Circular Reasoning for Epsilon-Delta Proof?epsilon-delta limit with multiple variablesScalar Multiplication of Limits $epsilon$ - $delta $ Proof$epsilon-delta$ definition for limits involving $infty$$epsilon-delta$ of $lim_xto4sqrtx=2$Second half of epsilon-delta limit proofWhy was $delta / sqrt 2$ used in this proof?Epsilon-delta definitions, inequality strict / non-strict?$epsilon - delta$ proof using inequalities which are true only for some intervals?Multivariable Epsilon Delta Limit ProofIs it possible for $x$ to appear in the definition of $delta$ in an $epsilon-delta$ proof of limit?
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Circular Reasoning for Epsilon-Delta Proof?
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Circular Reasoning for Epsilon-Delta Proof?
epsilon-delta limit with multiple variablesScalar Multiplication of Limits $epsilon$ - $delta $ Proof$epsilon-delta$ definition for limits involving $infty$$epsilon-delta$ of $lim_xto4sqrtx=2$Second half of epsilon-delta limit proofWhy was $delta / sqrt 2$ used in this proof?Epsilon-delta definitions, inequality strict / non-strict?$epsilon - delta$ proof using inequalities which are true only for some intervals?Multivariable Epsilon Delta Limit ProofIs it possible for $x$ to appear in the definition of $delta$ in an $epsilon-delta$ proof of limit?
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
$$lim_limitsx to 4 2x-5=3$$
In order to prove this limit, the epsilon-delta definition will be used.
$$|f(x)-L|<varepsilon$$ $$|x-a|<delta$$
In the proof, the above $2$ inequalities will be used to find how $delta$ is related to $varepsilon$ (e.g. $delta=epsilon/2$).
Then, this relationship between $delta$ and $varepsilon$ will be used show that for any $varepsilon > 0$,
$$|x-a|<delta$$
will result in
$$|f(x)-L|<varepsilon$$
Which seems weird to me, because it doesn't seem to be a good proof as there seems to be circular reasoning, but I am probably missing something here.
Am I missing out any details which prevents circular reasoning in this proof?
real-analysis epsilon-delta
asked 9 hours ago
helpmehelpme
286 bronze badges
$endgroup$
add a comment |
$begingroup$
$$lim_limitsx to 4 2x-5=3$$
In order to prove this limit, the epsilon-delta definition will be used.
$$|f(x)-L|<varepsilon$$ $$|x-a|<delta$$
In the proof, the above $2$ inequalities will be used to find how $delta$ is related to $varepsilon$ (e.g. $delta=epsilon/2$).
Then, this relationship between $delta$ and $varepsilon$ will be used show that for any $varepsilon > 0$,
$$|x-a|<delta$$
will result in
$$|f(x)-L|<varepsilon$$
Which seems weird to me, because it doesn't seem to be a good proof as there seems to be circular reasoning, but I am probably missing something here.
Am I missing out any details which prevents circular reasoning in this proof?
real-analysis epsilon-delta
asked 9 hours ago
helpmehelpme
286 bronze badges
$endgroup$
1
$begingroup$
At what point does the argument become circular? We choose $delta=f(epsilon)$ such that $|x-a|ltdeltaimplies|f(x)-L|ltepsilon$ for all $epsilongt0$. We usually require $delta$ to depend on $epsilon$ so that the implication is proven true for all $epsilongt0$ (otherwise it wouldn't always be true).
$endgroup$
– Peter Foreman
8 hours ago
$begingroup$
You remove all that work of finding $delta$ and put it on the margin so it doesn't come in the way of the actual proof.
$endgroup$
– ganeshie8
8 hours ago
2
$begingroup$
It's important to note that the logical order of a proof is not always the same as the conceptual order. Often in $epsilon-delta$ proofs we first do some side computations starting from the assumption that $|f(x) - L | < epsilon$ and use this to determine $delta$. But that's not part of the proof. In the proof, we are given $epsilon$ and we choose some $delta$ and then derive $|f(x) - L| < epsilon$ as the conclusion. This does sometimes have a side-effect of making the choice of $delta$ seem like magic since we've swept those computations under the rug, but it is the logical order.
$endgroup$
– Jair Taylor
8 hours ago
add a comment |
$begingroup$
$$lim_limitsx to 4 2x-5=3$$
In order to prove this limit, the epsilon-delta definition will be used.
$$|f(x)-L|<varepsilon$$ $$|x-a|<delta$$
In the proof, the above $2$ inequalities will be used to find how $delta$ is related to $varepsilon$ (e.g. $delta=epsilon/2$).
