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Proving a question on differentiability
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Here's the question I've been asked to prove:
Suppose that $f$ is differentiable at $c$ and that $f^prime (c)
ne 0$. Show that there exists a $delta > 0$ such that
$0<|x-c|<delta Rightarrow f(x)ne f(c)$.
Here's my attempt:
Assume the contrary that for every $delta > 0$, there is a $x_delta in textdom f$ such that $0<|x_delta-c|<delta$ with $f(x_delta)=f(c)$.
Now, we pick $varepsilon$ such that $0< varepsilon < |f^prime (c)|$. Since $f$ is differentiable at $c$, there is an $delta > 0$ such that if $0<|x-c|<delta$ then we have $left| fracf(x)-f(c)x-c - f^prime (c) right| <epsilon$. For this $delta$, there is $x_delta in textdom f$ such that $0<|x_delta-c|<delta$ with $f(x_delta)=f(c)$. Thus, we must have $left| fracf(x_delta)-f(c)x-c - f^prime (c) right| = left| f^prime (c) right|<epsilon$ which contradicts our choice of $varepsilon$.
Is this proof okay? Is there any possible direct proof? This question is asked in my analysis textbook before "The Mean Value Theorem" section.
real-analysis derivatives
asked 46 mins ago
Ashish K
596412
add a comment |Â
up vote
1
down vote
favorite
Here's the question I've been asked to prove:
Suppose that $f$ is differentiable at $c$ and that $f^prime (c)
ne 0$. Show that there exists a $delta > 0$ such that
$0<|x-c|<delta Rightarrow f(x)ne f(c)$.
Here's my attempt:
Assume the contrary that for every $delta > 0$, there is a $x_delta in textdom f$ such that $0<|x_delta-c|<delta$ with $f(x_delta)=f(c)$.
Now, we pick $varepsilon$ such that $0< varepsilon < |f^prime (c)|$. Since $f$ is differentiable at $c$, there is an $delta > 0$ such that if $0<|x-c|<delta$ then we have $left| fracf(x)-f(c)x-c - f^prime (c) right| <epsilon$. For this $delta$, there is $x_delta in textdom f$ such that $0<|x_delta-c|<delta$ with $f(x_delta)=f(c)$. Thus, we must have $left| fracf(x_delta)-f(c)x-c - f^prime (c) right| = left| f^prime (c) right|<epsilon$ which contradicts our choice of $varepsilon$.
Is this proof okay? Is there any possible direct proof? This question is asked in my analysis textbook before "The Mean Value Theorem" section.
real-analysis derivatives
asked 46 mins ago
Ashish K
596412
(You can simply usef'for the derivative in $TeX$.)
– Martin R
9 mins ago
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Here's the question I've been asked to prove:
Suppose that $f$ is differentiable at $c$ and that $f^prime (c)
ne 0$. Show that there exists a $delta > 0$ such that
$0<|x-c|<delta Rightarrow f(x)ne f(c)$.
Here's my attempt:
Assume the contrary that for every $delta > 0$, there is a $x_delta in textdom f$ such that $0<|x_delta-c|<delta$ with $f(x_delta)=f(c)$.
Now, we pick $varepsilon$ such that $0< varepsilon < |f^prime (c)|$. Since $f$ is differentiable at $c$, there is an $delta > 0$ such that if $0<|x-c|<delta$ then we have $left| fracf(x)-f(c)x-c - f^prime (c) right| <epsilon$. For this $delta$, there is $x_delta in textdom f$ such that $0<|x_delta-c|<delta$ with $f(x_delta)=f(c)$. Thus, we must have $left| fracf(x_delta)-f(c)x-c - f^prime (c) right| = left| f^prime (c) right|<epsilon$ which contradicts our choice of $varepsilon$.
Is this proof okay? Is there any possible direct proof? This question is asked in my analysis textbook before "The Mean Value Theorem" section.
real-analysis derivatives
asked 46 mins ago
Ashish K
596412
Here's the question I've been asked to prove:
Suppose that $f$ is differentiable at $c$ and that $f^prime (c)
ne 0$. Show that there exists a $delta > 0$ such that
$0<|x-c|<delta Rightarrow f(x)ne f(c)$.
Here's my attempt:
Assume the contrary that for every $delta > 0$, there is a $x_delta in textdom f$ such that $0<|x_delta-c|<delta$ with $f(x_delta)=f(c)$.
