
By need not exist, do you mean infinity?
$endgroup$
– K Split X
Jan 8 at 20:44
$begingroup$
No, I mean “not exist”. Like for the standard example $h(x) = x^2 sin(x)$ for $x neq 0$, $h(0)=0$, where $h'(x)$ will oscillate as $x to 0$, so that there is no limit (neither finite nor infinite).
$endgroup$
– Hans Lundmark
Jan 8 at 20:47
3
$begingroup$
Oops, that should have been $h(x)=x^2 sin(1/x)$, of course.
$endgroup$
– Hans Lundmark
Jan 8 at 20:53
2
$begingroup$
+1 for pointing out the exact flaw. Also the argument used by asker proves that if the limit of derivative exists then derivative is continuous. This is very special of derivatives that they can't have what is usually called simple / jump discontinuity.
$endgroup$
– Paramanand Singh
Jan 9 at 19:13
add a comment |
$begingroup$
Let $a=0$, $h(x) = begin{cases} x^2 sin {1 over x} , & x neq 0 \
0, & x = 0end{cases}$ on $K=(-1,1)$. Then $h$ is differentiable on $K$ but ${h(x) over x}$ has no limit as $x to 0$.
answered Jan 8 at 20:49
copper.hatcopper.hat
126k559160
$endgroup$
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
$begingroup$
L'Hospital's rule says that if $f'/g'$ converges to a limit $A$ (and some other assumptions are fulfilled), then also $f/g to A$.
In your “proof”, $displaystylelim_{x to a} h'(x)$ need not exist.
(But if it does, then the claim is true: Prove that $f'(a)=lim_{xrightarrow a}f'(x)$.)
answered Jan 8 at 20:42
Hans LundmarkHans Lundmark
35.3k564114
$endgroup$
$begingroup$
By need not exist, do you mean infinity?
$endgroup$
– K Split X
Jan 8 at 20:44
$begingroup$
No, I mean “not exist”. Like for the standard example $h(x) = x^2 sin(x)$ for $x neq 0$, $h(0)=0$, where $h'(x)$ will oscillate as $x to 0$, so that there is no limit (neither finite nor infinite).
$endgroup$
– Hans Lundmark
Jan 8 at 20:47
3
$begingroup$
Oops, that should have been $h(x)=x^2 sin(1/x)$, of course.
$endgroup$
– Hans Lundmark
Jan 8 at 20:53
2
$begingroup$
+1 for pointing out the exact flaw. Also the argument used by asker proves that if the limit of derivative exists then derivative is continuous. This is very special of derivatives that they can't have what is usually called simple / jump discontinuity.
$endgroup$
– Paramanand Singh
Jan 9 at 19:13
add a comment |
$begingroup$
L'Hospital's rule says that if $f'/g'$ converges to a limit $A$ (and some other assumptions are fulfilled), then also $f/g to A$.
In your “proof”, $displaystylelim_{x to a} h'(x)$ need not exist.
(But if it does, then the claim is true: Prove that $f'(a)=lim_{xrightarrow a}f'(x)$.)
answered Jan 8 at 20:42
Hans LundmarkHans Lundmark
35.3k564114
$endgroup$
$begingroup$
By need not exist, do you mean infinity?
$endgroup$
– K Split X
Jan 8 at 20:44
$begingroup$
No, I mean “not exist”. Like for the standard example $h(x) = x^2 sin(x)$ for $x neq 0$, $h(0)=0$, where $h'(x)$ will oscillate as $x to 0$, so that there is no limit (neither finite nor infinite).
$endgroup$
– Hans Lundmark
Jan 8 at 20:47
3
$begingroup$
Oops, that should have been $h(x)=x^2 sin(1/x)$, of course.
$endgroup$
– Hans Lundmark
Jan 8 at 20:53
2
$begingroup$
+1 for pointing out the exact flaw. Also the argument used by asker proves that if the limit of derivative exists then derivative is continuous. This is very special of derivatives that they can't have what is usually called simple / jump discontinuity.
$endgroup$
– Paramanand Singh
Jan 9 at 19:13
add a comment |
$begingroup$
L'Hospital's rule says that if $f'/g'$ converges to a limit $A$ (and some other assumptions are fulfilled), then also $f/g to A$.
In your “proof”, $displaystylelim_{x to a} h'(x)$ need not exist.
(But if it does, then the claim is true: Prove that $f'(a)=lim_{xrightarrow a}f'(x)$.)
answered Jan 8 at 20:42
Hans LundmarkHans Lundmark
35.3k564114
$endgroup$
L'Hospital's rule says that if $f'/g'$ converges to a limit $A$ (and some other assumptions are fulfilled), then also $f/g to A$.
In your “proof”, $displaystylelim_{x to a} h'(x)$ need not exist.
(But if it does, then the claim is true: Prove that $f'(a)=lim_{xrightarrow a}f'(x)$.)
answered Jan 8 at 20:42
Hans LundmarkHans Lundmark
35.3k564114
answered Jan 8 at 20:42
Hans LundmarkHans Lundmark
35.3k564114
answered Jan 8 at 20:42
Hans LundmarkHans Lundmark
35.3k564114
answered Jan 8 at 20:42
Hans LundmarkHans Lundmark
35.3k564114
35.3k564114
$begingroup$
By need not exist, do you mean infinity?