Then, this relationship between $delta$ and $varepsilon$ will be used show that for any $varepsilon > 0$,
$$|x-a|<delta$$
will result in
$$|f(x)-L|<varepsilon$$
Which seems weird to me, because it doesn't seem to be a good proof as there seems to be circular reasoning, but I am probably missing something here.
Am I missing out any details which prevents circular reasoning in this proof?
real-analysis epsilon-delta
asked 9 hours ago
helpmehelpme
286 bronze badges
$endgroup$
$$lim_limitsx to 4 2x-5=3$$
In order to prove this limit, the epsilon-delta definition will be used.
$$|f(x)-L|<varepsilon$$ $$|x-a|<delta$$
In the proof, the above $2$ inequalities will be used to find how $delta$ is related to $varepsilon$ (e.g. $delta=epsilon/2$).
Then, this relationship between $delta$ and $varepsilon$ will be used show that for any $varepsilon > 0$,
$$|x-a|<delta$$
will result in
$$|f(x)-L|<varepsilon$$
Which seems weird to me, because it doesn't seem to be a good proof as there seems to be circular reasoning, but I am probably missing something here.
Am I missing out any details which prevents circular reasoning in this proof?
real-analysis epsilon-delta
real-analysis epsilon-delta
asked 9 hours ago
helpmehelpme
286 bronze badges
asked 9 hours ago
helpmehelpme
286 bronze badges
edited 8 hours ago
helpme
asked 9 hours ago
helpmehelpme
286 bronze badges
asked 9 hours ago
helpmehelpme
286 bronze badges
asked 9 hours ago
helpmehelpme
286 bronze badges
286 bronze badges
1
$begingroup$
At what point does the argument become circular? We choose $delta=f(epsilon)$ such that $|x-a|ltdeltaimplies|f(x)-L|ltepsilon$ for all $epsilongt0$. We usually require $delta$ to depend on $epsilon$ so that the implication is proven true for all $epsilongt0$ (otherwise it wouldn't always be true).
$endgroup$
– Peter Foreman
8 hours ago
$begingroup$
You remove all that work of finding $delta$ and put it on the margin so it doesn't come in the way of the actual proof.
$endgroup$
– ganeshie8
8 hours ago
2
$begingroup$
It's important to note that the logical order of a proof is not always the same as the conceptual order. Often in $epsilon-delta$ proofs we first do some side computations starting from the assumption that $|f(x) - L | < epsilon$ and use this to determine $delta$. But that's not part of the proof. In the proof, we are given $epsilon$ and we choose some $delta$ and then derive $|f(x) - L| < epsilon$ as the conclusion. This does sometimes have a side-effect of making the choice of $delta$ seem like magic since we've swept those computations under the rug, but it is the logical order.
$endgroup$
– Jair Taylor
8 hours ago
add a comment |
1
$begingroup$
At what point does the argument become circular? We choose $delta=f(epsilon)$ such that $|x-a|ltdeltaimplies|f(x)-L|ltepsilon$ for all $epsilongt0$. We usually require $delta$ to depend on $epsilon$ so that the implication is proven true for all $epsilongt0$ (otherwise it wouldn't always be true).
$endgroup$
– Peter Foreman
8 hours ago
$begingroup$
You remove all that work of finding $delta$ and put it on the margin so it doesn't come in the way of the actual proof.
$endgroup$
– ganeshie8
8 hours ago
2
$begingroup$
It's important to note that the logical order of a proof is not always the same as the conceptual order. Often in $epsilon-delta$ proofs we first do some side computations starting from the assumption that $|f(x) - L | < epsilon$ and use this to determine $delta$. But that's not part of the proof. In the proof, we are given $epsilon$ and we choose some $delta$ and then derive $|f(x) - L| < epsilon$ as the conclusion. This does sometimes have a side-effect of making the choice of $delta$ seem like magic since we've swept those computations under the rug, but it is the logical order.
$endgroup$
– Jair Taylor
8 hours ago
1
1
$begingroup$
At what point does the argument become circular? We choose $delta=f(epsilon)$ such that $|x-a|ltdeltaimplies|f(x)-L|ltepsilon$ for all $epsilongt0$. We usually require $delta$ to depend on $epsilon$ so that the implication is proven true for all $epsilongt0$ (otherwise it wouldn't always be true).