Now, we pick $varepsilon$ such that $0< varepsilon < |f^prime (c)|$. Since $f$ is differentiable at $c$, there is an $delta > 0$ such that if $0<|x-c|<delta$ then we have $left| fracf(x)-f(c)x-c - f^prime (c) right| <epsilon$. For this $delta$, there is $x_delta in textdom f$ such that $0<|x_delta-c|<delta$ with $f(x_delta)=f(c)$. Thus, we must have $left| fracf(x_delta)-f(c)x-c - f^prime (c) right| = left| f^prime (c) right|<epsilon$ which contradicts our choice of $varepsilon$.
Is this proof okay? Is there any possible direct proof? This question is asked in my analysis textbook before "The Mean Value Theorem" section.
real-analysis derivatives
real-analysis derivatives
asked 46 mins ago
Ashish K
596412
asked 46 mins ago
Ashish K
596412
asked 46 mins ago
Ashish K
596412
asked 46 mins ago
Ashish K
596412
asked 46 mins ago
Ashish K
596412
596412
(You can simply usef'for the derivative in $TeX$.)
– Martin R
9 mins ago
add a comment |Â
(You can simply usef'for the derivative in $TeX$.)
– Martin R
9 mins ago
(You can simply use
f' for the derivative in $TeX$.)– Martin R
9 mins ago
(You can simply use
f' for the derivative in $TeX$.)– Martin R
9 mins ago
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
2
down vote
accepted
You're proof is fine. Here is a bit easier proof in my opinion: suppose that the statement you have to prove is wrong. Then for every $ninmathbbN$ there exists a point $x_n$ such that $0<|x_n-c|<frac1n$ and $f(x_n)=f(c)$. Then we got a sequence that satisfies $cne x_nto c$ but $fracf(x_n)-f(c)x_n-cto 0$. This is a contradiction because by Heine's definition of a limit we must have $fracf(x_n)-f(c)x_n-cto f'(c)$ which is not $0$.
answered 38 mins ago
Mark
3,024112
Thanks, it's easier using Heine's definition of limit. I wasn't aware of it
– Ashish K
31 mins ago
1
You're solution is also correct, I just prefer to use sequences instead of $epsilon-delta$ when possible. I believe it is a bit easier.
– Mark
29 mins ago
add a comment |Â
up vote
2
down vote
Here's my initial take on the problem:
If the hypothesis were not true, then look at a sequence of deltas, $delta_n = tfrac1n$. This means there is a sequence of $x_n to c$ as $nto infty$, with $f(x_n) = f(c)$ for all $n geq 1$. Hence
$$lim_nto infty fracf(x_n)-f(c)x_n - c = lim_nto infty 0 = 0,$$
which is a contradiction, since we know that $f'(c)neq 0$ and so any sequence converging to $f'(c)$ should approach this non-zero number.
answered 36 mins ago
Patch
1,4931125
add a comment |Â
up vote
2
down vote
Your proof is fine. With only a small modification you can make it
a direct argument:
For $0 < epsilon < |f'(c)|$ there is (as you said) a $delta > 0$ such that
for $0 < |x - c | < delta$
$$
left| fracf(x)-f(c)x-c - f'(c) right| < epsilon , .
$$
This implies (using the triangle inequality)
$$
left| fracf(x)-f(c)x-c right| > |f'(c)| - epsilon > 0
$$
and therefore $f(x) ne c$ for $0 < |x - c | < delta$.
answered 30 mins ago
Martin R
24.3k32844
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
You're proof is fine. Here is a bit easier proof in my opinion: suppose that the statement you have to prove is wrong. Then for every $ninmathbbN$ there exists a point $x_n$ such that $0<|x_n-c|<frac1n$ and $f(x_n)=f(c)$. Then we got a sequence that satisfies $cne x_nto c$ but $fracf(x_n)-f(c)x_n-cto 0$. This is a contradiction because by Heine's definition of a limit we must have $fracf(x_n)-f(c)x_n-cto f'(c)$ which is not $0$.
answered 38 mins ago
Mark
3,024112
Thanks, it's easier using Heine's definition of limit. I wasn't aware of it
– Ashish K
31 mins ago
1
You're solution is also correct, I just prefer to use sequences instead of $epsilon-delta$ when possible. I believe it is a bit easier.
