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mcastillo356
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- TL;DR Summary
- I've been given another and easier proof, but there is something I don't understand.
Hello, PF
This is Theorem 8 of Chapter 4 of the ninth edition of Calculus, by Robert A. Adams: "Existence of extreme values on open intervals". I have an alternative and easier proof, based on epsilon-delta arguments, but it's not mine, and I don't understand it completely.
The fact is that the original doubt was formulated in a Spanish forum. I quote the paragraph origin of the doubt:
"Since there must exist a number in such that
whenever
Similarly, there must exist a number in such that
whenever "
And this is the reply
"Actually what's in the book I don't see very well. Focusing on doubt, if definition of limit is simply applied appropriately for i.e. for real numbers:
. Definition of limit implies
Similarly,
, definition of limit implies
Specifically, for your doubts you can take '
Therefore it is known that the maximum of in the interval . And I think you can continue"
Question:
Let's place the function in the first quadrant, for simplifying the matter:
Why does ?. Similarly, why does ?
My attempt is not an attempt:
1- If limit exists, it must be
2-
3-
Hence, 2 and 3 are fixed numbers, but I can choose any . So I can state
the same way I could state
Greetings!
This is Theorem 8 of Chapter 4 of the ninth edition of Calculus, by Robert A. Adams: "Existence of extreme values on open intervals". I have an alternative and easier proof, based on epsilon-delta arguments, but it's not mine, and I don't understand it completely.
The fact is that the original doubt was formulated in a Spanish forum. I quote the paragraph origin of the doubt:
"Since
Similarly, there must exist a number
And this is the reply
"Actually what's in the book I don't see very well. Focusing on doubt, if definition of limit is simply applied appropriately for
Similarly,
Specifically, for your doubts you can take
Therefore it is known that the maximum of
Question:
Let's place the function in the first quadrant, for simplifying the matter:
Why does
My attempt is not an attempt:
1- If limit exists, it must be
2-
3-
Hence, 2 and 3 are fixed numbers, but I can choose any
the same way I could state
Greetings!
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