$endgroup$
– Peter Foreman
8 hours ago
$begingroup$
At what point does the argument become circular? We choose $delta=f(epsilon)$ such that $|x-a|ltdeltaimplies|f(x)-L|ltepsilon$ for all $epsilongt0$. We usually require $delta$ to depend on $epsilon$ so that the implication is proven true for all $epsilongt0$ (otherwise it wouldn't always be true).
$endgroup$
– Peter Foreman
8 hours ago
$begingroup$
You remove all that work of finding $delta$ and put it on the margin so it doesn't come in the way of the actual proof.
$endgroup$
– ganeshie8
8 hours ago
$begingroup$
You remove all that work of finding $delta$ and put it on the margin so it doesn't come in the way of the actual proof.
$endgroup$
– ganeshie8
8 hours ago
2
2
$begingroup$
It's important to note that the logical order of a proof is not always the same as the conceptual order. Often in $epsilon-delta$ proofs we first do some side computations starting from the assumption that $|f(x) - L | < epsilon$ and use this to determine $delta$. But that's not part of the proof. In the proof, we are given $epsilon$ and we choose some $delta$ and then derive $|f(x) - L| < epsilon$ as the conclusion. This does sometimes have a side-effect of making the choice of $delta$ seem like magic since we've swept those computations under the rug, but it is the logical order.
$endgroup$
– Jair Taylor
8 hours ago
$begingroup$
It's important to note that the logical order of a proof is not always the same as the conceptual order. Often in $epsilon-delta$ proofs we first do some side computations starting from the assumption that $|f(x) - L | < epsilon$ and use this to determine $delta$. But that's not part of the proof. In the proof, we are given $epsilon$ and we choose some $delta$ and then derive $|f(x) - L| < epsilon$ as the conclusion. This does sometimes have a side-effect of making the choice of $delta$ seem like magic since we've swept those computations under the rug, but it is the logical order.
$endgroup$
– Jair Taylor
8 hours ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
There is a thought process here which might feel circular, but the proof itself is not. Keep in mind that I'm allowed to use however silly a thought process I want - the only thing that matters is whether the actual proof produced at the end of the day is valid. If I got my inspiration for what $delta$ should be by rolling dice, well, that might not be something I should rely on in the future but that doesn't mean that I won't be able to turn that guess into a valid proof.
The general thought process in an $epsilon$/$delta$ argument is to start with the conclusion we want and try to "backsolve" for what choice of $delta$ (in terms of $epsilon$) would work. This does give a circular taste to the whole experience of discovering and subsequently writing the proof, since we seem to start at the end, go to the beginning, and then go back to the end. But the proof itself consists only of the second half of that development.
answered 8 hours ago
Noah SchweberNoah Schweber
138k10 gold badges164 silver badges313 bronze badges
$endgroup$
add a comment |
$begingroup$
Epsilon - delta proofs can seem circular when you first meet them, but they are certainly not.
The point of the proof is this: show that for any number $epsilon>0$ we could choose, there exists a corresponding $delta>0$ so that for every $x$ with $|x-a|<delta$ we have $|f(x) - L|<epsilon$.
To do this, we choose any arbitrary $epsilon$ and find a $delta$ which corresponds to that value of $epsilon$ to make is so that when we are within $delta$ of $a$ in the domain, we map to a point within our original "error tolerance" $epsilon$ of L.
The point is that we pick an arbitrary $epsilon$ and find a $delta$ which corresponds to it, thus demonstrating that no matter what $epsilon$ we pick, we can always find a $delta $ neighborhood with the properties we want. This isn't circular, it's just got multiple moving parts, one of which depends on the other
edited 1 min ago
svavil
2053 silver badges12 bronze badges
answered 8 hours ago
MathTrainMathTrain
8725 silver badges18 bronze badges
$endgroup$
add a comment |
$begingroup$
The definition of limit ask to prove that for any $$varepsilon >0$$ there exists a $$delta(varepsilon)$$ such that if $$0<|x-x_0|< delta$$ then $$|f(x)-l|<varepsilon$$
In your example, let be $$varepsilon >0$$ and consider the inequality
$$|(2x-5)-3|< varepsilon$$. You get
$$|2x-8|<varepsilon$$ which is satisfied for $$|x-4|<fracvarepsilon2$$. If you put $$delta(varepsilon)=fracvarepsilon2$$ you have done. Therefore there is no circular reasoning.
answered 8 hours ago
Luca Goldoni Ph.D.Luca Goldoni Ph.D.