– Mark
29 mins ago
add a comment |Â
up vote
2
down vote
accepted
You're proof is fine. Here is a bit easier proof in my opinion: suppose that the statement you have to prove is wrong. Then for every $ninmathbbN$ there exists a point $x_n$ such that $0<|x_n-c|<frac1n$ and $f(x_n)=f(c)$. Then we got a sequence that satisfies $cne x_nto c$ but $fracf(x_n)-f(c)x_n-cto 0$. This is a contradiction because by Heine's definition of a limit we must have $fracf(x_n)-f(c)x_n-cto f'(c)$ which is not $0$.
answered 38 mins ago
Mark
3,024112
Thanks, it's easier using Heine's definition of limit. I wasn't aware of it
– Ashish K
31 mins ago
1
You're solution is also correct, I just prefer to use sequences instead of $epsilon-delta$ when possible. I believe it is a bit easier.
– Mark
29 mins ago
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
You're proof is fine. Here is a bit easier proof in my opinion: suppose that the statement you have to prove is wrong. Then for every $ninmathbbN$ there exists a point $x_n$ such that $0<|x_n-c|<frac1n$ and $f(x_n)=f(c)$. Then we got a sequence that satisfies $cne x_nto c$ but $fracf(x_n)-f(c)x_n-cto 0$. This is a contradiction because by Heine's definition of a limit we must have $fracf(x_n)-f(c)x_n-cto f'(c)$ which is not $0$.
answered 38 mins ago
Mark
3,024112
You're proof is fine. Here is a bit easier proof in my opinion: suppose that the statement you have to prove is wrong. Then for every $ninmathbbN$ there exists a point $x_n$ such that $0<|x_n-c|<frac1n$ and $f(x_n)=f(c)$. Then we got a sequence that satisfies $cne x_nto c$ but $fracf(x_n)-f(c)x_n-cto 0$. This is a contradiction because by Heine's definition of a limit we must have $fracf(x_n)-f(c)x_n-cto f'(c)$ which is not $0$.
answered 38 mins ago
Mark
3,024112
answered 38 mins ago
Mark
3,024112
answered 38 mins ago
Mark
3,024112
answered 38 mins ago
Mark
3,024112
3,024112
Thanks, it's easier using Heine's definition of limit. I wasn't aware of it
– Ashish K
31 mins ago
1
You're solution is also correct, I just prefer to use sequences instead of $epsilon-delta$ when possible. I believe it is a bit easier.
– Mark
29 mins ago
add a comment |Â
Thanks, it's easier using Heine's definition of limit. I wasn't aware of it
– Ashish K
31 mins ago
1
You're solution is also correct, I just prefer to use sequences instead of $epsilon-delta$ when possible. I believe it is a bit easier.
– Mark
29 mins ago
Thanks, it's easier using Heine's definition of limit. I wasn't aware of it
– Ashish K
31 mins ago
Thanks, it's easier using Heine's definition of limit. I wasn't aware of it
– Ashish K
31 mins ago
1
1
You're solution is also correct, I just prefer to use sequences instead of $epsilon-delta$ when possible. I believe it is a bit easier.
– Mark
29 mins ago
You're solution is also correct, I just prefer to use sequences instead of $epsilon-delta$ when possible. I believe it is a bit easier.
– Mark
29 mins ago
add a comment |Â
up vote
2
down vote
Here's my initial take on the problem:
If the hypothesis were not true, then look at a sequence of deltas, $delta_n = tfrac1n$. This means there is a sequence of $x_n to c$ as $nto infty$, with $f(x_n) = f(c)$ for all $n geq 1$. Hence
$$lim_nto infty fracf(x_n)-f(c)x_n - c = lim_nto infty 0 = 0,$$
which is a contradiction, since we know that $f'(c)neq 0$ and so any sequence converging to $f'(c)$ should approach this non-zero number.
answered 36 mins ago
Patch
1,4931125
add a comment |Â
up vote
2
down vote
Here's my initial take on the problem:
If the hypothesis were not true, then look at a sequence of deltas, $delta_n = tfrac1n$. This means there is a sequence of $x_n to c$ as $nto infty$, with $f(x_n) = f(c)$ for all $n geq 1$. Hence
$$lim_nto infty fracf(x_n)-f(c)x_n - c = lim_nto infty 0 = 0,$$
which is a contradiction, since we know that $f'(c)neq 0$ and so any sequence converging to $f'(c)$ should approach this non-zero number.
