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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 ##\displaystyle\lim_{x \to{a^{+}}}{f(x)}=L## there must exist a number ##x_1## in ##(a,u)## such that

##f(x)<f(u)## whenever ##a<x<x_1##

Similarly, there must exist a number ##x_2## in ##(u,b)## such that

##f(x)<f(u)## whenever ##x_2<x<b##"

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 ##a,b\in{R}## i.e. for real numbers:

##\exists{\epsilon} |~ 0<\epsilon<f(u)-L##. Definition of limit implies ##\exists{\delta}>0 |~ if \ x\in{(a,a+\delta)}\Rightarrow{\left |{f(x)-L}\right |<\epsilon}\Rightarrow{f(x)<L+\epsilon<f(u)}##

Similarly,

##\exists{\epsilon'} |~ 0<\epsilon'<f(u)-M## , definition of limit implies ##\exists{\delta'}>0 |~ if \ x\in{(b-\delta',b)}\Rightarrow{\left |{f(x)-M}\right |<\epsilon'}\Rightarrow{f(x)<M+\epsilon'<f(u)}##

Specifically, for your doubts you can take ##x_1=a+\delta \wedge x_2=b-\delta##'

Therefore it is known that the maximum of ##f## in the interval ##[a+\delta,b-\delta']##. And I think you can continue"

Question:

Let's place the function in the first quadrant, for simplifying the matter:

Why does ##\exists{\epsilon} |~ 0<\epsilon<f(u)-L## ?. Similarly, why does ##\exists{\epsilon'} |~ 0<\epsilon'<f(u)-M##?

My attempt is not an attempt:

1- If limit exists, it must be ##\forall{\epsilon>0}##

2- ##f(u)-L>0##

3- ##f(u)-M>0##

Hence, 2 and 3 are fixed numbers, but I can choose any ##\epsilon##. So I can state

##\exists{\epsilon} |~ 0<\epsilon<f(u)-L##

##\exists{\epsilon'} |~0<\epsilon'<f(u)-M##

the same way I could state

##\exists{\epsilon} |~ 0<f(u)-L<\epsilon##

##\exists{\epsilon'} |~ 0<f(u)-M<\epsilon'##

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 ##\displaystyle\lim_{x \to{a^{+}}}{f(x)}=L## there must exist a number ##x_1## in ##(a,u)## such that

##f(x)<f(u)## whenever ##a<x<x_1##

Similarly, there must exist a number ##x_2## in ##(u,b)## such that

##f(x)<f(u)## whenever ##x_2<x<b##"

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 ##a,b\in{R}## i.e. for real numbers:

##\exists{\epsilon} |~ 0<\epsilon<f(u)-L##. Definition of limit implies ##\exists{\delta}>0 |~ if \ x\in{(a,a+\delta)}\Rightarrow{\left |{f(x)-L}\right |<\epsilon}\Rightarrow{f(x)<L+\epsilon<f(u)}##

Similarly,

##\exists{\epsilon'} |~ 0<\epsilon'<f(u)-M## , definition of limit implies ##\exists{\delta'}>0 |~ if \ x\in{(b-\delta',b)}\Rightarrow{\left |{f(x)-M}\right |<\epsilon'}\Rightarrow{f(x)<M+\epsilon'<f(u)}##

Specifically, for your doubts you can take ##x_1=a+\delta \wedge x_2=b-\delta##'

Therefore it is known that the maximum of ##f## in the interval ##[a+\delta,b-\delta']##. And I think you can continue"

Question:

Let's place the function in the first quadrant, for simplifying the matter:

Why does ##\exists{\epsilon} |~ 0<\epsilon<f(u)-L## ?. Similarly, why does ##\exists{\epsilon'} |~ 0<\epsilon'<f(u)-M##?

My attempt is not an attempt:

1- If limit exists, it must be ##\forall{\epsilon>0}##

2- ##f(u)-L>0##

3- ##f(u)-M>0##

Hence, 2 and 3 are fixed numbers, but I can choose any ##\epsilon##. So I can state

##\exists{\epsilon} |~ 0<\epsilon<f(u)-L##

##\exists{\epsilon'} |~0<\epsilon'<f(u)-M##

the same way I could state

##\exists{\epsilon} |~ 0<f(u)-L<\epsilon##

##\exists{\epsilon'} |~ 0<f(u)-M<\epsilon'##

Greetings!

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