213 bronze badges
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
There is a thought process here which might feel circular, but the proof itself is not. Keep in mind that I'm allowed to use however silly a thought process I want - the only thing that matters is whether the actual proof produced at the end of the day is valid. If I got my inspiration for what $delta$ should be by rolling dice, well, that might not be something I should rely on in the future but that doesn't mean that I won't be able to turn that guess into a valid proof.
The general thought process in an $epsilon$/$delta$ argument is to start with the conclusion we want and try to "backsolve" for what choice of $delta$ (in terms of $epsilon$) would work. This does give a circular taste to the whole experience of discovering and subsequently writing the proof, since we seem to start at the end, go to the beginning, and then go back to the end. But the proof itself consists only of the second half of that development.
answered 8 hours ago
Noah SchweberNoah Schweber
138k10 gold badges164 silver badges313 bronze badges
$endgroup$
add a comment |
$begingroup$
There is a thought process here which might feel circular, but the proof itself is not. Keep in mind that I'm allowed to use however silly a thought process I want - the only thing that matters is whether the actual proof produced at the end of the day is valid. If I got my inspiration for what $delta$ should be by rolling dice, well, that might not be something I should rely on in the future but that doesn't mean that I won't be able to turn that guess into a valid proof.
The general thought process in an $epsilon$/$delta$ argument is to start with the conclusion we want and try to "backsolve" for what choice of $delta$ (in terms of $epsilon$) would work. This does give a circular taste to the whole experience of discovering and subsequently writing the proof, since we seem to start at the end, go to the beginning, and then go back to the end. But the proof itself consists only of the second half of that development.
answered 8 hours ago
Noah SchweberNoah Schweber
138k10 gold badges164 silver badges313 bronze badges
$endgroup$
add a comment |
$begingroup$
There is a thought process here which might feel circular, but the proof itself is not. Keep in mind that I'm allowed to use however silly a thought process I want - the only thing that matters is whether the actual proof produced at the end of the day is valid. If I got my inspiration for what $delta$ should be by rolling dice, well, that might not be something I should rely on in the future but that doesn't mean that I won't be able to turn that guess into a valid proof.
The general thought process in an $epsilon$/$delta$ argument is to start with the conclusion we want and try to "backsolve" for what choice of $delta$ (in terms of $epsilon$) would work. This does give a circular taste to the whole experience of discovering and subsequently writing the proof, since we seem to start at the end, go to the beginning, and then go back to the end. But the proof itself consists only of the second half of that development.
answered 8 hours ago
Noah SchweberNoah Schweber
138k10 gold badges164 silver badges313 bronze badges
$endgroup$
There is a thought process here which might feel circular, but the proof itself is not. Keep in mind that I'm allowed to use however silly a thought process I want - the only thing that matters is whether the actual proof produced at the end of the day is valid. If I got my inspiration for what $delta$ should be by rolling dice, well, that might not be something I should rely on in the future but that doesn't mean that I won't be able to turn that guess into a valid proof.
The general thought process in an $epsilon$/$delta$ argument is to start with the conclusion we want and try to "backsolve" for what choice of $delta$ (in terms of $epsilon$) would work. This does give a circular taste to the whole experience of discovering and subsequently writing the proof, since we seem to start at the end, go to the beginning, and then go back to the end. But the proof itself consists only of the second half of that development.
answered 8 hours ago
Noah SchweberNoah Schweber
138k10 gold badges164 silver badges313 bronze badges
answered 8 hours ago
Noah SchweberNoah Schweber
138k10 gold badges164 silver badges313 bronze badges
answered 8 hours ago
Noah SchweberNoah Schweber
138k10 gold badges164 silver badges313 bronze badges
answered 8 hours ago
Noah SchweberNoah Schweber
138k10 gold badges164 silver badges313 bronze badges
138k10 gold badges164 silver badges313 bronze badges
add a comment |
add a comment |
$begingroup$
Epsilon - delta proofs can seem circular when you first meet them, but they are certainly not.
The point of the proof is this: show that for any number $epsilon>0$ we could choose, there exists a corresponding $delta>0$ so that for every $x$ with $|x-a|<delta$ we have $|f(x) - L|<epsilon$.
