Epsilon Delta Proof of Limits Being Equal
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I have a question about this Epsilon-Delta Proof of Limits Being Equal
Why did the person that answered this assume that δ=δ0?
Thank you!
calculus limits
asked Nov 23 at 18:01
noam Azulay
11
add a comment |
up vote
-3
down vote
favorite
I have a question about this Epsilon-Delta Proof of Limits Being Equal
Why did the person that answered this assume that δ=δ0?
Thank you!
calculus limits
asked Nov 23 at 18:01
noam Azulay
11
He chose $delta = delta_0$. This can be assumed because he originally assumed $lim_{x rightarrow 0} f(x) = L$ which ensures such $delta$ exists.
– Hongyu Wang
Nov 23 at 18:13
2
No, the answerer didn't assume it, they simply set $delta$ to be $delta_0$. To do an $varepsilon$-$delta$ proof, you are given an $varepsilon$, and you need to find a $delta$ satisfying certain conditions depending on $varepsilon$ and $delta$. In the answer, the answerer first fixes some $varepsilon$, and then obtain a $delta_0$ using the limit, and after that take $delta$ to be the $delta_0$ obtained previously.
– Alex Vong
Nov 23 at 18:19
3
Why is this question urgent, exactly?
– Noah Schweber
Nov 23 at 18:25
1
He's saying you can use the same $delta$ for the second limit as you did for the first.
– saulspatz
Nov 23 at 18:35
add a comment |
up vote
-3
down vote
favorite
up vote
-3
down vote
favorite
I have a question about this Epsilon-Delta Proof of Limits Being Equal
Why did the person that answered this assume that δ=δ0?
Thank you!
calculus limits
asked Nov 23 at 18:01
noam Azulay
11
I have a question about this Epsilon-Delta Proof of Limits Being Equal
Why did the person that answered this assume that δ=δ0?
Thank you!
calculus limits
calculus limits
asked Nov 23 at 18:01
noam Azulay
11
asked Nov 23 at 18:01
noam Azulay
11
edited Nov 23 at 18:37
asked Nov 23 at 18:01
noam Azulay
11
asked Nov 23 at 18:01
noam Azulay
11
asked Nov 23 at 18:01
noam Azulay
11
11
He chose $delta = delta_0$. This can be assumed because he originally assumed $lim_{x rightarrow 0} f(x) = L$ which ensures such $delta$ exists.
– Hongyu Wang
Nov 23 at 18:13
2
No, the answerer didn't assume it, they simply set $delta$ to be $delta_0$. To do an $varepsilon$-$delta$ proof, you are given an $varepsilon$, and you need to find a $delta$ satisfying certain conditions depending on $varepsilon$ and $delta$. In the answer, the answerer first fixes some $varepsilon$, and then obtain a $delta_0$ using the limit, and after that take $delta$ to be the $delta_0$ obtained previously.
– Alex Vong
Nov 23 at 18:19
3
Why is this question urgent, exactly?
– Noah Schweber
Nov 23 at 18:25
1
He's saying you can use the same $delta$ for the second limit as you did for the first.
– saulspatz
Nov 23 at 18:35
add a comment |
He chose $delta = delta_0$. This can be assumed because he originally assumed $lim_{x rightarrow 0} f(x) = L$ which ensures such $delta$ exists.
– Hongyu Wang
Nov 23 at 18:13
2
No, the answerer didn't assume it, they simply set $delta$ to be $delta_0$. To do an $varepsilon$-$delta$ proof, you are given an $varepsilon$, and you need to find a $delta$ satisfying certain conditions depending on $varepsilon$ and $delta$. In the answer, the answerer first fixes some $varepsilon$, and then obtain a $delta_0$ using the limit, and after that take $delta$ to be the $delta_0$ obtained previously.
– Alex Vong
Nov 23 at 18:19
3
Why is this question urgent, exactly?
– Noah Schweber
Nov 23 at 18:25
1
He's saying you can use the same $delta$ for the second limit as you did for the first.
– saulspatz
Nov 23 at 18:35
He chose $delta = delta_0$. This can be assumed because he originally assumed $lim_{x rightarrow 0} f(x) = L$ which ensures such $delta$ exists.
– Hongyu Wang
Nov 23 at 18:13
He chose $delta = delta_0$. This can be assumed because he originally assumed $lim_{x rightarrow 0} f(x) = L$ which ensures such $delta$ exists.
– Hongyu Wang
Nov 23 at 18:13
2
2
No, the answerer didn't assume it, they simply set $delta$ to be $delta_0$. To do an $varepsilon$-$delta$ proof, you are given an $varepsilon$, and you need to find a $delta$ satisfying certain conditions depending on $varepsilon$ and $delta$. In the answer, the answerer first fixes some $varepsilon$, and then obtain a $delta_0$ using the limit, and after that take $delta$ to be the $delta_0$ obtained previously.
– Alex Vong
Nov 23 at 18:19
No, the answerer didn't assume it, they simply set $delta$ to be $delta_0$. To do an $varepsilon$-$delta$ proof, you are given an $varepsilon$, and you need to find a $delta$ satisfying certain conditions depending on $varepsilon$ and $delta$. In the answer, the answerer first fixes some $varepsilon$, and then obtain a $delta_0$ using the limit, and after that take $delta$ to be the $delta_0$ obtained previously.
– Alex Vong
Nov 23 at 18:19
3
3
Why is this question urgent, exactly?
– Noah Schweber
Nov 23 at 18:25
Why is this question urgent, exactly?
– Noah Schweber
Nov 23 at 18:25
1
1
He's saying you can use the same $delta$ for the second limit as you did for the first.
– saulspatz
Nov 23 at 18:35
He's saying you can use the same $delta$ for the second limit as you did for the first.
– saulspatz
Nov 23 at 18:35
add a comment |
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He chose $delta = delta_0$. This can be assumed because he originally assumed $lim_{x rightarrow 0} f(x) = L$ which ensures such $delta$ exists.
– Hongyu Wang
Nov 23 at 18:13
2
No, the answerer didn't assume it, they simply set $delta$ to be $delta_0$. To do an $varepsilon$-$delta$ proof, you are given an $varepsilon$, and you need to find a $delta$ satisfying certain conditions depending on $varepsilon$ and $delta$. In the answer, the answerer first fixes some $varepsilon$, and then obtain a $delta_0$ using the limit, and after that take $delta$ to be the $delta_0$ obtained previously.
– Alex Vong
Nov 23 at 18:19
3
Why is this question urgent, exactly?
– Noah Schweber
Nov 23 at 18:25
1
He's saying you can use the same $delta$ for the second limit as you did for the first.
– saulspatz
Nov 23 at 18:35