$endgroup$
– K Split X
Jan 8 at 20:44
$begingroup$
No, I mean “not exist”. Like for the standard example $h(x) = x^2 sin(x)$ for $x neq 0$, $h(0)=0$, where $h'(x)$ will oscillate as $x to 0$, so that there is no limit (neither finite nor infinite).
$endgroup$
– Hans Lundmark
Jan 8 at 20:47
3
$begingroup$
Oops, that should have been $h(x)=x^2 sin(1/x)$, of course.
$endgroup$
– Hans Lundmark
Jan 8 at 20:53
2
$begingroup$
+1 for pointing out the exact flaw. Also the argument used by asker proves that if the limit of derivative exists then derivative is continuous. This is very special of derivatives that they can't have what is usually called simple / jump discontinuity.
$endgroup$
– Paramanand Singh
Jan 9 at 19:13
add a comment |
$begingroup$
By need not exist, do you mean infinity?
$endgroup$
– K Split X
Jan 8 at 20:44
$begingroup$
No, I mean “not exist”. Like for the standard example $h(x) = x^2 sin(x)$ for $x neq 0$, $h(0)=0$, where $h'(x)$ will oscillate as $x to 0$, so that there is no limit (neither finite nor infinite).
$endgroup$
– Hans Lundmark
Jan 8 at 20:47
3
$begingroup$
Oops, that should have been $h(x)=x^2 sin(1/x)$, of course.
$endgroup$
– Hans Lundmark
Jan 8 at 20:53
2
$begingroup$
+1 for pointing out the exact flaw. Also the argument used by asker proves that if the limit of derivative exists then derivative is continuous. This is very special of derivatives that they can't have what is usually called simple / jump discontinuity.
$endgroup$
– Paramanand Singh
Jan 9 at 19:13
$begingroup$
By need not exist, do you mean infinity?
$endgroup$
– K Split X
Jan 8 at 20:44
$begingroup$
By need not exist, do you mean infinity?
$endgroup$
– K Split X
Jan 8 at 20:44
$begingroup$
No, I mean “not exist”. Like for the standard example $h(x) = x^2 sin(x)$ for $x neq 0$, $h(0)=0$, where $h'(x)$ will oscillate as $x to 0$, so that there is no limit (neither finite nor infinite).
$endgroup$
– Hans Lundmark
Jan 8 at 20:47
$begingroup$
No, I mean “not exist”. Like for the standard example $h(x) = x^2 sin(x)$ for $x neq 0$, $h(0)=0$, where $h'(x)$ will oscillate as $x to 0$, so that there is no limit (neither finite nor infinite).
$endgroup$
– Hans Lundmark
Jan 8 at 20:47
3
3
$begingroup$
Oops, that should have been $h(x)=x^2 sin(1/x)$, of course.
$endgroup$
– Hans Lundmark
Jan 8 at 20:53
$begingroup$
Oops, that should have been $h(x)=x^2 sin(1/x)$, of course.
$endgroup$
– Hans Lundmark
Jan 8 at 20:53
2
2
$begingroup$
+1 for pointing out the exact flaw. Also the argument used by asker proves that if the limit of derivative exists then derivative is continuous. This is very special of derivatives that they can't have what is usually called simple / jump discontinuity.
$endgroup$
– Paramanand Singh
Jan 9 at 19:13
$begingroup$
+1 for pointing out the exact flaw. Also the argument used by asker proves that if the limit of derivative exists then derivative is continuous. This is very special of derivatives that they can't have what is usually called simple / jump discontinuity.
$endgroup$
– Paramanand Singh
Jan 9 at 19:13
add a comment |
$begingroup$
Let $a=0$, $h(x) = begin{cases} x^2 sin {1 over x} , & x neq 0 \
0, & x = 0end{cases}$ on $K=(-1,1)$. Then $h$ is differentiable on $K$ but ${h(x) over x}$ has no limit as $x to 0$.
answered Jan 8 at 20:49
copper.hatcopper.hat
126k559160
$endgroup$
add a comment |
$begingroup$
Let $a=0$, $h(x) = begin{cases} x^2 sin {1 over x} , & x neq 0 \
0, & x = 0end{cases}$ on $K=(-1,1)$. Then $h$ is differentiable on $K$ but ${h(x) over x}$ has no limit as $x to 0$.
answered Jan 8 at 20:49
copper.hatcopper.hat
126k559160
$endgroup$
add a comment |
$begingroup$
Let $a=0$, $h(x) = begin{cases} x^2 sin {1 over x} , & x neq 0 \
0, & x = 0end{cases}$ on $K=(-1,1)$. Then $h$ is differentiable on $K$ but ${h(x) over x}$ has no limit as $x to 0$.
answered Jan 8 at 20:49
copper.hatcopper.hat
126k559160
$endgroup$
Let $a=0$, $h(x) = begin{cases} x^2 sin {1 over x} , & x neq 0 \
0, & x = 0end{cases}$ on $K=(-1,1)$. Then $h$ is differentiable on $K$ but ${h(x) over x}$ has no limit as $x to 0$.
answered Jan 8 at 20:49
copper.hatcopper.hat
126k559160
edited Jan 8 at 20:55
answered Jan 8 at 20:49
copper.hatcopper.hat
126k559160
answered Jan 8 at 20:49
copper.hatcopper.hat
126k559160
answered Jan 8 at 20:49
copper.hatcopper.hat
126k559160
126k559160
add a comment |
add a comment |
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$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Aloizio Macedo♦
Jan 10 at 5:58