To do this, we choose any arbitrary $epsilon$ and find a $delta$ which corresponds to that value of $epsilon$ to make is so that when we are within $delta$ of $a$ in the domain, we map to a point within our original "error tolerance" $epsilon$ of L.
The point is that we pick an arbitrary $epsilon$ and find a $delta$ which corresponds to it, thus demonstrating that no matter what $epsilon$ we pick, we can always find a $delta $ neighborhood with the properties we want. This isn't circular, it's just got multiple moving parts, one of which depends on the other
edited 1 min ago
svavil
2053 silver badges12 bronze badges
answered 8 hours ago
MathTrainMathTrain
8725 silver badges18 bronze badges
$endgroup$
add a comment |
$begingroup$
Epsilon - delta proofs can seem circular when you first meet them, but they are certainly not.
The point of the proof is this: show that for any number $epsilon>0$ we could choose, there exists a corresponding $delta>0$ so that for every $x$ with $|x-a|<delta$ we have $|f(x) - L|<epsilon$.
To do this, we choose any arbitrary $epsilon$ and find a $delta$ which corresponds to that value of $epsilon$ to make is so that when we are within $delta$ of $a$ in the domain, we map to a point within our original "error tolerance" $epsilon$ of L.
The point is that we pick an arbitrary $epsilon$ and find a $delta$ which corresponds to it, thus demonstrating that no matter what $epsilon$ we pick, we can always find a $delta $ neighborhood with the properties we want. This isn't circular, it's just got multiple moving parts, one of which depends on the other
edited 1 min ago
svavil
2053 silver badges12 bronze badges
answered 8 hours ago
MathTrainMathTrain
8725 silver badges18 bronze badges
$endgroup$
add a comment |
$begingroup$
Epsilon - delta proofs can seem circular when you first meet them, but they are certainly not.
The point of the proof is this: show that for any number $epsilon>0$ we could choose, there exists a corresponding $delta>0$ so that for every $x$ with $|x-a|<delta$ we have $|f(x) - L|<epsilon$.
To do this, we choose any arbitrary $epsilon$ and find a $delta$ which corresponds to that value of $epsilon$ to make is so that when we are within $delta$ of $a$ in the domain, we map to a point within our original "error tolerance" $epsilon$ of L.
The point is that we pick an arbitrary $epsilon$ and find a $delta$ which corresponds to it, thus demonstrating that no matter what $epsilon$ we pick, we can always find a $delta $ neighborhood with the properties we want. This isn't circular, it's just got multiple moving parts, one of which depends on the other
edited 1 min ago
svavil
2053 silver badges12 bronze badges
answered 8 hours ago
MathTrainMathTrain
8725 silver badges18 bronze badges
$endgroup$
Epsilon - delta proofs can seem circular when you first meet them, but they are certainly not.
The point of the proof is this: show that for any number $epsilon>0$ we could choose, there exists a corresponding $delta>0$ so that for every $x$ with $|x-a|<delta$ we have $|f(x) - L|<epsilon$.
To do this, we choose any arbitrary $epsilon$ and find a $delta$ which corresponds to that value of $epsilon$ to make is so that when we are within $delta$ of $a$ in the domain, we map to a point within our original "error tolerance" $epsilon$ of L.
The point is that we pick an arbitrary $epsilon$ and find a $delta$ which corresponds to it, thus demonstrating that no matter what $epsilon$ we pick, we can always find a $delta $ neighborhood with the properties we want. This isn't circular, it's just got multiple moving parts, one of which depends on the other
edited 1 min ago
svavil
2053 silver badges12 bronze badges
answered 8 hours ago
MathTrainMathTrain
8725 silver badges18 bronze badges
edited 1 min ago
svavil
2053 silver badges12 bronze badges
edited 1 min ago
svavil
2053 silver badges12 bronze badges
edited 1 min ago
svavil
2053 silver badges12 bronze badges
2053 silver badges12 bronze badges
answered 8 hours ago
MathTrainMathTrain
8725 silver badges18 bronze badges
answered 8 hours ago
MathTrainMathTrain
8725 silver badges18 bronze badges
answered 8 hours ago
MathTrainMathTrain
8725 silver badges18 bronze badges
8725 silver badges18 bronze badges
add a comment |
add a comment |
$begingroup$
The definition of limit ask to prove that for any $$varepsilon >0$$ there exists a $$delta(varepsilon)$$ such that if $$0<|x-x_0|< delta$$ then $$|f(x)-l|<varepsilon$$
In your example, let be $$varepsilon >0$$ and consider the inequality
$$|(2x-5)-3|< varepsilon$$. You get
$$|2x-8|<varepsilon$$ which is satisfied for $$|x-4|<fracvarepsilon2$$. If you put $$delta(varepsilon)=fracvarepsilon2$$ you have done. Therefore there is no circular reasoning.
answered 8 hours ago
Luca Goldoni Ph.D.Luca Goldoni Ph.D.