answered 36 mins ago
Patch
1,4931125
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Here's my initial take on the problem:
If the hypothesis were not true, then look at a sequence of deltas, $delta_n = tfrac1n$. This means there is a sequence of $x_n to c$ as $nto infty$, with $f(x_n) = f(c)$ for all $n geq 1$. Hence
$$lim_nto infty fracf(x_n)-f(c)x_n - c = lim_nto infty 0 = 0,$$
which is a contradiction, since we know that $f'(c)neq 0$ and so any sequence converging to $f'(c)$ should approach this non-zero number.
answered 36 mins ago
Patch
1,4931125
Here's my initial take on the problem:
If the hypothesis were not true, then look at a sequence of deltas, $delta_n = tfrac1n$. This means there is a sequence of $x_n to c$ as $nto infty$, with $f(x_n) = f(c)$ for all $n geq 1$. Hence
$$lim_nto infty fracf(x_n)-f(c)x_n - c = lim_nto infty 0 = 0,$$
which is a contradiction, since we know that $f'(c)neq 0$ and so any sequence converging to $f'(c)$ should approach this non-zero number.
answered 36 mins ago
Patch
1,4931125
answered 36 mins ago
Patch
1,4931125
answered 36 mins ago
Patch
1,4931125
answered 36 mins ago
Patch
1,4931125
1,4931125
add a comment |Â
add a comment |Â
up vote
2
down vote
Your proof is fine. With only a small modification you can make it
a direct argument:
For $0 < epsilon < |f'(c)|$ there is (as you said) a $delta > 0$ such that
for $0 < |x - c | < delta$
$$
left| fracf(x)-f(c)x-c - f'(c) right| < epsilon , .
$$
This implies (using the triangle inequality)
$$
left| fracf(x)-f(c)x-c right| > |f'(c)| - epsilon > 0
$$
and therefore $f(x) ne c$ for $0 < |x - c | < delta$.
answered 30 mins ago
Martin R
24.3k32844
add a comment |Â
up vote
2
down vote
Your proof is fine. With only a small modification you can make it
a direct argument:
For $0 < epsilon < |f'(c)|$ there is (as you said) a $delta > 0$ such that
for $0 < |x - c | < delta$
$$
left| fracf(x)-f(c)x-c - f'(c) right| < epsilon , .
$$
This implies (using the triangle inequality)
$$
left| fracf(x)-f(c)x-c right| > |f'(c)| - epsilon > 0
$$
and therefore $f(x) ne c$ for $0 < |x - c | < delta$.
answered 30 mins ago
Martin R
24.3k32844
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Your proof is fine. With only a small modification you can make it
a direct argument:
For $0 < epsilon < |f'(c)|$ there is (as you said) a $delta > 0$ such that
for $0 < |x - c | < delta$
$$
left| fracf(x)-f(c)x-c - f'(c) right| < epsilon , .
$$
This implies (using the triangle inequality)
$$
left| fracf(x)-f(c)x-c right| > |f'(c)| - epsilon > 0
$$
and therefore $f(x) ne c$ for $0 < |x - c | < delta$.
answered 30 mins ago
Martin R
24.3k32844
Your proof is fine. With only a small modification you can make it
a direct argument:
For $0 < epsilon < |f'(c)|$ there is (as you said) a $delta > 0$ such that
for $0 < |x - c | < delta$
$$
left| fracf(x)-f(c)x-c - f'(c) right| < epsilon , .
$$
This implies (using the triangle inequality)
$$
left| fracf(x)-f(c)x-c right| > |f'(c)| - epsilon > 0
$$
and therefore $f(x) ne c$ for $0 < |x - c | < delta$.
answered 30 mins ago
Martin R
24.3k32844
edited 10 mins ago
answered 30 mins ago
Martin R
24.3k32844
answered 30 mins ago
Martin R
24.3k32844
answered 30 mins ago
Martin R
24.3k32844
24.3k32844
add a comment |Â
add a comment |Â
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(You can simply use
f'for the derivative in $TeX$.)– Martin R
9 mins ago