213 bronze badges
$endgroup$
add a comment |
$begingroup$
The definition of limit ask to prove that for any $$varepsilon >0$$ there exists a $$delta(varepsilon)$$ such that if $$0<|x-x_0|< delta$$ then $$|f(x)-l|<varepsilon$$
In your example, let be $$varepsilon >0$$ and consider the inequality
$$|(2x-5)-3|< varepsilon$$. You get
$$|2x-8|<varepsilon$$ which is satisfied for $$|x-4|<fracvarepsilon2$$. If you put $$delta(varepsilon)=fracvarepsilon2$$ you have done. Therefore there is no circular reasoning.
answered 8 hours ago
Luca Goldoni Ph.D.Luca Goldoni Ph.D.
213 bronze badges
$endgroup$
add a comment |
$begingroup$
The definition of limit ask to prove that for any $$varepsilon >0$$ there exists a $$delta(varepsilon)$$ such that if $$0<|x-x_0|< delta$$ then $$|f(x)-l|<varepsilon$$
In your example, let be $$varepsilon >0$$ and consider the inequality
$$|(2x-5)-3|< varepsilon$$. You get
$$|2x-8|<varepsilon$$ which is satisfied for $$|x-4|<fracvarepsilon2$$. If you put $$delta(varepsilon)=fracvarepsilon2$$ you have done. Therefore there is no circular reasoning.
answered 8 hours ago
Luca Goldoni Ph.D.Luca Goldoni Ph.D.
213 bronze badges
$endgroup$
The definition of limit ask to prove that for any $$varepsilon >0$$ there exists a $$delta(varepsilon)$$ such that if $$0<|x-x_0|< delta$$ then $$|f(x)-l|<varepsilon$$
In your example, let be $$varepsilon >0$$ and consider the inequality
$$|(2x-5)-3|< varepsilon$$. You get
$$|2x-8|<varepsilon$$ which is satisfied for $$|x-4|<fracvarepsilon2$$. If you put $$delta(varepsilon)=fracvarepsilon2$$ you have done. Therefore there is no circular reasoning.
answered 8 hours ago
Luca Goldoni Ph.D.Luca Goldoni Ph.D.
213 bronze badges
answered 8 hours ago
Luca Goldoni Ph.D.Luca Goldoni Ph.D.
213 bronze badges
answered 8 hours ago
Luca Goldoni Ph.D.Luca Goldoni Ph.D.
213 bronze badges
answered 8 hours ago
Luca Goldoni Ph.D.Luca Goldoni Ph.D.
213 bronze badges
213 bronze badges
add a comment |
add a comment |
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1
$begingroup$
At what point does the argument become circular? We choose $delta=f(epsilon)$ such that $|x-a|ltdeltaimplies|f(x)-L|ltepsilon$ for all $epsilongt0$. We usually require $delta$ to depend on $epsilon$ so that the implication is proven true for all $epsilongt0$ (otherwise it wouldn't always be true).
$endgroup$
– Peter Foreman
8 hours ago
$begingroup$
You remove all that work of finding $delta$ and put it on the margin so it doesn't come in the way of the actual proof.
$endgroup$
– ganeshie8
8 hours ago
2
$begingroup$
It's important to note that the logical order of a proof is not always the same as the conceptual order. Often in $epsilon-delta$ proofs we first do some side computations starting from the assumption that $|f(x) - L | < epsilon$ and use this to determine $delta$. But that's not part of the proof. In the proof, we are given $epsilon$ and we choose some $delta$ and then derive $|f(x) - L| < epsilon$ as the conclusion. This does sometimes have a side-effect of making the choice of $delta$ seem like magic since we've swept those computations under the rug, but it is the logical order.
$endgroup$
– Jair Taylor
8 hours ago